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Let $U$ be an open and connected subset of $\mathbb C$.

$f:U \to [-\infty, \infty)$ is said to be subharmonic if its upper semi continuous and satisfy local mean value inequality.

I want example of discontinuous subharmonic function. Even standard example of $log |f|$ where $f$ is holomorphic is also continuous.

Thanks.

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Say $a_n>0$ and $\sum a_n<\infty$. Define $$u(z)=\sum a_n\log|z-1/n|.$$ Then $u$ is subharmonic; you could verify this directly or note that it's the logarithmic potential of a finite measure. But if $a_n\to0$ fast enough then $u(0)>-\infty$, hence $u$ is not continuous.

(If I have my inequalities straight then you can get a real-valued example by considering $\phi\circ u$ for an appropriate convex function $\phi$, for example $\phi(t)=e^t$.)

Edit: To answer questions that came up in the comments: No, there's no such example on $\Bbb R$, because there subharmonic is the same as convex, and convex functions are continuous.

Why is convexity equivalent to subharmonicity on $\Bbb R$? We don't need to cite Seirpinski, it's easy. Of course $u$ is convex if $$u(tx+(1-t)y)\le tu(x)+(1-t)u(y)\quad(0\le t\le 1)\quad(*).$$

The defininig inequality for subharmonic functions is $(*)$ for $t=1/2$, "midpoint convex". Since for example $$\frac12(x+\frac12(x+y))=\frac34x+\frac14y$$it's easy to see that midpoint convexity implies (*) for dyadic rationals $t$. But we're not assuming $u$ is continuous...

Heh: Recall that if $u$ is subharmmonic, $v$ is continuous in a closed disk and harmonic in the interior, and $u\le v$ on the boundary then $u\le v$ in the disk. The proof of that works just as well in one variable, and if you think about it you see that it says exactly that a subharmonic function on $\Bbb R$ satisfies $(*)$.

So, on the line, midpoint convex plus usc, ie subharmonic, implies convex. The converse is clear.

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  • $\begingroup$ Thanks for the answer! Can I use simillar technique to construct discontinuous subharmonic function on interval in $\mathbb R$? Because I read somewhere that on $\mathbb R$, convex iff subharmonic. $\endgroup$
    – Mayuresh L
    Jun 5, 2018 at 17:43
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    $\begingroup$ That would depend on the definition. For a function defined on $\Bbb R$ yes it does seem that "convex" is the most natural analogue to "subharmonic"; as you must know, a convex function on $\Bbb R$ is continuous. $\endgroup$ Jun 5, 2018 at 17:46
  • $\begingroup$ Yes convex function on $\mathbb R$ is continuous. If I take definition on interval to be same as above i.e. upper semi continuous and satisfy MVT inequality. Is then subharmonicity equivalent to convex? $\endgroup$
    – Mayuresh L
    Jun 5, 2018 at 17:50
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    $\begingroup$ I also got it. MVT inequality is equivalent to mid point convex and according to Wikipedia, Sierpinski theorem says if function is Lebesgue measurable then mid point convex implies convex. We know that upper semi continuous function is Lebesgue measurable. Hence subharmonic iff convex. Thank you very much for your help. $\endgroup$
    – Mayuresh L
    Jun 5, 2018 at 19:24
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    $\begingroup$ Or see edit..... $\endgroup$ Jun 5, 2018 at 20:27

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