Taking derivatives inside of $\inf$

Question

For convex $f$, I am to show that

$$g(x) \equiv \inf_{\alpha \gt 0} \frac{f(\alpha x)}{\alpha}$$

is convex.¹

The given answer recasts $g$ as a perspective transformation of $f$. But I used the following approach instead:

$$\begin{align} g(x) & = \inf_{\alpha \gt 0} \frac{f(\alpha x)}{\alpha} \\ g'(x) & = \inf_{\alpha \gt 0} \frac{\alpha f'(\alpha x)}{\alpha} \\& = \inf_{\alpha \gt 0} f'(\alpha x) \\ g''(x) & = \inf_{\alpha \gt 0} \alpha f''(\alpha x) = 0 \end{align}$$

Since $g'' \geq 0$ for all $x$, $g$ is convex.

Is this reasoning sound? Am I allowed to take derivatives "inside" of $\inf$ like this?

Please note that while there are other questions on this site asking to show the convexity of this function, my question is distinct, as it concerns the validity of the new proof given above.

1. Stephen Boyd and Lieven Vandenberghe, Convex Optimization, exercise 3.31c.

Answer 1

Your proof is not correct, because $$g(x)= \inf_{\alpha \gt 0} \frac{f(\alpha x)}{\alpha}\not\Rightarrow g'(x)= \inf_{\alpha \gt 0} f'(\alpha x).$$ Take $f(x)=x^2$ then $$g(x)=\inf_{\alpha \gt 0} \frac{f(\alpha x)}{\alpha}=\inf_{\alpha \gt 0} \frac{\alpha^2 x^2}{\alpha}=\inf_{\alpha \gt 0} \alpha x^2=0\implies g'(x)=0.$$ On the other hand, $$\inf_{\alpha \gt 0} f'(\alpha x)=\inf_{\alpha \gt 0} 2(\alpha x)=\begin{cases} 0&\text{if $x\geq 0$,}\\ -\infty&\text{if $x<0$.}\\ \end{cases}$$ Another example: let $f(x)=(x+1)^2$. For $x>0$, $$g(x)=\inf_{\alpha \gt 0} \frac{f(\alpha x)}{\alpha}=\inf_{\alpha \gt 0} \left(\alpha x^2 +2x+\frac{1}{\alpha}\right)=4x\implies g'(x)=4.$$ On the other hand, $$\inf_{\alpha \gt 0} f'(\alpha x)=\inf_{\alpha \gt 0} 2(\alpha x+1)=2.$$