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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 $f''(x) \geq 0$)?

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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.

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  • $\begingroup$ Thank you for your note. But what s your idea about concluding that $f$ is convex because $f''(ax+b)\geq 0$? $\endgroup$ – Milad Jan 28 at 1:05
  • $\begingroup$ How do you know that $f''$ exists? $\endgroup$ – copper.hat Jan 28 at 1:09
  • $\begingroup$ For simplicity, we may assume that both $f$ and $h$ are differentiable functions. $\endgroup$ – Milad Jan 28 at 1:11
  • $\begingroup$ If $f,h$ are twice differentiable, then if $a=0$ you can say nothing about $f$ and if $a \neq 0$ then you see that $f''(y) \ge 0$ for all $y$. $\endgroup$ – copper.hat Jan 28 at 1:13
  • $\begingroup$ Right, but $f''(y)$ is not eqaul to $f''(x)$, so the quation is can we conclude that $f$ is convex? $\endgroup$ – Milad Jan 28 at 1:17

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