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Does anyone know, is $h\left(z,w\right):=\frac{\left|zw\right|}{\left|z\right|+\left|w\right|}$ for $z$ and $w$ in the unit disk $\mathbb{D}$ of the complex plane a (pluri)subharmonic function? Thanks!

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2 Answers 2

I didn't see an easy non-computational solution, but write $$h(z,w) = \frac{\sqrt{z\bar z}\sqrt{w\bar w}}{\sqrt{z \bar z}+\sqrt{w \bar w}}$$ and start computing. (I admit using Maple, since I'm lazy). We get $$\frac{\partial^2 h}{\partial z \partial\bar z} = \frac14\frac{|z|^2\big(|w|-|z|\big)}{|w|\,\big(|z|+|w|\big)^3},$$ $$\frac{\partial^2 h}{\partial w \partial\bar w} = \frac14\frac{|w|^2\big(|z|-|w|\big)}{|z|\,\big(|z|+|w|\big)^3}$$ and $$\frac{\partial^2 h}{\partial z \partial\bar w} = \frac12\frac{z \bar w}{\big(|z|+|w|\big)^3}.$$

Hence (after lots of tedious algebra) $$\Delta h = 4\left( \frac{\partial^2 h}{\partial z \partial\bar z} + \frac{\partial^2 h}{\partial w \partial\bar w} \right) = \frac{\big(|z|-|w|\big)^2(|z|^2+|zw|+|w|^2)}{|zw|\big(|z|+|w|\big)^3} \ge 0$$ so $h$ is subharmonic (outside the coordinate axis, and it's easy to check that it's subharmonic even there). On the other hand $$\frac{\partial^2 h}{\partial z \partial\bar z}\frac{\partial^2 h}{\partial w \partial\bar w} - \frac{\partial^2 h}{\partial z \partial\bar w}\frac{\partial^2 h}{\partial w \partial\bar z} = -\frac{1}{16} \frac{|zw|}{(|z|+|w|)^4}$$ which is negative for most points in the bidisc. This shows that $h$ is not plurisubharmonic.

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I don't have an easier proof of subharmonicity than the one by mrf, but here is the non-psh part. Note that $h$ is polyradial: it depends only on $|z|$ and $|w|$. Such a function is plurisubharmonic (on a Reinhardt region where $zw\ne 0$) if and only if $h(e^x,e^y)$ is a convex function of $(x,y)$: see Proposition 1.14.40 here. Looking at the line $y=-x$, we get $h(e^x,e^{-x})=1/(2\cosh x)$ which is certainly not convex near $0$, having strict maximum at $0$. Well, the line $y=-x$ does not actually meet the region $\{x<0,y<0\}$ which we are to consider; but since convexity is preserved by pointwise limits, it must fail on the line $y=-\epsilon-x$ for all small $\epsilon>0$.

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