Proving an inequality involving a concave function I have encountered the following inequality, but I'm struggling to prove it:
Suppose $f:\mathbb{R} \to \mathbb{R}$ is a concave function and $\theta > 1$. Then
$$ \theta^k f(x) \geq f(\theta^k x)$$
for all $k = 1,2,\dots$.
I tried to use Jensen's inequality, but got nowhere. I also tried to come up with a counter-example, but all my attempts have failed. Can I get a hint?
 A: Some counterexamples and thought:
What about $f(x) = ax-b$ for $a,b>0$? Then $\theta^k f(x) = \theta^k ax - b\theta^k < \theta^k ax - b = f(\theta^k x)$.
Similarly $g(x) = -x^2 - b$ for $b>0$.
Then inequality $\theta^k g(x) = -\theta^k x^2 - \theta^k b \le - \theta^{2k}x^2 - b = g(\theta^k x)$ is equivalent to
$x^2\theta^k(\theta^k -1) \le \theta^kb - b$, which doesn't hold for every $x$.
The problem in both cases was the value at $0$. Indeed, let $f$ be concave. Then by concativity, for any $\lambda \in (0,1)$, $x,y \in \mathbb R$ we have inequality:
$$ f(\lambda x+ (1-\lambda)y) \ge \lambda f(x) + (1-\lambda)f(y)$$
We want to prove $a f(x) \ge f(ax)$ for $a>1, x \in \mathbb R$. Taking $\lambda = \frac{1}{a} \in (0,1)$, $x = at$ for some $t \in \mathbb R$ and $y = 0$ we get:
$$ af(t) \ge f(at) + (a-1)f(0)$$
As we see, with $f(0) \ge 0$ it holds, when $f(0) < 0$ we have potential problems as showed in counterexamples above. I don't want to say that it does not hold for $f$ concave such that $f(0) < 0$, because I do not have a proof (but I'll be glad to see one), but taking arbitrary concave $f$ and looking at $f_M = f - M$, your inequality is equivalent to $$ af_M(t) \ge f_M(at) $$ so that $$ af(t) - aM \ge f(t) - M $$ and finally: $$ \frac{a}{M}f(t) - a \ge \frac{1}{M}f(t) - 1$$
Taking arbitrary $t \in \mathbb R, a>1$, we can take $M$ big enough so that it does not hold (since $a>1$ and moreover, terms with $\frac{1}{M}$ tend to $0$). In other words, we showed that if we substract enough from arbitrary concave function, then your inequality does not hold.
Edit: Actually it's simplier than I thought. If $f(0)<0$ then letting $x=0$ in your inequality we get $\theta^k f(0) \ge f(0)$ which is false. Hence your inequality holds for concave functions $f$ if and only if $f(0) \ge 0$.
