# A theorem about convex function

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

## 1 Answer

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.

• Thank you for your note. But what s your idea about concluding that $f$ is convex because $f''(ax+b)\geq 0$? – Milad Jan 28 at 1:05
• How do you know that $f''$ exists? – copper.hat Jan 28 at 1:09
• For simplicity, we may assume that both $f$ and $h$ are differentiable functions. – Milad Jan 28 at 1:11
• 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 $y$. – copper.hat Jan 28 at 1:13
• Right, but $f''(y)$ is not eqaul to $f''(x)$, so the quation is can we conclude that $f$ is convex? – Milad Jan 28 at 1:17