I have the following result. Let $X\subset \mathbb{R}^d$ containing the origin.
Let $f: X\rightarrow [0,\infty)$ be a differentiable concave function. Then for any $x\in\mathbb{R}^d$ and $c\geq 1$ such that $cx\in X$, we have $f(cx)\leq cf(x)$.
To prove the statement, first note that
$0\leq f(0) \leq f(x) - \langle\nabla f(x),x\rangle \Rightarrow \langle\nabla f(x),x\rangle \leq f(x) $
Now
$f(cx)\leq f(x) + \langle \nabla f(x), (c-1)x\rangle. \leq f(x) + (c-1)f(x) = cf(x)$
Now suppose $f: \mathbb{R}^d\rightarrow [0,\infty)$ be a differentiable convex function. Does there exists a similar relation between $f(cx) $ and $f(x)$, where $c\in\mathbb{R}$?
Thanks
If you were to have a general relation between $f(cx)$ and $cf(x)$, it would have to be $$ f(cx)\ge cf(x) $$ since this is true about $f(x)=\|x\|^2$: $$ f(cx)=\|cx\|^2=c^2\|x\|^2=c^2f(x)\ge cf(x), $$ since $c\ge 1$.
Now consider the shifted version (in two dimensions, for simplicity) $$ g(x_1,x_2)=\|(x_1-1,x_2-1)\|^2 $$ Take the point $x=(1-\sqrt{2}/2, 1-\sqrt{2}/2)$ and $c=(2+\sqrt{2})/(2-\sqrt{2})>1$. Then you can easily check that $$ g(x)=1, g(cx)=1 $$ and, therefore, $g(cx)<cg(x)$.
Conclusion: there is no such general property.