Fall 2014 - Problem 1

Lp spaces

Show that

A = \set{f \in L^3\p{\R} \st \int_\R \abs{f\p{x}}^2 \,\diff{x} < \infty}

is a Borel subset of $L^3\p{\R^3}$ .

Consider $\func{\phi_N}{L^3\p{\R}}{\R}$ given by

\phi_N\p{f} = \int_{-N}^N \abs{f\p{x}}^2 \,\diff{x}.

Observe that

\begin{aligned} \abs{\phi_N\p{f} - \phi_N\p{g}} &\leq \int_{-N}^N \abs{f\p{x}}^2 - \abs{g\p{x}}^2 \,\diff{x} \\ &\leq \int_{-N}^N \abs{f\p{x} - g\p{x}}\p{\abs{f\p{x}} + \abs{g\p{x}}} \,\diff{x} \\ &\leq \norm{f - g}_{L^3}\norm{\abs{f} + \abs{g}}_{L^{3/2}\p{\br{-N,N}}}. \end{aligned}

Notice further that

\begin{aligned} \norm{\abs{f} + \abs{g}}_{L^{3/2}\p{\br{-N,N}}}^{2/3} &= \int \p{\abs{f\p{x}} + \abs{g\p{x}}}^{3/2} \chi_{\br{-N,N}}\p{x} \,\diff{x} \\ &\leq \norm{\abs{f} + \abs{g}}_{L^3}^{3/4} \norm{\chi_{\br{-N,N}}}_{L^{4/3}}, \end{aligned}

so $\phi_N$ is continuous. Hence,

A = \bigcup_{k=1}^\infty \bigcap_{N=1}^\infty \phi_N^{-1}\p{\br{0, k}},

so $A$ is Borel.