Find the value of $f\left(\frac{1}{4}\right)$.

Question

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:

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

2. Can we solve it using a general approach?

I would appreciate your help. Thank you.

Answer 1

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, $$f\left(\frac14\right) = f\left(\frac{1}{2^2}\right) = \frac12$$

Answer 2

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 $f(x)=\sqrt{x}$.