Question About Concave Functions 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 $ \Bbb 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 l
Laplacian exists)
Thanks!
 A: 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 $(7)$.
A: 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.
A: 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.
