Prove $\int_{0}^1 f(x^2)\,\mathrm{d}x \le f\left(\frac{1}{3}\right)$ an unspecified $f$

Question

Let $f(x)$ be 2 times differentiable in $[0,1]$, $f''(x)\le0, x \in [0,1].$ Prove $$\int_{0}^1 f(x^2)\,\mathrm{d}x \le f\left(\frac{1}{3}\right).$$

First, because $x=\frac{1}{3} \in [0,1] \Rightarrow x^2 \in [0,1].$

I would start by using Taylor in $x_0=\dfrac{1}{3}$: $$f(x^2)=f\left(\frac{1}{3}\right)+f'\left(\frac{1}{3}\right)\left(x^2-\frac{1}{3}\right)+\frac{f''\left(\frac{1}{3}\right)}{2}\left(x^2-\frac{1}{3}\right)^{\!2}.$$

However, this is where I'm stuck. I thought about integrating both sides, but that seemed a dead end to me.

Answer 1

If you know Jensen, then Jack D'Aurizio's proof is the easiest way to go, but your approach would have worked too. You didn't quite have Taylor right. Taylor gives

$$f(x^2)=f(1/3)+f'(1/3)(x^2-1/3)+f''(c_x)(x^2-1/3)^2/2.$$

Because $f''\le 0,$ the above is $\le f(1/3)+f'(1/3)(x^2-1/3).$ Thus

$$\int_0^1f(x)\,dx \le \int_0^1[f(1/3)+f'(1/3)(x^2-1/3)]\,dx = f(1/3).$$

Answer 2

If $f''\leq0$ then $f$ if concave and Jensen's inequality gives $$ \int_{0}^{1}f(x^2)\,dx \leq f\left(\int_{0}^{1}x^2\,dx\right).$$