Let $f$ be a periodic function, $\mathcal{C}^1$ on $\mathbb{R}$ such that: $$\displaystyle\int_0^{2 \pi} f(t) \, dt = 0$$

$$f(2 \pi) = f(0)$$

Prove that $$\forall t \in [0,2 \pi]: \int_0^{2 \pi} |f(t)|^2 dt \leq \int_0^{2 \pi} |f'(t)|^2 dt$$

How can we prove this please. I don't have any idea.

  • $\begingroup$ You don't need the absolute value. $\endgroup$ – Git Gud Jun 9 '13 at 17:26
  • $\begingroup$ @GitGud may be OP talks about complex valued function $f$ $\endgroup$ – Norbert Jun 9 '13 at 17:26
  • $\begingroup$ @Norbert I considered it, but the "on $\Bbb R$" part leads me to believe everything is real. Can't be sure if she means the domain or the range. $\endgroup$ – Git Gud Jun 9 '13 at 17:28
  • $\begingroup$ @GitGud "on $\mathbb{R}$" (as opposed to, I guess, "to $\mathbb{R}$") usually means that the domain is $\mathbb{R}$. $\endgroup$ – Omnomnomnom Jun 9 '13 at 17:29
  • $\begingroup$ Did you want $dt$ where you wrote $dx$ in that last integral? $\endgroup$ – Michael Hardy Jun 9 '13 at 19:17

Hint: Expand $f$ in Fourier series.

  • $\begingroup$ yes, the domain is $\mathbb{R}.$and this exercice is on chapitre 'Fourier series" $\endgroup$ – lili Jun 9 '13 at 17:45
  • $\begingroup$ So how we can solve the exercice please $\endgroup$ – lili Jun 9 '13 at 17:52
  • $\begingroup$ have you an idea ? please $\endgroup$ – lili Jun 9 '13 at 18:49
  • $\begingroup$ @lili you have $$f=\sum\limits_{n\in\mathbb{Z}}c_n(f)e^{int}\qquad f'=\sum\limits_{n\in\mathbb{Z}}nc_n(f)e^{int}$$ I leave it to you to verify that $$\int\limits_0^{2\pi} |f(t)|^2dt=\sum\limits_{n\in\mathbb{Z}}|c_n(f)|^2\qquad\int\limits_0^{2\pi} |f'(t)|^2dt=\sum\limits_{n\in\mathbb{Z}}n^2 |c_n(f)|^2$$ Since $\int\limits_0^{2\pi}f(t)dt=0$, then $c_0(f)$, so $$\int\limits_0^{2\pi} |f'(t)|^2dt-\int\limits_0^{2\pi} |f(t)|^2dt=\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}(n^2-1)|c_n(f)|^2\geq 0$$ $\endgroup$ – Norbert Jun 9 '13 at 19:29

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.