I'm interested in the following question: If $f$ is a convex function and $t\geq 1$, can we show that $f(tx) \geq tf(x)$? Apparently this is supposed to be a direct consequence of (one of) the definitions of convex functions, but I tried applying them without any success. Note that we can assume here that $f$ is twice continuously differentiable, but we do not know its derivative's exact form. Is there some kind of inequality definition of convex functions using the second derivative of $f$ or some other argument that can help me?
Is there an extra assumption on $f$ that can help?
If $f$ is convex and $t$ is greater or equal to 1, is f(tx) greater or equal to $tf(x)$
convex optimizationconvex-analysis
Best Answer
Since $t \geqslant 1$, we have $0 < \frac{1}{t} \leqslant 1$. Now, since $f$ is convex: $$\frac{1}{t} f(tx) + \frac{t-1}{t}f(0) \geqslant f(\frac{tx}{t} + 0) \implies f(tx)+(t-1)f(0) \geqslant tf(x)$$
If $f(0) \leqslant 0$, then $f(tx) \geqslant f(tx) + (t-1)f(0) \geqslant tf(x)$, and the claim holds true.
If $f(0)>0$, the claim must be false, with the counterexample $x=0$, since $tf(0) > f(0)$ for any $t>1$.