Prove this integral inequality

Question

Prove this assuming $f$ is integratable: $$\int_{-\pi}^\pi\vert f(t)\vert \, dt\leq \sqrt{2\pi}\sqrt{\int_{-\pi}^\pi\vert f(t)\vert^2}\, dt =2\pi \Vert f\Vert.$$ I tried to square both sides and use the absolute value integral inequality, however that wasn't the right way. Can anyone help?

Answer 1

By using Holder's inequality one writes: $$ \int_{-\pi}^{\pi}|f(t)|dt \leq \Bigg(\int_{-\pi}^{\pi}1^{2}dt\Bigg)^{1/2}\Bigg(\int_{-\pi}^{\pi}|f(t)|^{2}dt\Bigg)^{1/2}. $$