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!
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 $g(t)=t$.
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 $$\Vert f\Vert_2=1, \ \ \ f_n'(0)=i n^{3/2} 2\sqrt 2/\sqrt 3 \pi$$
