The answer is "yes". We have $$f(x)=\int_{-\omega}^\omega g(t)e^{2\pi itx}dt,$$ so $$|f'(x)|^2\leq\left(\int_{-\omega}^\omega |t||g(t)|dt\right)^2\leq\frac{2\omega^3}{3}\| g\|_2,$$ by Cauchy-Bounyakovski-Schwarz inequality. It remains to notice that $\| g\|_2=\| f\|_2$ according to Parseval . Equality when $g(t)=t$.

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 $g(t)=t$.

The answer is "yes". We have $$f(x)=\int_{-\omega}^\omega g(t)e^{2\pi itx}dt,$$ so $$|f'(x)|^2\leq\left(\int_{-\omega}^\omega |t||g(t)|dt\right)^2\leq\omega^2\| g\|_2,$$ by Cauchy-Bounyakovski-Schwarz inequality. It remains to notice that $\| g\|_2=\| f\|_2$ according to Parseval .

The answer is "yes". We have $$f(x)=\int_{-\omega}^\omega g(t)e^{2\pi itx}dt,$$ so $$|f'(x)|^2\leq\left(\int_{-\omega}^\omega |t||g(t)|dt\right)^2\leq\frac{2\omega^3}{3}\| g\|_2,$$ by Cauchy-Bounyakovski-Schwarz inequality. It remains to notice that $\| g\|_2=\| f\|_2$ according to Parseval . Equality when $g(t)=t$.