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It easy to prove that no non-constant positive concave function exists (for example by integrating: $ u'' \leq 0 \to u' \leq c \to u \leq cx+c_2 $ and since $u>0$ , we obviously get a contradiction.

Can this result be generalized to $ \mathbb{R}^2 $ and the laplacian? Is there an easy way to see this? (i.e.- no non-constant positive real-valued function with non-positive laplacian exists)

Thanks!

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    You want that to be strictly concave. Otherwise $f(x)\equiv 1$ is positive and concave. ($c=0$)2012-07-02
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    Also, you assume that the function is twice diferentiable. Not all concave functions are differentiable everywhere.2012-07-02
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    Another implicit assumption is that the function is defined on $\mathbb{R}$. For example the function $f:(0,1)\to\mathbb{R},x\mapsto x(1-x)$ is both concave and positive.2012-07-02
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    @celtschk: you meant concave.2012-07-02
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    But without demanding strictly concave, and by assuming that I don't want constant functions, my claim will be true? i.e.- concave, positive, and non-constant Thanks !2012-07-02
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    I received an answer for this below... Thanks ! I'll be glad to receive some guidance regarding the laplacian case!2012-07-02
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    See the answer by @copper.hat — in a nutshell, if you look at the function along a straight line, you'll get a one-dimensional function. Since that one cannot be positive unless constant, the whole function cannot be either (because it has a negative value on that line if not constant on it, and for a non-constant function you'll always find a line along which it is not constant).2012-07-02
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    Nonpositive Laplacian follows from superharmonic: the value at a point is $\ge$ the average around a circle centered there. Equivalent if twice-differentiable. Then (considering the plane to be the complex numbers) $\log|f(z)|$ is harmonic.2012-07-02

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