Finding the number of continuous functions

Question

Question:

Find the number of continuous function(s) $f:[0, 1]\to\mathbb{R}$ satisfying $$\int_0^1f(x)\text{d}x=\frac{1}{3}+\int_0^1f^2(x^2)\text{d}x$$

My approach:

I put $x^2=t$, giving $2x\text{d}x=\text{d}t$, but I am not able to find/ proceed further. Can anyone help please?

Answer 1

You should do the same but inversely: $$\int ^{1}_{0} f(x) dx = \int ^{1}_{0} f(t^2) 2tdt = \int ^{1}_{0} 2x f(x^2) dx$$ So we have: $$\int ^{1}_{0} 2x f(x^2) dx = \frac{1}{3} +\int ^{1}_{0} f^{2}\left( x^{2}\right) dx \Longrightarrow \int ^{1}_{0} f^{2}\left( x^{2}\right) dx - \int ^{1}_{0} 2x f(x^2) dx + \frac{1}{3} = 0$$ $$\Longrightarrow 0 = \int ^{1}_{0} \left(f^{2}\left( x^{2}\right) - 2x f(x^2) \right)dx + \frac{1}{3} = \int ^{1}_{0} \left(\left(f\left( x^{2}\right) - x\right)^2 -x^2 \right)dx + \frac{1}{3} = $$ $$ = \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx - \int ^{1}_{0}x^2dx + \frac{1}{3} = \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx - \left[\frac{x^3}{3}\right]_0^1 + \frac{1}{3}$$ $$\Longrightarrow \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx = 0$$ Since left side is a non negative function, it must be zero everywhere: $$\left(f\left( x^{2}\right) - x\right)^2 = 0 \Longrightarrow f\left( x^{2}\right) - x = 0 \Longrightarrow f\left( x^{2}\right) = x \Longrightarrow f(x) = \sqrt x $$

Answer 2

Using A.M G.M Inequality

$\displaystyle (f(x^2))^2+x^2\geq 2xf(x^2)$

$\displaystyle \int^{1}_{0}\bigg((f(x^2)^2+x^2\bigg)dx\geq \int^1_02xf(x^2)dx$

$\displaystyle \int^1_0(f(x^2))^2dx+\frac{1}{3}\geq \int^1_0f(x)dx$

Equality hold when

$\displaystyle (f(x))^2=x^2\Longrightarrow f(x^2)=\pm x$