Find all continuous functions $f:[0,1]\rightarrow \mathbb{R}$ that satisfy: $\int_0^1 f(x)dx=1/3 + \int_0^1 f^2(x^2)dx$ (Note that $f^2(x)=f(x)\cdot f(x)$ and not composition.)
Since both integrals are defined, derivation is out of the question. I tried integrating the second integral by parts but reached something chaotic, so I'm more than sure there's a catch to this I'm not getting. Any hints will be appreciated.
 A: You can solve it with calculus of variations. Let us define a functional:
$$I(f) = \int_0^1 f(x)-f(x^2)^2 \ dx.$$
Then:
$$I(f+h) = \int_0^1 f(x)-f(x^2)^2 + h(x)-h(x^2)^2 -2f(x^2)h(x^2) \ dx = I(f) + \int_0^1 h(x)-2f(x^2)h(x^2) \ dx + o(h),$$
so, with a change of variables,
$$\delta I (f) (h) = \int_0^1 h(x)-2f(x^2)h(x^2) \ dx = \int_0^1 h(x) \left(1-\frac{f(x)}{\sqrt{x}}\right) \ dx.$$
Obviously, $f(x) = \sqrt{x}$ is a critical point, and it turns out that $I(\sqrt{x}) = 1/3$. To see that it's a maximum, it is sufficient to look at the second derivative, or more simply to factorize the whole qudratic form $I$:
$$I(\sqrt{x}+h(x)) = \frac{1}{3} -\int_0^1 h(x^2)^2 \ dx,$$
so for continuous $f(x) = \sqrt{x}+h(x)$, we have $I(f) = 1/3$ if and only if $h = 0$, so if and only if $f(x)=\sqrt{x}$.
A: The answer is $ f(x)=\sqrt{x}$. Indeed, for every continuous function $ f$ we have
$$\eqalign{\int_0^1\left( f(t)-\sqrt{t}\right)^2\,\frac{1}{2\sqrt{t}} dt&=
\int_0^1f^2(t)\frac{dt}{2\sqrt{t}} -\int_0^1f(t)dt+\frac12\int_0^1\sqrt{t}dt\cr
&=\int_0^1f^2(x^2)dx-\int_0^1f(t)dt+\frac13}
$$
thus the assumption is equivalent to
$$
\int_0^1\left( f(t)-\sqrt{t}\right)^2\,\frac{1}{2\sqrt{t}} dt=0
$$
Since the integrand is positive and continuous on $(0,1]$, the announced
 answer follows.
A: This I just found out using some easy trial-error method::
 Take $g(x)=x^2$ .Check that $\int _0^1 g(x)dx =1/3$ So ONE solution may be found out like ..
$$\int _0^xf(y)\ dy=\int _0^xg(y)\ dy+\int _0^x(f(y^2))^2 dy$$
Now differentiate both sides w.r.t $x$ and you get$$f(x)=x^2+f^2(x^2)$$
Obviously this wont get you the entire set of solutions but still something is better than nothing....:-)
