Find the value of $f\left(\frac{1}{4}\right)$. We have a continuous function $f$ on $[0,1]$ such that $$\int_0^{1} f(x) dx=\frac{1}{3}+ \int_0^{1} f^2(x^2) dx$$ Then find $f(1/4)$.
I tried to find such function. Suppose that $f(x)=a+bx$, now we wish to find the values of $a,b$ if such $f$ exists and satifies the given condition. Let $f$ satisfy the given equality, we get, $$a+\frac{b}{2}=a+ab+\frac{b^2}{3}+\frac{1}{3}$$
Now, since $a,b$ are arbitrary real numbers satisfying the second equality. We take $a=0$, we get no real $b$ and when $b=1$, we get $a=-1/6$. 
Therefore, our function $f(x)=\frac{-1}{6}+x$, which is clearly continuous on $[0,1]$. Now, $f\left(\frac{1}{4}\right)= 1/12$.
My Question:


*

*We will be getting different values of $f\left(\frac{1}{4}\right)$ for different functions satisfying the given condition. Is this method correct? 

*Can we solve it using a general approach?
I would appreciate your help. Thank you.
 A: Let $u = x^2$, notice
$$\begin{align}\int_0^1 (f(x^2) - x)^2 dx
&= \int_0^1 (f(x^2)^2 - 2x f(x^2) + x^2 ) dx\\
&= \int_0^1 f(x^2)^2 dx - \int_0^1 f(u) du + \frac13\\
&= 0\end{align}$$
and what's in the integrand on LHS is non-negative continuous, we have
$$f(x^2) = x\quad\text{ for }\quad x \in [0,1]$$
In particular,
$$f\left(\frac14\right) = f\left(\frac{1}{2^2}\right) = \frac12$$
A: To me, starting from the given equation looks less like magic (although I have nothing against slick answers).
$$
\begin{align}
\int_0^1f(x)\,\mathrm{d}x
&=\frac13+\int_0^1f^2\!\left(x^2\right)\mathrm{d}x\tag1\\
&=\frac13+\frac12\int_0^1f^2(x)\,\frac{\mathrm{d}x}{\sqrt{x}}\tag2\\
0&=\frac13+\frac12\int_0^1\left(f^2(x)-2\sqrt{x}f(x)\right)\frac{\mathrm{d}x}{\sqrt{x}}\tag3\\
&=\frac12\int_0^1\left(f^2(x)-2\sqrt{x}f(x)+x\right)\frac{\mathrm{d}x}{\sqrt{x}}\tag4\\
&=\frac12\int_0^1\left(f(x)-\sqrt{x}\right)^2\frac{\mathrm{d}x}{\sqrt{x}}\tag5
\end{align}
$$
Explanation:
$(1)$: given equation
$(2)$: substitute $x\mapsto\sqrt{x}$
$(3)$: subtract $\int_0^1f(x)\,\mathrm{d}x$ from both sides
$(4)$: complete the square and notice $\frac12\int_0^1x\,\frac{\mathrm{d}x}{\sqrt{x}}=\frac13$
$(5)$: Binomial Theorem
Since $f$ is continuous, equation $(5)$ says that $f(x)=\sqrt{x}$.
