Talk:Subharmonic function

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The following comment was left in the article proper by Szilard.revesz (talk · contribs):

For definition of subharmonic functions on Riemannian manifolds, the defining inequalities between the subharmonic function $f$ and the harmonic function $f_1$ should be the reverse, in accordance with the definition given above 8and in accordance with the heuristical content of the name SUBharmonic).

I'm not sure what he means precisely, possibly due to notational confusion. Ray^Talk 19:23, 10 November 2010 (UTC)[reply]

One stated that for a holomorphic function ${\displaystyle f(z)}$ the function ${\displaystyle \varphi (z)=\log |f(z)|}$ is a subharmonic function. I think I have a counterexample, since this is only valid for functions which do not have a local maximum in ${\displaystyle G}$.

Let ${\displaystyle f(z)=z^{2}}$ and ${\displaystyle G=D(0,{\tfrac {1}{2}})}$, a closed disk centered at 0 with radius a half, then

${\displaystyle 0\leq {\frac {1}{2\pi }}\int _{0}^{2\pi }\log |r^{2}e^{2i\theta }|d\theta =2\log r<0}$, if ${\displaystyle 0<r<1}$

Am I wrong? —Preceding unsigned comment added by 94.212.22.34 (talk) 15:41, 20 January 2011 (UTC)[reply]

I think you confused ${\displaystyle f(0)}$ and ${\displaystyle \log |f(0)|}$ in your counterexample. If you interpret ${\displaystyle \log(0)=-\infty }$, (as one is supposed to do), the inequality really holds. Vigfus (talk) 20:23, 3 August 2012 (UTC)[reply]

But in the section Examples it is also stated, that for an analytic function ${\displaystyle f}$ the function ${\displaystyle \log |f|}$ is a subharmonic function. This is definitely not true. Consider for example the identity function ${\displaystyle f(x)=x}$ on ${\displaystyle \mathbb {R} ^{1}}$. Then ${\displaystyle \Delta \log |f(x)|=-x^{-2}}$, which is negative in general. — Preceding unsigned comment added by 131.152.55.74 (talk) 17:37, 11 December 2020 (UTC)[reply]

Let's suppose that ${\displaystyle u:\Omega \subseteq \mathbb {R} ^{2}\longrightarrow \mathbb {R} }$ is a harmonic function non-constant,and ${\displaystyle \Omega \,}$ is an open simply connected set.I want to prove that there is not ${\displaystyle z_{0}\in \Omega }$ with ${\displaystyle |u(z)|\leq |u(z_{0})|\,\forall z\in \Omega }$.We observe this equality is equivalent to ${\displaystyle u(z)^{2}\leq u(z_{0})^{2}\,\forall z\in \Omega }$.Since ${\displaystyle u^{2}\,}$ is subharmonic,we find a contradiction with the maximum principle for subharmonic functions.

¿Is it correct?

Nawiks (talk) 00:34, 11 January 2012 (UTC)[reply]