(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.

  • $\begingroup$ where is this problem from? a journal? a book? $\endgroup$ – DeepSea Jun 21 '16 at 14:51
  • $\begingroup$ It's from my real analysis exam. $\endgroup$ – Alex B. Jun 21 '16 at 14:53
  • $\begingroup$ I don't get this at all. If we take almost any continuous function with a free parameter or two then we should be able to vary the parameters to satisfy the condition. Eg $Ae^{kx}$, polynomials etc. $\endgroup$ – almagest Jun 21 '16 at 15:10

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.


You can solve it with calculus of variations. Let us define a functional:

$$I(f) = \int_0^1 f(x)-f(x^2)^2 \ dx.$$


$$I(f+h) = \int_0^1 f(x)-f(x^2)^2 + h(x)-h(x^2)^2 -2f(x^2)h(x^2) \ dx = I(f) + \int_0^1 h(x)-2f(x^2)h(x^2) \ dx + o(h),$$

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


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....:-)

  • $\begingroup$ Shouldn't the "y" inside the third integral be squared? $\endgroup$ – Alex B. Jun 21 '16 at 15:15
  • $\begingroup$ @Alexandru-OctavianBadea Yup! $\endgroup$ – Qwerty Jun 21 '16 at 15:17

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