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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. ($c=0$) –  tomasz Jul 2 '12 at 19:35
    
Also, you assume that the function is twice diferentiable. Not all concave functions are differentiable everywhere. –  tomasz Jul 2 '12 at 19:37
    
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. –  celtschk Jul 2 '12 at 19:57
    
@celtschk: you meant concave. –  copper.hat Jul 2 '12 at 19:59
    
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 ! –  joshua Jul 2 '12 at 20:12

3 Answers 3

There is no non-constant positive concave function on all of $\mathbb{R}$.

If $f$ is a concave function, then the difference ratio $$ \frac{f(x)-f(y)}{x-y} $$ is a non-increasing function of both $x$ and $y$.

Suppose there are $x$ and $y$ so that $f(x)<f(y)$ and let $M=\dfrac{f(x)-f(y)}{x-y}$.


If $x<y$, then $M>0$ and for all $z< x$, we have that $$ \frac{f(z)-f(y)}{z-y}\ge\frac{f(x)-f(y)}{x-y}=M\tag{1} $$ therefore, $$ f(z)\le f(y)+(z-y)M\tag{2} $$ and $(2)$ says that for all $z< y-\frac{f(y)}{M}$, we have $f(z)<0$.


If $x>y$, then $M<0$ and for all $z>x$, we have that $$ \frac{f(z)-f(y)}{z-y}\le\frac{f(x)-f(y)}{x-y}=M\tag{3} $$ therefore, $$ f(z)\le f(y)+(z-y)M\tag{4} $$ and $(4)$ says that for all $z>y-\frac{f(y)}{M}$, we have $f(z)<0$.


Thus no matter whether $x>y$ or $x<y$, we have that $f$ cannot stay positive.

Superharmonic Functions

A superharmonic function is one whose Laplacian is non-positive everywhere. I believe this is what you mean by "the Laplacian case".

Since the case of harmonic functions can be handled using Harnack's Inequality, let's consider the case of strictly superharmonic functions; that is, ones for which $\int_{\mathbb{R}^n}\Delta f(y)\,\mathrm{d}y\lt0$.

Let $B(r,x)=\{y\in\mathbb{R}^n:|y-x|\le r\}$. Then, by the Divergence Theorem, $$ \int_{B(r,x)}\Delta f(y)\,\mathrm{d}y=\int_{\partial B(r,x)}\nabla f(y)\cdot\vec{n}\,\mathrm{d}\sigma(y)\tag{5} $$ The right hand side of $(5)$ is the measure of $\partial B(r,x)$ times the rate of change of $m_f(r,x)$, the mean value of $f$ over $\partial B(r,x)$. The left hand side of $(5)$ is strictly negative past a certain $r_0$. Rewriting this, we get for $r\ge r_0$, $$ \omega_{n-1}r^{n-1}\frac{\partial m_f(r,x)}{\partial r}\le C\lt0\tag{6} $$ which leads to $$ \frac{\partial m_f(r,x)}{\partial r}\le\frac{C}{\omega_{n-1}r^{n-1}}\tag{7} $$ For $n\le2$, $(7)$ says that there can be no lower bound for $m_f(r,x)$, which means there can be no lower bound for $f$.

According to item $8)$ here, a superharmonic function on the whole of $\mathbb{R}^2$ that is bounded below is constant. However, the same is not true on $\mathbb{R}^n$ for $n\ge3$. This agrees with $(7)$.

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Thanks a lot! I'll update my assumptions... What about the laplacian case? Thanks ! –  joshua Jul 2 '12 at 20:14
    
That's excatly what I mean ! Do you know of any possible way to prove this? Thanks ! –  joshua Jul 3 '12 at 5:56
    
@joshua: I've added a bit which explains the claim made in item $8)$ here. –  robjohn Jul 3 '12 at 18:05
    
Thanks a lot ! I'll read it depply on weekend ! Hope I'll figure it out ! Thanks Again! –  joshua Jul 3 '12 at 19:28
    
OK. I have now read your answer, and I found several things difficult: First of all, can you please explain why does your final conclousion is only true for $n \leq 2 $ ? I can't figure this out. In addition, can you please explain again how did you deduce what you said about the RHS of (5) ? Have you got any idea about standard PDE methods of proving this? Something like denoting $ v^\epsilon = u + \epsilon log(\sqrt{x^2 + y^2} ) $ , where $u$ is a subharmonic function, and estaimating $\epsilon$ or something? Thanks again ! –  joshua Jul 4 '12 at 20:00

Concavity (or convexity) is essentially a 'one dimensional' property in that you only need to consider behavior on a line. The ambient space is somewhat irrelevant (at least for concavity/convexity issues).

It is straightforward to establish that if $f$ is a concave real-valued function on a vector space, and bounded below, then $f$ is a constant. (In your question, the lower bound is $0$.)

To see this, suppose $f(x_1) > f(x_2)$, and consider the (concave) function $\phi(t) = f(t x_2+(1-t)x_1)$. Then taking $t>1$, by concavity, we have $$\phi(\frac{1}{t} t + (1-\frac{1}{t}) 0) \geq \frac{1}{t} \phi(t) + (1-\frac{1}{t}) \phi(0).$$ Multiplying across by $t$, and rearranging gives: $$\phi(0) + t (\phi(1)-\phi(0)) \geq \phi(t).$$ Since $\phi(1)-\phi(0) = f(x_2)-f(x_1) < 0$, and $t>1$ was arbitrary, we see that $\phi$ is unbounded below, which is a contradiction. Hence $f$ is a constant.

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Thanks ! But what about the Laplacian case? (i.e- I don't demand concavity, but non-positive laplacian) –  joshua Jul 2 '12 at 20:13
    
But shouldn't the concavity condition only give that inequality for $0 < 1/t \le 1$, that is $t \ge 1$? Of course that doesn't invalidate the argument. –  celtschk Jul 2 '12 at 20:14
    
Oops, thanks for catching that, I did mean $t>1$. –  copper.hat Jul 2 '12 at 20:37

Let $u$ strictly concave and twice diferentiable in $\mathbb{R^{2}}$ então $v(x) = u(x,0)$ is strictly concave and twice diferentiable in $\mathbb{R}$. Hence assume negative value.

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