L2 norm in measurable set

Question

Let $f\in L^2(E)$ and $E$ be a measurable set. Prove that $$\|f\|_{L^2(E)}=\sup\left\{\int_E f(x)g(x)\,dx:\|g\|_{L^2(E)}=1\right\}.$$

Answer 1

Following the advice of Dirk. Observe \begin{align} \left|\int_E f(x)g(x)\ dx \right|\leq \| f\|_{L^2(E)} \|g\|_{L^2(E)} = \| f\|_{L^2(E)} \end{align} which means \begin{align} \sup\left\{ \int_E fg\ dx \mid \|g\|_{L^2(E)}=1 \right\} \leq \| f\|_{L^2(E)}. \end{align} Now, let us pick $g$ given by \begin{align} g(x) = \frac{f(x)}{\| f\|_{L^2(E)}} \end{align} then we see $\|g\|_{L^2(E)}=1$ and \begin{align} \int_E g(x)f(x)\ dx = \frac{1}{\|f\|_{L^2(E)}}\int_{E} |f(x)|^2\ dx = \|f\|_{L^2(E)}. \end{align} Hence we have \begin{align} \sup\left\{ \int_E fg\ dx \mid \|g\|_{L^2(E)}=1 \right\} = \| f\|_{L^2(E)} \end{align} since the maximum is attained.