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I'm trying to answer this problem: Consider a real function f, bandlimited by frequency $\omega$, which satisfy $$\int_{-\infty}^\infty f(x)^2dx=c.$$ (For pure mathematicians: "bandlimited" means that it Fourier transform is supported on $[-\omega,\omega]$.)

Is the derivative of this function limited in absolute value? That is, is there an expression $A(\omega,c)$ for which $|f'(x)|<A(\omega,c)$ for all x?

Thank for the help!

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closed as off-topic by Christian Remling, Lucia, Yemon Choi, Alex Degtyarev, Ryan Budney Jul 19 '15 at 7:47

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    $\begingroup$ Not sure that "bandlimited" is a math term. Can you give a definition? $\endgroup$ – Alex Degtyarev Jul 18 '15 at 11:05
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    $\begingroup$ In future, a question at this level might belong better on math.stackexchange.com $\endgroup$ – Yemon Choi Jul 18 '15 at 19:43
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The answer is "yes". We have $$f(x)=\int_{-\omega}^\omega g(t)e^{2\pi itx}dt,$$ so $$|f'(x)|^2\leq\left(2\pi\int_{-\omega}^\omega |t||g(t)|dt\right)^2\leq\frac{8\pi^2\omega^3}{3}\| g\|_2,$$ by Cauchy-Bounyakovski-Schwarz inequality. It remains to notice that $\| g\|_2=\| f\|_2$ according to Parseval . Equality when $g(t)=t$.

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  • $\begingroup$ Could you think of a function that satisfy the equality (that is, has the maximum-possible derivative value)? $\endgroup$ – Ron Jul 18 '15 at 13:45
  • $\begingroup$ The inequality comes from the Cauchy inequality. Equality holds when the functions are equal. So take $g(t)=t$. $\endgroup$ – Alexandre Eremenko Jul 18 '15 at 18:59
  • $\begingroup$ Isn't there a $(2\pi)^2$ missing? $\endgroup$ – Ron Jul 19 '15 at 7:13
  • $\begingroup$ I think the derivative of f(x) result with $2\pi$ inside the parentheses of the middle term of the inequality (See also answer by Josep) $\endgroup$ – Ron Jul 19 '15 at 13:08
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I'm answering your second question.

If $f(x)=\int_{-M}^M\hat f(w) e^{2\pi i wx}dw$ in $L^2(R)$, with $f\in L^2(R),$ we have

$\vert f'(x)\vert=\vert\int_{-M}^{M} 2\pi iw\hat f(w) e^{2\pi i wx}dw\vert\leq 2\pi \Vert f\Vert_2 \sqrt{2M^3/3}$

If you take now $$f_n(w)={{\sqrt 3}\over {\sqrt 2 n^{3/2}}}({{ -n \cos(2 n \pi x)}\over{\pi x}} + {{\sin(2 n \pi x)}\over{2 \pi^2 x^2}})$$ we have $$f_n\in L^2(R), \ \hat f_n (t)= i t \chi_{[-n,n]}(t)$$ Moreover $$\Vert f\Vert_2=1, \ \ \ f_n'(0)=i n^{3/2} 2\sqrt 2/\sqrt 3 \pi$$

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  • $\begingroup$ How does this differ from what Alexandre Eremenko already said in his updated answer? $\endgroup$ – Yemon Choi Jul 18 '15 at 19:44
  • $\begingroup$ i didn't see his updated answer, I only saw his first answer. $\endgroup$ – user75485 Jul 18 '15 at 19:56
  • $\begingroup$ Is it possible that there's a missing sqrt for f morm in the second equation? $\endgroup$ – Ron Jul 19 '15 at 7:11
  • $\begingroup$ Also, did you mean $f_n(x)?$ $\endgroup$ – Ron Jul 19 '15 at 7:18

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