concave function is positive on convex set I have seen the following sentence somewhere, but I can not prove it.
Let $f$ be a nonnegative concave function  on a convex domain $D$. Assume that $f$ achieves a positive value at some point $x\in D$. Then $f(y)>0$ for all $y\in D^o$.
 A: I don't think this question is that complicated, maybe you have overthought a little bit.
So let $x_0 \in D $ be that positive point, that is $f(x_0)>0$
For any point $y$ in the interior of $D$, we can always choose and point $z \in D^{o}$ ( altogether with a $\lambda \in (0,1)$) such that:
$$(1-\lambda)x_0 +\lambda z = y$$
( Note that $\dfrac{y- (1-\lambda)x_0}{\lambda} \longrightarrow y$ when $\lambda \longrightarrow 1^-$)

Hence :
$$ (1-\lambda)f(x_0) +\lambda f(z) \le f(y)$$
Which is a contradiction if $f(y)=0$ because the $LHS$ is positive( due to the positivity of $f(x_0)$
A: Hint: Let $x$ be a point in $D$ for which $f(x) > 0$, and suppose (for the purpose of contradiction) that $y \in D^o$ is such that $f(y) = 0$. Note that for $t \geq 1$, we have
$$
f((1 - t)x + ty) \leq (1-t)f(x) + tf(y).
$$
A: I presume we are working in $\mathbb R^n$.
As the statement is already valid for the point $y$ itself, choose another point $x\in D^{o}, x\ne y$. The open ball $B(x,\epsilon)$ is contained in $D$ for some $\epsilon > 0$.
Now, look at the line going through $x$ and $y$: $l(t)=x+t(y-x)$. For small negative values of $t$ (say $-\delta<t\le 0$) we will still have $l(t)\in B(x,\epsilon)$ because $|l(t)-x|=|t||y-x|$ and by making $|t|\lt\delta=\frac{\epsilon}{|y-x|}$ this will be true.
Find one such point $z=l(t)\in B(x,\epsilon)\subseteq D$ for a small negative $t$ so we have $z, x, y$ colinear, with $x$ between $y$ and $z$, and as both $y$ and $z$ belong to $D$, the whole segment between $y$ and $z$ is contained in $D$ (convexity). With a suitable re-parameterisation of the line $xyz$ (new parameter: $\tau$) we can have $x=y\tau+z(1-\tau)$ with $0\lt\tau\lt 1$. We know that $f(z)\ge 0$ and $f(y)>0$, so $f(x)=f(y\tau+z(1-\tau))\ge\text{ (concavity) }\tau f(y)+(1-\tau)f(z)>0$ - since at least the first term $\tau f(y)$ is $>0$ (and the second is $\ge 0$).
