Let $f$ be a continuous, real-valued function on $[0,1]$.

Define $F$ with $F(0)=f(0)$ and $$F(x)= \frac{1}{x}\int_0^x f(y) dy$$ for $x \in (0,1].$

Show that

$$\int\limits_0^1 F^2(x) dx \leq 4\int\limits_0^1 f^2(x) dx.$$

I think you have to solve the problem with partial integration and the Cauchy-Schwarz inequality.

$$ \int\limits_0^1 \left(\frac{1}{x}\int_0^x f(y) dy\right)^2 dx =\int\limits_0^1 \frac{1}{x^2}\left(\int_0^x f(y) dy\right)^2 dx$$ Partial integration with $ u(x)=x^{-2}, u'(x)=-x^{-1},v(x)=(\int_0^x f(y) dy)^2, v'(x)=2\int_0^x f(y) dy\cdot f(x)\\$ $$ \begin{align} &=-\frac{(\int_0^x f(y) dy)^2}x\Bigg|_0^1-\int_0^1\left(-\frac{2}{x}\int_0^x f(y) dy\cdot f(x)\right)dx\\ &=-F^2(1)+2\int_0^1\frac{\int_0^x f(y) dy}{x}\cdot f(x)dx\\ &\le -F^2(1)+2\,\left(\int_0^1\frac{(\int_0^x f(y) dy)^2}{x^2}dx\int_0^1f^2(x)dx\right)^{1/2}\\ &= -F^2(1)+2\,\left(\int_0^1F^2(x)dx\int_0^1f^2(x)dx\right)^{1/2}\\ \end{align} $$

But I didn't know how to proceed from here.

  • $\begingroup$ Why is $(\int_0^1f(x)dx)^2 = F^2(1) = 0$ in your third line? $\endgroup$ – JJR May 17 '17 at 21:49
  • $\begingroup$ Oh, I simply forgot to write it down, I will correct it. But it isn't $F^2(1)$ because $F(x)$ is already defined otherwise, right? $\endgroup$ – Niko B. May 17 '17 at 21:55
  • $\begingroup$ I thought that by definition $F(x)|_{x=1} =\frac{1}{x}\int_0^xf(y)dy|_{x=1} = \int_0^1f(y)dy$ $\endgroup$ – JJR May 17 '17 at 21:59
  • $\begingroup$ Ah, I see, thanks! $\endgroup$ – Niko B. May 17 '17 at 22:01
  • $\begingroup$ Note that this a special case of Exercise 14, Chapter 3 in Rudin's "Real & Complex Analysis". $\endgroup$ – copper.hat May 19 '17 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.