If $f'' \ge 0$, for $a, b \in (0, 1)$, prove that $f(ab)\le af(b)$

Question

I tried to use the definition of the concave function, but I failed to transform it into the inequality that has to be proved. The original question is:

If $f'' \ge 0, f(0)=0$, and $f$ is differentiable, $\forall a, b \in (0, 1)$, prove that $f(ab)\le a f(b)$

I thought that if $f$ is concave up, the following satisfies: $$\frac{f(b)-f(0)}{b-0}\le\frac{f(b-h)-f(b)}{-h}$$ where $0\le h\le b$. I still don't know how this satisfies mathematically. When we change the inequality by multiplying $-h$: $$f(b-h)\le f(b)-\frac{h}{b}f(b)=\left(1-\frac{h}{b}\right)f(b)$$ When we put $h=(1-a)b$, $$f(ab)\le a f(b)$$

Is this the right procedure? Furthermore, I want to know more precise proof. Could you please give me some key points to this question? Thanks for your advice.

Answer 1

Since $f''$ is nonnegative, $f$ is a convex function. So we can use Jensen's inequality:

For any $x,y$ and for any $t\in[0,1]$, we have $$f(tx+(1-t)y)\le tf(x)+(1-t)f(y).$$

To get the desired inequality, take $t=a$, $x=b$ and $y=0$.