SCVEAMER · VERSION 1.4 · ENGLISH EXAMPLES

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Formulas appear from the start. To reveal the formula only after observing, use \exploreframe[formula=hide,lang=en]{…}{…}{…}. Use the answer button to show the verdict and explanation.
In SH −log mode, the surface shows u and the tested function is −log(u). The ID selects the mode automatically.
Subharmonic?std-b01
u(z) = zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The complex Hessian is nonnegative throughout the unit disc.
\exploreframe[lang=en]{sh}{std-b01}{Explore the function}
Subharmonic?std-b02
u(z) = −zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The complex Hessian is negative somewhere in the unit disc, so the function is not subharmonic.
\exploreframe[lang=en]{sh}{std-b02}{Explore the function}
Subharmonic?std-b03
u(z) = (z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

This function is harmonic: its complex Hessian vanishes identically, so it is also subharmonic.
\exploreframe[lang=en]{sh}{std-b03}{Explore the function}
Subharmonic?std-b04
u(z) = (z3 − z3)/(2i)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

This function is harmonic: its complex Hessian vanishes identically, so it is also subharmonic.
\exploreframe[lang=en]{sh}{std-b04}{Explore the function}
Subharmonic?std-b05
u(z) = (zz)2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The complex Hessian is nonnegative throughout the unit disc.
\exploreframe[lang=en]{sh}{std-b05}{Explore the function}
Subharmonic?std-b06
u(z) = zz − 2(zz)2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

u_{z z̄}=1−8|z|² is negative when |z|>1/√8. Look for the failure away from the origin.
\exploreframe[lang=en]{sh}{std-b06}{Explore the function}
Subharmonic?std-b07
u(z) = 2(zz)2 − zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

u_{z z̄}=8|z|²−1 is negative close to the origin (|z|<1/√8).
\exploreframe[lang=en]{sh}{std-b07}{Explore the function}
Subharmonic?std-b08
u(z) = zz + (z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The complex Hessian is nonnegative throughout the unit disc.
\exploreframe[lang=en]{sh}{std-b08}{Explore the function}
Subharmonic?std-b09
u(z) = zz − (z3 + z3)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The complex Hessian is nonnegative throughout the unit disc.
\exploreframe[lang=en]{sh}{std-b09}{Explore the function}
Subharmonic?std-b10
u(z) = zz(1 + (z + z)/2)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

u_{z z̄}=1+2 Re(z), which is negative in the part Re(z)<−1/2.
\exploreframe[lang=en]{sh}{std-b10}{Explore the function}
Subharmonic?std-b11
u(z) = zz + (zz)3/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The complex Hessian is nonnegative throughout the unit disc.
\exploreframe[lang=en]{sh}{std-b11}{Explore the function}
Subharmonic?std-b12
u(z) = −(z + z)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

This function is harmonic: its complex Hessian vanishes identically, so it is also subharmonic.
\exploreframe[lang=en]{sh}{std-b12}{Explore the function}
Subharmonic?std-b13
u(z) = zz(1 − zz)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

u_{z z̄}=1−4|z|² is negative for |z|>1/2.
\exploreframe[lang=en]{sh}{std-b13}{Explore the function}
Subharmonic?std-b14
u(z) = log(1 + zz)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The complex Hessian is nonnegative throughout the unit disc.
\exploreframe[lang=en]{sh}{std-b14}{Explore the function}
Subharmonic?std-b15
u(z) = −log(1 + zz)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The complex Hessian is negative somewhere in the unit disc, so the function is not subharmonic.
\exploreframe[lang=en]{sh}{std-b15}{Explore the function}
Subharmonic?std-b16
u(z) = (z3 + z3)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

This function is harmonic: its complex Hessian vanishes identically, so it is also subharmonic.
\exploreframe[lang=en]{sh}{std-b16}{Explore the function}
Subharmonic?std-b17
u(z) = zz · (z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

u_{z z̄}=3(Re(z)²−Im(z)²) is negative where |Im(z)|>|Re(z)|.
\exploreframe[lang=en]{sh}{std-b17}{Explore the function}
Subharmonic?std-i01
u(z) = max{log|z|, −2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of log|z| and a constant is subharmonic. Its positive Laplacian measure is supported on |z| = e⁻².
\exploreframe[lang=en]{sh}{std-i01}{Explore the function}
Subharmonic?std-i02
u(z) = max{0, (z+z)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of 0 and Re z is subharmonic; its positive Laplacian measure lies on Re z = 0.
\exploreframe[lang=en]{sh}{std-i02}{Explore the function}
Subharmonic?std-i03
u(z) = min{0, (z+z)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The minimum of two harmonic functions has a negative Laplacian measure along Re z = 0.
\exploreframe[lang=en]{sh}{std-i03}{Explore the function}
Subharmonic?std-i04
u(z) = |(z+z)/2|

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

|Re z| is the maximum of two harmonic functions; the positive measure is concentrated on Re z = 0.
\exploreframe[lang=en]{sh}{std-i04}{Explore the function}
Subharmonic?std-i05
u(z) = −|(z+z)/2|

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

−|Re z| has a negative Laplacian measure on its crease.
\exploreframe[lang=en]{sh}{std-i05}{Explore the function}
Subharmonic?std-i06
u(z) = max{0, (z2 + z2)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of two harmonic functions is subharmonic; the crease lies on Re(z²)=0.
\exploreframe[lang=en]{sh}{std-i06}{Explore the function}
Subharmonic?std-i07
u(z) = max{zz, (z+z)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

Both |z|² and Re z are subharmonic; so is their maximum.
\exploreframe[lang=en]{sh}{std-i07}{Explore the function}
Subharmonic?std-i08
u(z) = min{zz, (z+z)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Although the smooth pieces have nonnegative Hessian, a negative singular measure occurs along the switching curve.
\exploreframe[lang=en]{sh}{std-i08}{Explore the function}
Subharmonic?std-i09
u(z) = |(z+z)/2| − 0.8zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The crease carries a positive measure, but u_{z z̄}=−0.8 away from it. At a=1/2 and r=0.08 the circle misses the crease, and J_u(a,r)=−0.00512<0.
\exploreframe[lang=en]{sh}{std-i09}{Explore the function}
Subharmonic?std-i10
u(z) = max{0, (z3 + z3)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of 0 and Re(z³) is subharmonic.
\exploreframe[lang=en]{sh}{std-i10}{Explore the function}
Subharmonic?std-i11
u(z) = −max{0, (z2 + z2)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

A negative Laplacian measure appears on the crease.
\exploreframe[lang=en]{sh}{std-i11}{Explore the function}
Subharmonic?std-i12
u(z) = max{zz, 1/4}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of |z|² and a constant is subharmonic; a positive singular measure lies on |z|=1/2.
\exploreframe[lang=en]{sh}{std-i12}{Explore the function}
Subharmonic?std-a01
u(z) = Re ∑k=0∞ 0.55k(z/1.03)2k

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

This is the real part of a holomorphic lacunary series, hence harmonic. The circle |z|=1.03 is a natural boundary; the picture uses a finite partial sum.
\exploreframe[lang=en]{sh}{std-a01}{Explore the function}
Subharmonic?std-a02
u(z) = Re ∑k=0∞ 0.55k(z/1.03)2k + 2.4zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A harmonic function plus 2.4|z|² has a positive constant Hessian.
\exploreframe[lang=en]{sh}{std-a02}{Explore the function}
Subharmonic?std-a03
u(z) = Re ∑k=0∞ 0.55k(z/1.03)2k − 2.4zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Despite the complex appearance, its Hessian comes from −2.4|z|² and is negative.
\exploreframe[lang=en]{sh}{std-a03}{Explore the function}
Subharmonic?std-a04
u(z) = gP(z),   P(z)=z2−2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

g_P is the escape-rate Green function of P(z)=z²−2. It is subharmonic and its Laplacian measure is supported on the Julia set [−2,2]. The rendering is iterative and approximate.
\exploreframe[lang=en]{sh}{std-a04}{Explore the function}
Subharmonic?std-a05
u(z) = −gP(z),   P(z)=z2−2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

−g_P has a negative Laplacian measure on the Julia set and is not subharmonic. Its rendering uses iteration.
\exploreframe[lang=en]{sh}{std-a05}{Explore the function}
Subharmonic?std-a06
u(z) = |Re(z/(1 − 0.75z))|

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The absolute value of the real part of a holomorphic function is the maximum of two harmonic functions.
\exploreframe[lang=en]{sh}{std-a06}{Explore the function}
Subharmonic?std-a07
u(z) = −|Re(z/(1 − 0.75z))|

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Re(z/(1−0.75z)) changes sign in Δ; −|Re f| has a negative Laplacian measure along its zero set.
\exploreframe[lang=en]{sh}{std-a07}{Explore the function}
Subharmonic?std-a08
u(z) = Re(1/(1 − 0.75z))

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

This function is harmonic: its complex Hessian vanishes identically, so it is also subharmonic.
\exploreframe[lang=en]{sh}{std-a08}{Explore the function}
Subharmonic?std-a09
u(z) = max{0, Re(z/(1 − 0.75z))}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of 0 and a harmonic function is subharmonic.
\exploreframe[lang=en]{sh}{std-a09}{Explore the function}
Subharmonic?log-b01
u(z) = 1 − zz

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

The Hessian of −log(1−|z|²) is positive and grows near the boundary.
\exploreframe[lang=en]{sh}{log-b01}{Explore the function}
Subharmonic?log-b02
u(z) = 1 + zz

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

log(1+|z|²) is subharmonic, but its negative is not.
\exploreframe[lang=en]{sh}{log-b02}{Explore the function}
Subharmonic?log-b03
u(z) = exp((z+z)/2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log(exp(Re z)) = −Re z is harmonic.
\exploreframe[lang=en]{sh}{log-b03}{Explore the function}
Subharmonic?log-b04
u(z) = exp(−zz)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u = |z|².
\exploreframe[lang=en]{sh}{log-b04}{Explore the function}
Subharmonic?log-b05
u(z) = exp(zz)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u = −|z|².
\exploreframe[lang=en]{sh}{log-b05}{Explore the function}
Subharmonic?log-b06
u(z) = 2 + (z2 + z2)/2

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

u is positive and harmonic; −log u is subharmonic for any positive harmonic u.
\exploreframe[lang=en]{sh}{log-b06}{Explore the function}
Subharmonic?log-b07
u(z) = 2 − (z2 + z2)/2

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

The negative logarithm of a positive harmonic function is subharmonic.
\exploreframe[lang=en]{sh}{log-b07}{Explore the function}
Subharmonic?log-b08
u(z) = 2 + (zz)2

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

Its Hessian is negative away from the origin.
\exploreframe[lang=en]{sh}{log-b08}{Explore the function}
Subharmonic?log-b09
u(z) = 2 + (z+z)/2

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

u=2+Re z is positive and harmonic, so −log u is subharmonic.
\exploreframe[lang=en]{sh}{log-b09}{Explore the function}
Subharmonic?log-b10
u(z) = 2 − zz

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

The Hessian of −log(2−|z|²) is 2/(2−|z|²)², which is positive.
\exploreframe[lang=en]{sh}{log-b10}{Explore the function}
Subharmonic?log-b11
u(z) = exp((z2 + z2)/2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u=−Re(z²) is harmonic.
\exploreframe[lang=en]{sh}{log-b11}{Explore the function}
Subharmonic?log-b12
u(z) = exp(−(z2 + z2)/2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

This function is harmonic: its complex Hessian vanishes identically, so it is also subharmonic.
\exploreframe[lang=en]{sh}{log-b12}{Explore the function}
Subharmonic?log-i01
u(z) = max{1, 1.3+(z+z)/2}

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u is the minimum of two subharmonic functions; the mean-value inequality fails at the switching line.
\exploreframe[lang=en]{sh}{log-i01}{Explore the function}
Subharmonic?log-i02
u(z) = min{1, 1.3+(z+z)/2}

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u=max{0,−log(1.3+Re z)}, the maximum of two subharmonic functions.
\exploreframe[lang=en]{sh}{log-i02}{Explore the function}
Subharmonic?log-i03
u(z) = exp(|(z+z)/2|)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u=−|Re z| has a negative Laplacian measure on its crease.
\exploreframe[lang=en]{sh}{log-i03}{Explore the function}
Subharmonic?log-i04
u(z) = exp(−|(z+z)/2|)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u=|Re z| is the maximum of two harmonic functions.
\exploreframe[lang=en]{sh}{log-i04}{Explore the function}
Subharmonic?log-i05
u(z) = max{1.5−zz, 1.4}

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u is the minimum of two subharmonic functions, with a negative singular measure on the crease.
\exploreframe[lang=en]{sh}{log-i05}{Explore the function}
Subharmonic?log-i06
u(z) = min{1.5−zz, 1.4}

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u is the maximum of two subharmonic functions.
\exploreframe[lang=en]{sh}{log-i06}{Explore the function}
Subharmonic?log-a01
u(z) = exp(−gP(z),   P(z)=z2−2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u=g_P is a polynomial Green function and is subharmonic. The graph is approximate.
\exploreframe[lang=en]{sh}{log-a01}{Explore the function}
Subharmonic?log-a02
u(z) = exp(gP(z),   P(z)=z2−2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u=−g_P has a negative Laplacian measure on the Julia set.
\exploreframe[lang=en]{sh}{log-a02}{Explore the function}
Subharmonic?log-a03
u(z) = exp(Re ∑k=0∞ 0.55k(z/1.03)2k)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u is the negative real part of a holomorphic function and hence harmonic. The rendering is truncated.
\exploreframe[lang=en]{sh}{log-a03}{Explore the function}
Subharmonic?log-a04
u(z) = exp(Re ∑k=0∞ 0.55k(z/1.03)2k+1.4zz)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u=−Re f−1.4|z|² has a negative Hessian.
\exploreframe[lang=en]{sh}{log-a04}{Explore the function}
Subharmonic?log-a05
u(z) = exp(Re ∑k=0∞ 0.55k(z/1.03)2k−1.4zz)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u=−Re f+1.4|z|² has a positive Hessian.
\exploreframe[lang=en]{sh}{log-a05}{Explore the function}
Subharmonic?std-i13
u(z) = zz + 3(z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

u=4x²−2y² is strongly concave in one direction, yet u_{z z̄}=1>0. Subharmonicity depends on the sum of directional curvatures.
\exploreframe[lang=en]{sh}{std-i13}{Explore the function}
Subharmonic?std-i14
u(z) = −zz + 3(z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

u=2x²−2y² is strongly convex in one direction yet u_{z z̄}=−1. Small-circle averages are below the center value.
\exploreframe[lang=en]{sh}{std-i14}{Explore the function}
Subharmonic?std-i15
u(z) = zz − 3(zz)2 + 3(zz)3

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The graph rises monotonically with t=|z|² because F′(t)=(1−3t)²≥0, but u_{z z̄}=(1−3t)(1−9t)<0 for 1/9<t<1/3.
\exploreframe[lang=en]{sh}{std-i15}{Explore the function}
Subharmonic?std-i16
u(z) = zz + 2(z3 + z3)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The harmonic term 2 Re(z³) contributes no Hessian. Despite the peaks and troughs, u_{z z̄}=1.
\exploreframe[lang=en]{sh}{std-i16}{Explore the function}
Subharmonic?std-i17
u(z) = −zz + 2(z3 + z3)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

2 Re(z³) is harmonic, leaving the negative Hessian of −|z|², even near the origin.
\exploreframe[lang=en]{sh}{std-i17}{Explore the function}
Subharmonic?std-i18
u(z) = 1.8zz + 2(z4+z4)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A fourth-degree harmonic perturbation creates deep valleys, but the complex Hessian remains 1.8>0.
\exploreframe[lang=en]{sh}{std-i18}{Explore the function}
Subharmonic?std-i19
u(z) = −1.8zz + 2(z4+z4)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The degree-four harmonic component does not change the Hessian: −1.8|z|² makes the function non-subharmonic.
\exploreframe[lang=en]{sh}{std-i19}{Explore the function}
Subharmonic?std-i20
u(z) = −log(1−zz/1.122)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A potential for the Poincaré metric of the disc of radius 1.12 (up to normalization). Smooth and strictly subharmonic on Δ.
\exploreframe[lang=en]{sh}{std-i20}{Explore the function}
Subharmonic?std-i21
u(z) = max{(z+z)/2, Re(ωz), Re(ω2z)}, ω=e2πi/3

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of three harmonic functions; positive Laplacian measure lies on three radiating creases.
\exploreframe[lang=en]{sh}{std-i21}{Explore the function}
Subharmonic?std-i22
u(z) = min{(z+z)/2, Re(ωz), Re(ω2z)}, ω=e2πi/3

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The minimum of three harmonic functions gives negative Laplacian measure along switching lines.
\exploreframe[lang=en]{sh}{std-i22}{Explore the function}
Subharmonic?std-i23
u(z) = max{log|z−0.48|, log|z+0.48|}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of two logarithmic potentials is subharmonic. Each pole is masked by the other branch; the switching set carries positive measure.
\exploreframe[lang=en]{sh}{std-i23}{Explore the function}
Subharmonic?std-i24
u(z) = min{max(log|z−0.48|,−2.4),max(log|z+0.48|,−2.4)}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Each truncated logarithmic potential is subharmonic, but taking their minimum creates a negative singular measure on the switching locus.
\exploreframe[lang=en]{sh}{std-i24}{Explore the function}
Subharmonic?std-i25
u(z) = log(|z2−0.4|2 + |z−0.25i|2 + 0.262)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

For holomorphic f=(z²−0.4,z−0.25i,0.26), log‖f‖² is subharmonic. The nonzero constant component makes it strictly subharmonic.
\exploreframe[lang=en]{sh}{std-i25}{Explore the function}
Subharmonic?std-i26
u(z) = max{0, zz+2(z2 + z2)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

|z|²+2Re(z²) is strictly subharmonic, and its maximum with 0 remains subharmonic despite its saddle-shaped trough.
\exploreframe[lang=en]{sh}{std-i26}{Explore the function}
Subharmonic?std-i27
u(z) = min{0, zz+2(z2 + z2)/2}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

On one side u=0, on the other u=|z|²+2Re(z²). Taking the minimum introduces a negative measure at their switching set.
\exploreframe[lang=en]{sh}{std-i27}{Explore the function}
Subharmonic?std-i28
u(z) = |(z+z)/2| + 0.8zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The crease has positive measure, and the smooth region has complex Hessian 0.8>0; the whole function is strictly subharmonic.
\exploreframe[lang=en]{sh}{std-i28}{Explore the function}
Subharmonic?std-a10
u(z) = gz²−1(2.85z)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

Escape-rate Green function for P(w)=w²−1, composed with w=2.85z. It is subharmonic; its positive equilibrium measure lives on the rescaled Julia set.
\exploreframe[lang=en]{sh}{std-a10}{Explore the function}
Subharmonic?std-a11
u(z) = gz²−0.12256+0.74486i(2.85z)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A Green function for P(w)=w²−0.12256+0.74486i. Its fractal appearance changes, but its Laplacian measure remains positive.
\exploreframe[lang=en]{sh}{std-a11}{Explore the function}
Subharmonic?std-a12
u(z) = −gz²−1(2.85z)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Negating the Green function makes the equilibrium measure on the Julia set negative.
\exploreframe[lang=en]{sh}{std-a12}{Explore the function}
Subharmonic?std-a13
u(z) = 3gz²−1(2.85z)+1.8zz+2.8(z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A positive Green measure plus a quadratic form with Hessian 1.8. The saddle (4.6x²−1.0y²) has a concave direction, yet the sum is strictly subharmonic.
\exploreframe[lang=en]{sh}{std-a13}{Explore the function}
Subharmonic?std-a14
u(z) = 3gz²−1(2.85z)−1.8zz+2.8(z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The saddle (1.0x²−4.6y²) has a convex direction, but G is harmonic off the Julia set, where u_{z z̄}=−1.8.
\exploreframe[lang=en]{sh}{std-a14}{Explore the function}
Subharmonic?std-a15
u(z) = 3gz²−0.12256+0.74486i(2.85z)+1.8zz+2.8 Re(eiπ/4z2)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A positive Julia-set measure plus a rotated harmonic saddle and 1.8|z|²: strictly subharmonic despite complex directional curvature.
\exploreframe[lang=en]{sh}{std-a15}{Explore the function}
Subharmonic?std-a16
u(z) = 3gz²−0.12256+0.74486i(2.85z)−1.8zz+2.8 Re(eiπ/4z2)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Green is harmonic off the Julia set, leaving the negative constant Hessian −1.8.
\exploreframe[lang=en]{sh}{std-a16}{Explore the function}
Subharmonic?std-a17
u(z) = −4gz²−1(2.85z)+1.8zz+2(z2 + z2)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Off the Julia set the Hessian is +1.8, but on it there is −4 times the equilibrium measure. Positivity on the smooth part does not suffice.
\exploreframe[lang=en]{sh}{std-a17}{Explore the function}
Subharmonic?std-a18
u(z) = gz²−1(2.85z)+1.8zz+2(z3 + z3)/2

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A cubic harmonic term produces strong waves, but the Hessian remains 1.8 plus the positive Green measure.
\exploreframe[lang=en]{sh}{std-a18}{Explore the function}
Subharmonic?std-a19
u(z) = min{2gz²−1(2.85z), 0.65}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Capping a Green function from above creates a negative Laplacian measure along a regular level curve. This contrasts with a maximum.
\exploreframe[lang=en]{sh}{std-a19}{Explore the function}
Subharmonic?std-a20
u(z) = max{2gz²−1(2.85z), 0.65}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

The maximum of a Green function and a constant is subharmonic, adding positive measure on a level curve.
\exploreframe[lang=en]{sh}{std-a20}{Explore the function}
Subharmonic?std-a21
u(z) = max{2gz²−1(2.85z),1.1zz}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

Both the Green function and 1.1|z|² are subharmonic, so their maximum is, too; a switching crease carries positive measure.
\exploreframe[lang=en]{sh}{std-a21}{Explore the function}
Subharmonic?std-a22
u(z) = log(1+|P3(1.7z)|2), P(w)=w2−1

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

Logarithmic norm of an iterated polynomial. For holomorphic F, log(1+|F|²) has Hessian |F′|²/(1+|F|²)²≥0.
\exploreframe[lang=en]{sh}{std-a22}{Explore the function}
Subharmonic?std-a23
u(z) = −log(1+|P3(1.7z)|2), P(w)=w2−1+1.5zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

Negate the logarithmic norm and add 1.5|z|². The Hessian is 1.5−|F′|²/(1+|F|²)², negative in regions near suitable zeros of F.
\exploreframe[lang=en]{sh}{std-a23}{Explore the function}
Subharmonic?log-i07
u(z) = exp(zz+(z2 + z2)/2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

u=exp(2x²) looks strongly convex, yet −log u=−2x² has Hessian −1.
\exploreframe[lang=en]{sh}{log-i07}{Explore the function}
Subharmonic?log-i08
u(z) = exp(−zz+(z2 + z2)/2)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

u=exp(−2y²) curves strongly downwards, while −log u=2y² is strictly subharmonic.
\exploreframe[lang=en]{sh}{log-i08}{Explore the function}
Subharmonic?log-i09
u(z) = 1−zz/1.122

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u is a Poincaré metric potential on the disc of radius 1.12, strictly subharmonic.
\exploreframe[lang=en]{sh}{log-i09}{Explore the function}
Subharmonic?log-i10
u(z) = exp(|(z+z)/2|−1.3zz)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: No.

−log u=1.3|z|²−|Re z|. The smooth Hessian is positive, but the crease carries a negative Laplacian measure.
\exploreframe[lang=en]{sh}{log-i10}{Explore the function}
Subharmonic?log-i11
u(z) = exp(−|(z+z)/2|−1.3zz)

Property tested: Subharmonicity of −log u

Show verdict and explanation

Subharmonicity of −log u: Yes.

−log u=1.3|z|²+|Re z|, with positive contributions both on the smooth regions and the crease.
\exploreframe[lang=en]{sh}{log-i11}{Explore the function}
Subharmonic?std-i29
u(z) = .24 P.91(z; .2)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

P_q(z;θ)=Re[(1+q e^(−iθ)z/1.055)/(1−q e^(−iθ)z/1.055)]. The Poisson solution of a Dirichlet problem on the disc of radius 1.055. Sharp boundary features do not change its vanishing complex Hessian.
\exploreframe[lang=en]{sh}{std-i29}{Explore the function}
Subharmonic?std-i30
u(z) = .24 P.87(z; .65) − .26 P.87(z; 3.45)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

P_q(z;θ)=Re[(1+q e^(−iθ)z/1.055)/(1−q e^(−iθ)z/1.055)]. Two sharp boundary peaks with opposite signs: a difference of harmonic Poisson extensions.
\exploreframe[lang=en]{sh}{std-i30}{Explore the function}
Subharmonic?std-i31
u(z) = (4/π) Im artanh(z/1.055)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

Poisson extension of boundary values +1 on the upper semicircle and −1 on the lower semicircle; harmonic inside.
\exploreframe[lang=en]{sh}{std-i31}{Explore the function}
Subharmonic?std-i32
u(z) = .75 S(z) + .45 P.78(z; 1.85) − .35 P.80(z; 4.25)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

S(z)=(4/π)Im artanh(z/1.055), P_q(z;θ)=Re[(1+q e^(−iθ)z/1.055)/(1−q e^(−iθ)z/1.055)]. Combining sharp steps and boundary peaks still gives a harmonic function.
\exploreframe[lang=en]{sh}{std-i32}{Explore the function}
Subharmonic?std-i33
u(z) = .7 Re((z/1.028)3) + .36 Re((z/1.028)11) − .27 Im((z/1.028)16)

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

A finite Fourier/Poisson extension with rapid boundary oscillations: a real linear combination of real and imaginary parts of holomorphic polynomials.
\exploreframe[lang=en]{sh}{std-i33}{Explore the function}
Subharmonic?std-i34
u(z) = .55·(.24 P.91(z; .2)) + 2.3[(zz)2 − .65zz]

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

P_q(z;θ)=Re[(1+q e^(−iθ)z/1.055)/(1−q e^(−iθ)z/1.055)]. The Poisson part is harmonic, but the added radial term has Hessian 2.3(4|z|²−0.65), negative on |z|<√(0.65/4).
\exploreframe[lang=en]{sh}{std-i34}{Explore the function}
Subharmonic?std-i35
u(z) = .5 S(z) − 2.2zz

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

S(z)=(4/π)Im artanh(z/1.055). The step-like Poisson extension is harmonic; adding −2.2|z|² makes its complex Hessian identically −2.2.
\exploreframe[lang=en]{sh}{std-i35}{Explore the function}
Subharmonic?std-a24
u(z) = log KAt(.88,.88), t = 1 + 0.15z

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

Annular fibers A_t={0.25<|w|<2.4e^{-φ(t)}}, φ(t)=.24|z|²+.14Re(z³)+.16Re(t⁻⁷−1). Pseudoconvex family and holomorphic section w=.88 give subharmonic log Bergman kernel; evaluated numerically via a Laurent basis.
\exploreframe[lang=en]{sh}{std-a24}{Explore the function}
Subharmonic?std-a25
u(z) = log KBt(.88,.88), t = 1 + 0.15z

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: Yes.

B_t={0.25<|w|<2.4e^{-ψ(t)}}, ψ(t)=.18|z|²+.24Re(z⁵)−.13Re(t⁻⁸−1). Despite complex oscillation, the log Bergman kernel along a holomorphic section of a pseudoconvex family is subharmonic.
\exploreframe[lang=en]{sh}{std-a25}{Explore the function}
Subharmonic?std-a26
u(z) = log KA(.40+.47zz, .40+.47zz), A={0.25<|w|<1.9}

Property tested: Subharmonicity of u

Show verdict and explanation

Subharmonicity of u: No.

The logarithm of a fixed annulus Bergman kernel is psh, but evaluation at w=.40+.47|z|² is NOT along a holomorphic section. The complex Hessian is negative at the origin.
\exploreframe[lang=en]{sh}{std-a26}{Explore the function}
Plurisubharmonic?N01
u(z, w) = |z|² + |w|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The Levi form is diag(1,1), positive definite.
\exploreframe[lang=en]{psh}{N01}{Explore the function}
Plurisubharmonic?N02
u(z, w) = |w|² − |z|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

The Levi curvature in the z-direction is −1, despite subharmonic w-slices.
\exploreframe[lang=en]{psh}{N02}{Explore the function}
Plurisubharmonic?N03
u(z, w) = |z + w|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The squared modulus of a holomorphic function; its Levi matrix is [[1,1],[1,1]] ≥ 0.
\exploreframe[lang=en]{psh}{N03}{Explore the function}
Plurisubharmonic?N04
u(z, w) = −|z + w|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

On z=w=ζ the restriction is −4|ζ|², not subharmonic.
\exploreframe[lang=en]{psh}{N04}{Explore the function}
Plurisubharmonic?N05
u(z, w) = Re(zw)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The real part of the holomorphic function zw is pluriharmonic.
\exploreframe[lang=en]{psh}{N05}{Explore the function}
Plurisubharmonic?N06
u(z, w) = |z|² + 2|w|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The Levi matrix is diag(1,2), positive definite.
\exploreframe[lang=en]{psh}{N06}{Explore the function}
Plurisubharmonic?N07
u(z, w) = |z|² − |w|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

The Levi curvature is −1 along the w-direction.
\exploreframe[lang=en]{psh}{N07}{Explore the function}
Plurisubharmonic?N08
u(z, w) = |z − w|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The squared modulus of z−w has a positive-semidefinite Levi matrix.
\exploreframe[lang=en]{psh}{N08}{Explore the function}
Plurisubharmonic?N09
u(z, w) = |z|² + Re(w²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

Re(w²) is pluriharmonic and contributes nothing to the Levi form.
\exploreframe[lang=en]{psh}{N09}{Explore the function}
Plurisubharmonic?N10
u(z, w) = −|z|² + Re(w²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

The w-slices are harmonic, but the z-direction has Levi curvature −1.
\exploreframe[lang=en]{psh}{N10}{Explore the function}
Plurisubharmonic?A01
u(z, w) = |z|² + |w|² − 1.8Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The Levi matrix is [[1,−0.9],[−0.9,1]], with eigenvalues 1±0.9. Both coordinate slices are strictly subharmonic; the mixed term decides positivity.
\exploreframe[lang=en]{psh}{A01}{Explore the function}
Plurisubharmonic?A02
u(z, w) = |z|² + |w|² − 2.2Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

The Levi matrix is [[1,−1.1],[−1.1,1]], with eigenvalues 1±1.1. Both coordinate slices are strictly subharmonic; the mixed term decides positivity.
\exploreframe[lang=en]{psh}{A02}{Explore the function}
Plurisubharmonic?A03
u(z, w) = |z|² + |w|² − 4Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

The Levi matrix is [[1,−2],[−2,1]], with eigenvalues 1±2. Both coordinate slices are strictly subharmonic; the mixed term decides positivity.
\exploreframe[lang=en]{psh}{A03}{Explore the function}
Plurisubharmonic?A04
u(z, w) = |z|² + |w|² − 2Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The Levi matrix is [[1,−1],[−1,1]], with eigenvalues 1±1. Both coordinate slices are strictly subharmonic; the mixed term decides positivity.
\exploreframe[lang=en]{psh}{A04}{Explore the function}
Plurisubharmonic?A05
u(z, w) = |w − z²|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The squared modulus of w−z² is psh; its Levi matrix has determinant zero.
\exploreframe[lang=en]{psh}{A05}{Explore the function}
Plurisubharmonic?A06
u(z, w) = |z|² + |w|² − 2Re(zw)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

Re(zw) is pluriharmonic, so the Levi matrix remains diag(1,1).
\exploreframe[lang=en]{psh}{A06}{Explore the function}
Plurisubharmonic?A07
u(z, w) = 0.2|z|² + |w|² − 1.2Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

Both coordinate sections are strictly subharmonic, yet det(H)=0.2−0.6²=−0.16.
\exploreframe[lang=en]{psh}{A07}{Explore the function}
Plurisubharmonic?A08
u(z, w) = 0.4|z|² + |w|² − 0.8Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

Positive diagonal entries and determinant 0.4−0.4²=0.24>0.
\exploreframe[lang=en]{psh}{A08}{Explore the function}
Plurisubharmonic?A09
u(z, w) = |zw|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The squared modulus of the holomorphic function zw has positive-semidefinite Levi form.
\exploreframe[lang=en]{psh}{A09}{Explore the function}
Plurisubharmonic?A10
u(z, w) = −|zw|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

Fixing w=1/2 gives −|z|²/4, which is not subharmonic.
\exploreframe[lang=en]{psh}{A10}{Explore the function}
Plurisubharmonic?M01
u(z, w) = log(1 + |z|² + |w|²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

log(|1|²+|z|²+|w|²) is psh; its Levi form is positive definite.
\exploreframe[lang=en]{psh}{M01}{Explore the function}
Plurisubharmonic?M02
u(z, w) = −log(1 + |z|² + |w|²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

At the origin its Levi form is −I, hence it is not psh.
\exploreframe[lang=en]{psh}{M02}{Explore the function}
Plurisubharmonic?M03
u(z, w) = |z|² + |w|² − 3|zw|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

At w=0.7 one has u_{z z̄}=1−3(0.7)²<0.
\exploreframe[lang=en]{psh}{M03}{Explore the function}
Plurisubharmonic?M04
u(z, w) = |z|² + |w|² + |zw|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

A sum of squared moduli of the holomorphic functions z, w, and zw.
\exploreframe[lang=en]{psh}{M04}{Explore the function}
Plurisubharmonic?M05
u(z, w) = |w − z²|² − 0.2|z|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

At the origin u_{z z̄}=−0.2, despite the prominent squared-modulus term.
\exploreframe[lang=en]{psh}{M05}{Explore the function}
Plurisubharmonic?M06
u(z, w) = |w − z²|² + 0.2|z|²

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

A sum of squared moduli; the Levi determinant is identically 0.2.
\exploreframe[lang=en]{psh}{M06}{Explore the function}
Plurisubharmonic?M07
u(z, w) = |z|⁴ + |w|⁴ − 0.4Re(z w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

Both coordinate slices are subharmonic, yet H(0,0)=[[0,−0.2],[−0.2,0]] has a negative eigenvalue.
\exploreframe[lang=en]{psh}{M07}{Explore the function}
Plurisubharmonic?M08
u(z, w) = |z|⁴ + |w|⁴

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The Levi form is diag(4|z|²,4|w|²)≥0, degenerate at the origin.
\exploreframe[lang=en]{psh}{M08}{Explore the function}
Plurisubharmonic?M09
u(z, w) = |z|² + |w|² + 3Re(z² w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

Both coordinate slices are strictly subharmonic, but the Levi eigenvalues are 1±3|z|.
\exploreframe[lang=en]{psh}{M09}{Explore the function}
Plurisubharmonic?M10
u(z, w) = |z|² + |w|² + Re(z² w̄)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The Levi eigenvalues are 1±|z|, both positive throughout Δ×Δ.
\exploreframe[lang=en]{psh}{M10}{Explore the function}
Plurisubharmonic?S01
u(z, w) = max(|z|², |w|²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The maximum of two psh functions is psh, despite the crease at |z|=|w|.
\exploreframe[lang=en]{psh}{S01}{Explore the function}
Plurisubharmonic?S02
u(z, w) = min(|z|², |w|²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

Taking a minimum does not preserve psh: a complex disc crossing the switching crease violates the mean inequality.
\exploreframe[lang=en]{psh}{S02}{Explore the function}
Plurisubharmonic?S03
u(z, w) = max(|z + w|², |z − w|²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: Yes.

The maximum of |z+w|² and |z−w|² is psh, including across the switching locus.
\exploreframe[lang=en]{psh}{S03}{Explore the function}
Plurisubharmonic?S04
u(z, w) = min(|z + w|², |z − w|²)

Property tested: Plurisubharmonicity of u

Show verdict and explanation

Plurisubharmonicity of u: No.

Although both branches are psh, their minimum has a negative circle-mean gap across the switching locus.
\exploreframe[lang=en]{psh}{S04}{Explore the function}
Ryokuinzun01
Ω = { (z,w) : |z| < 1, |w| < 0.90 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A fixed circular fiber.
\exploreframe[lang=en]{ryokuinzu}{n01}{Explore the domain}
Ryokuinzun02
Ω = { (z,w) : |z| < 1, |w| < exp(−0.65|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

The fiber shrinks away from the origin.
\exploreframe[lang=en]{ryokuinzu}{n02}{Explore the domain}
Ryokuinzun03
Ω = { (z,w) : |z| < 1, |w| < exp(+0.65|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The fiber expands away from the origin.
\exploreframe[lang=en]{ryokuinzu}{n03}{Explore the domain}
Ryokuinzun04
Ω = { (z,w) : |z| < 1, |w| < exp(−0.8 Re(z²)) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Growth differs by direction, but the potential is harmonic.
\exploreframe[lang=en]{ryokuinzu}{n04}{Explore the domain}
Ryokuinzun05
Ω = { (z,w) : |z| < 1, |w| < exp(0.5|z|² − 0.6 Re(z²)) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A harmonic contribution does not repair a negative complex Hessian.
\exploreframe[lang=en]{ryokuinzu}{n05}{Explore the domain}
Ryokuinzun06
Ω = { (z,w) : |z| < 1, |w − 0.65z| < 0.90 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A disc translating holomorphically.
\exploreframe[lang=en]{ryokuinzu}{n06}{Explore the domain}
Ryokuinzun07
Ω = { (z,w) : |z| < 1, |w − 0.65z| < exp(−0.40|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A moving disc with a shrinking radius.
\exploreframe[lang=en]{ryokuinzu}{n07}{Explore the domain}
Ryokuinzun08
Ω = { (z,w) : |z| < 1, |w − 0.65z| < exp(+0.45|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A moving disc with a growing radius.
\exploreframe[lang=en]{ryokuinzu}{n08}{Explore the domain}
Ryokuinzun09
Ω = { (z,w) : |z| < 1, |w − 0.55z²| < 0.85 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A curved path of disc centers.
\exploreframe[lang=en]{ryokuinzu}{n09}{Explore the domain}
Ryokuinzun10
Ω = { (z,w) : |z| < 1, |w| < exp(−0.55|z|⁴) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

The radius has weak curvature at the origin.
\exploreframe[lang=en]{ryokuinzu}{n10}{Explore the domain}
Ryokuinzun11
Ω = { (z,w) : |z| < 1, |w| < exp(+0.55|z|⁴) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A fourth-order obstruction, invisible to second order at z = 0.
\exploreframe[lang=en]{ryokuinzu}{n11}{Explore the domain}
Ryokuinzua01
Ω = { (z,w) : |z| < 1, |w| < exp(−[0.30|z|² + 0.85 Re(z²)]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Real convexity is not complex subharmonicity.
\exploreframe[lang=en]{ryokuinzu}{a01}{Explore the domain}
Ryokuinzua02
Ω = { (z,w) : |z| < 1, |w| < exp(−[−0.35|z|² + 0.85 Re(z²)]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The visually dominant harmonic term disguises a negative Hessian.
\exploreframe[lang=en]{ryokuinzu}{a02}{Explore the domain}
Ryokuinzua03
Ω = { (z,w) : |z| < 1, |w − 0.45z² − 0.25iz| < exp(−[0.35|z|² + 0.45 Re(z³)]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A polynomially moving center and an anisotropic radius.
\exploreframe[lang=en]{ryokuinzu}{a03}{Explore the domain}
Ryokuinzua04
Ω = { (z,w) : |z| < 1, |w − 0.45z² − 0.25iz| < exp(−[−0.38|z|² + 0.45 Re(z³)]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

Change only the sign of the complex curvature.
\exploreframe[lang=en]{ryokuinzu}{a04}{Explore the domain}
Ryokuinzua05
Ω = { (z,w) : |z| < 1, |w| < exp(−[0.15|z|² + 0.23|z|⁴]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Positive radial curvature, strengthening towards the edge.
\exploreframe[lang=en]{ryokuinzu}{a05}{Explore the domain}
Ryokuinzua06
Ω = { (z,w) : |z| < 1, |w| < exp(−[0.30|z|² − 0.24|z|⁴]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The curvature changes sign towards the boundary.
\exploreframe[lang=en]{ryokuinzu}{a06}{Explore the domain}
Ryokuinzua07
Ω = { (z,w) : |z| < 1, |w| < exp(−[−0.12|z|² + 0.28|z|⁴]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

Most of the disc behaves correctly, but the center does not.
\exploreframe[lang=en]{ryokuinzu}{a07}{Explore the domain}
Ryokuinzua08
Ω = { (z,w) : |z| < 1, |w| < 1.05, |w − 0.72z| < 0.88 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Intersect a fixed disc with a holomorphically moving disc.
\exploreframe[lang=en]{ryokuinzu}{a08}{Explore the domain}
Ryokuinzua09
Ω = { (z,w) : |z| < 1, |w − 0.60z| < 0.96, |w + 0.55z| < 0.96 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A lens formed by two moving discs.
\exploreframe[lang=en]{ryokuinzu}{a09}{Explore the domain}
Ryokuinzua10
Ω = { (z,w) : |z| < 1, |w| < exp(+0.55|z|²), |w − 0.65z| < 1.27 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A growing disc is cut by a second disc. A bad boundary patch survives.
\exploreframe[lang=en]{ryokuinzu}{a10}{Explore the domain}
Ryokuinzua11
Ω = { (z,w) : |z| < 1, |w| < 1.30, |w − 0.40z²| < exp(−0.27|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A lens with nonlinear motion and a shrinking constraint.
\exploreframe[lang=en]{ryokuinzu}{a11}{Explore the domain}
Ryokuinzum01
Ω = { (z,w) : |z| < 1, 0.30 < |w| < 1.25 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Each fiber is the same annulus.
\exploreframe[lang=en]{ryokuinzu}{m01}{Explore the domain}
Ryokuinzum02
Ω = { (z,w) : |z| < 1, 0.30 exp(0.18|z|²) < |w| < 1.23 exp(−0.20|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Both boundaries bend in the pseudoconvex direction.
\exploreframe[lang=en]{ryokuinzu}{m02}{Explore the domain}
Ryokuinzum03
Ω = { (z,w) : |z| < 1, 0.30 exp(−0.45|z|²) < |w| < 1.20 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The inner boundary bends in the wrong direction.
\exploreframe[lang=en]{ryokuinzu}{m03}{Explore the domain}
Ryokuinzum04
Ω = { (z,w) : |z| < 1, 0.30 < |w| < 1.20 exp(+0.50|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The outer boundary has negative curvature.
\exploreframe[lang=en]{ryokuinzu}{m04}{Explore the domain}
Ryokuinzum05
Ω = { (z,w) : |z| < 1, 0.30 exp(0.25|z|²) < |w − 0.38z| < 1.22 exp(−0.10|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A holomorphically translating annulus.
\exploreframe[lang=en]{ryokuinzu}{m05}{Explore the domain}
Ryokuinzum06
Ω = { (z,w) : |z| < 1, |w − 0.38z³ − 0.25iz²| < exp(−[0.12|z|² + 0.22|z|⁴ + 0.70 Re(z³)]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Strong directional deformation with positive complex curvature.
\exploreframe[lang=en]{ryokuinzu}{m06}{Explore the domain}
Ryokuinzum07
Ω = { (z,w) : |z| < 1, |w − 0.38z³ − 0.25iz²| < exp(−[0.38|z|² − 0.28|z|⁴ + 0.70 Re(z³)]) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The complex curvature switches sign near the rim.
\exploreframe[lang=en]{ryokuinzu}{m07}{Explore the domain}
Ryokuinzum08
Ω = { (z,w) : |z| < 1, |w − 0.45z| < 1.05, |w + 0.40z| < 1.04, |w − 0.30i| < 1.12 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Three intersecting discs with holomorphic centers.
\exploreframe[lang=en]{ryokuinzu}{m08}{Explore the domain}
Ryokuinzum09
Ω = { (z,w) : |z| < 1, |w| < exp(+0.45|z|²), |w − 0.50z| < 1.28, |w + 0.35z| < 1.28 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

Two additional constraints obscure a locally nonpseudoconvex boundary.
\exploreframe[lang=en]{ryokuinzu}{m09}{Explore the domain}
Ryokuinzun12
Ω = { (z,w) : |z| < 1, |w| < 0.90, (|z| > 0.38 or |w| < 0.33) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A classical Hartogs figure: a thin central core and an outer collar.
\exploreframe[lang=en]{ryokuinzu}{n12}{Explore the domain}
Ryokuinzun13
Ω = { (z,w) : |z| < 1, |w| < 0.80 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A polydisc with a smaller w-radius.
\exploreframe[lang=en]{ryokuinzu}{n13}{Explore the domain}
Ryokuinzun14
Ω = { (z,w) : |z| < 1, |Re w|<0.64, |Im w|<0.56, Re(z+w)<1.05, Im(w−z)<1.00 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

An intersection of real affine half-spaces and the base disc is convex.
\exploreframe[lang=en]{ryokuinzu}{n14}{Explore the domain}
Ryokuinzun15
Ω = { (z,w) : |z| < 1, |Re w|<0.70, |Im w|<0.60, (Re z>0.08 or Re w<−0.08) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A re-entrant real-polyhedral corner is nonpseudoconvex.
\exploreframe[lang=en]{ryokuinzu}{n15}{Explore the domain}
Ryokuinzun16
Ω = { (z,w) : |z| < 1, |w|<0.95, |w² − 0.72z|<0.42 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

An analytic polyhedron with possibly disconnected fibers.
\exploreframe[lang=en]{ryokuinzu}{n16}{Explore the domain}
Ryokuinzun17
Ω = { (z,w) : |z| < 1, u=(z−w)/√2, v=(z+w)/√2, |u|<0.62, |v|<0.62 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

The unitary image of a scaled bidisc. The slices are lenses.
\exploreframe[lang=en]{ryokuinzu}{n17}{Explore the domain}
Ryokuinzua12
Ω = { (z,w) : |z| < 1, u=(z−w)/√2, v=(z+w)/√2, |u|,|v|<0.62, (|u|>0.29 or |v|<0.23) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A genuinely unitary rotation of a Hartogs figure.
\exploreframe[lang=en]{ryokuinzu}{a12}{Explore the domain}
Ryokuinzua13
Ω = { (z,w) : |z| < 1, |w−0.36z²|<0.76 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A polynomial automorphism of a polydisc with holomorphically moving discs.
\exploreframe[lang=en]{ryokuinzu}{a13}{Explore the domain}
Ryokuinzua14
Ω = { (z,w) : |z| < 1, |w−0.52z²|<0.90, (|z|>0.38 or |w−0.52z²|<0.33) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A Hartogs figure under the automorphism (z,v)↦(z,v+0.52z²).
\exploreframe[lang=en]{ryokuinzu}{a14}{Explore the domain}
Ryokuinzua15
Ω = { (z,w) : |z| < 1, |w−0.28z²|<0.95, |(w−0.28z²)²−0.72z|<0.42 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

The analytic polyhedron after a quadratic shear in w.
\exploreframe[lang=en]{ryokuinzu}{a15}{Explore the domain}
Ryokuinzua16
Ω = { (z,w) : |z| < 1, |u|<0.62, |v|<0.62, u=cos(π/6)z−sin(π/6)w, v=sin(π/6)z+cos(π/6)w }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

Another unitary image of a polydisc with a less symmetric rotation.
\exploreframe[lang=en]{ryokuinzu}{a16}{Explore the domain}
Ryokuinzua17
Ω = { (z,w) : |z| < 1, |z|<1, |Re(w−0.22z²)|<0.64, |Im(w−0.22z²)|<0.56, Re(z+w−0.22z²)<1.05, Im(w−0.22z²−z)<1.00 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A quadratic holomorphic shear of a real convex polyhedral domain.
\exploreframe[lang=en]{ryokuinzu}{a17}{Explore the domain}
Ryokuinzum10
Ω = { (z,w) : |z| < 1, u=z−0.42w², v=w, |u|,|v|<0.62, (|u|>0.29 or |v|<0.23) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A Hartogs figure under a triangular polynomial automorphism mixing the base.
\exploreframe[lang=en]{ryokuinzu}{m10}{Explore the domain}
Ryokuinzum11
Ω = { (z,w) : |z| < 1, u=z−0.38w², v=w, |u|<0.62, |v|<0.62, |v²−0.72u|<0.42 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

An analytic polyhedron with quadratic mixing of parameter and fiber.
\exploreframe[lang=en]{ryokuinzu}{m11}{Explore the domain}
Ryokuinzum12
Ω = { (z,w) : |z| < 1, 0<|z|<1, |w−exp(i t)|²<1−η(t), t=log(|z/0.16|²), η(t)=0.60·max(|t|−2.20,0)⁴ }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A smooth pseudoconvex Diederich–Fornæss-type worm with rotating disc fibers.
\exploreframe[lang=en]{ryokuinzu}{m12}{Explore the domain}
Ryokuinzum13
Ω = { (z,w) : |z| < 1, u=cos(π/6)z−sin(π/6)w, v=sin(π/6)z+cos(π/6)w, |u|,|v|<0.62, (|u|>0.29 or |v|<0.23) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A less symmetric unitary rotation of a Hartogs figure.
\exploreframe[lang=en]{ryokuinzu}{m13}{Explore the domain}
Ryokuinzum14
Ω = { (z,w) : |z| < 1, |z|<1, |Re(w−0.30z²)|<0.70, |Im(w−0.30z²)|<0.60, (Re z>0.08 or Re(w−0.30z²)<−0.08) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

A biholomorphic shear of a nonpseudoconvex, nonconvex real-polyhedral domain.
\exploreframe[lang=en]{ryokuinzu}{m14}{Explore the domain}
Ryokuinzum15
Ω = { (z,w) : |z| < 1, |z|<1, |w|<0.73, |w³−0.50z|<0.20 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A cubic analytic polyhedron with changing fiber connectivity.
\exploreframe[lang=en]{ryokuinzu}{m15}{Explore the domain}
Ryokuinzun18
Ω = { (z,w) : |z| < 1, (|z| < 0.48 and |w| < 1) or (0.40 < |w| < 1) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The standard Hartogs figure: the central disc becomes an annulus outside |z| = 0.48.
\exploreframe[lang=en]{ryokuinzu}{n18}{Explore the domain}
Ryokuinzun19
Ω = { (z,w) : |z| < 1, 0.30 exp(−0.20|z|²) < |w| < 1.20 exp(+0.24|z|²) }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: No.

The annular band thickens outwards on both sides; the inner log-radius is not subharmonic.
\exploreframe[lang=en]{ryokuinzu}{n19}{Explore the domain}
Ryokuinzun20
Ω = { (z,w) : |z| < 1, 0.30 < |w − 0.30z| < 1.10 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

A constant annulus translating holomorphically.
\exploreframe[lang=en]{ryokuinzu}{n20}{Explore the domain}
Ryokuinzulecture-ball
Ω = { (z,w) ∈ ℂ² : |z|² + |w|² < 1 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

The unit ball in ℂ². Its fiber over z is the disc of radius √(1−|z|²).
\exploreframe[lang=en]{ryokuinzu}{lecture-ball}{Explore the domain}
Ryokuinzulecture-polydisc
Ω = { (z,w) ∈ ℂ² : |z| < 1, |w| < 1 }

Property tested: Pseudoconvexity of the domain

Show verdict and explanation

Pseudoconvexity of the domain: Yes.

The unit bidisc Δ × Δ. Every fiber is the unit disc.
\exploreframe[lang=en]{ryokuinzu}{lecture-polydisc}{Explore the domain}

All 194 registered examples are included: SH 106, PSH 34 and Ryokuinzu 54.