Fall 2020 - Problem 6

calculus, Lp spaces

Prove that the following inequality is valid for all odd $C^1$ functions $\func{f}{\br{-1,1}}{\R}$ :

\int_{-1}^1 \abs{f\p{x}}^2 \,\diff{x} \leq \int_{-1}^1 \abs{f'\p{x}}^2 \,\diff{x}.

By odd, we mean that $f\p{-x} = -f\p{x}$ .

Notice that because $f$ is odd, we immediately have $f\p{0} = 0$ . By the fundamental theorem of calculus and Cauchy-Schwarz,

\begin{aligned} \int_{-1}^1 \abs{f\p{x}}^2 \,\diff{x} &= \int_{-1}^1 \p{\int_0^x f'\p{t} \,\diff{t}}^2 \,\diff{x} \\ &\leq \int_{-1}^1 \p{\int_{-1}^1 \chi_{\br{0,x}} \,\diff{t}} \p{\int_{-1}^1 \abs{f'\p{t}}^2 \,\diff{t}} \,\diff{x} \\ &\leq \p{\int_{-1}^1 \abs{x} \,\diff{x}} \p{\int_{-1}^1 \abs{f'\p{t}}^2 \,\diff{t}} \\ &= \int_{-1}^1 \abs{f'\p{x}}^2 \,\diff{x}. \end{aligned}