Assume that function $h(x)=f(ax+b)$ is a convex function. What can we say about the convexity of function $f(x)$?
My notes:
By taking the second derivative from both sides of eqaution $h(x)=f(ax+b)$ with respect to $x$, we have:
$h''(x)=d^2h(x)/dx^2=a^2 \times d^2f(ax+b)/d(ax+b)^2=a^2\times f''(ax+b)$
Since $h(x)$ is convex, $h''(x) \geq 0$.
Also, $a^2 \geq 0$. So, by taking into account the above-mentioned points, we can conclude that:
$f''(ax+b)\geq 0$
Here, the question is can we conclude that $f(x)$ is convex or not (Note that we do not know $f''(x) \geq 0$)?
You don't need derivatives.
If $a=0$ then $h$ is constant and you can say nothing about $f$ (other than its value at $b$).
If $a \neq 0$ then $f(x) = h({1 \over a} y -{b \over a})$ which is a convex function composed with an affine function and hence convex.