Guess what: the in-class exercise loop has winding number 1 now! That means I’m up again.
The exercise was to prove that \( \|M_g\| = m = \sup \left\{ {|g(x)| \colon x \in [0,2 \pi] } \right\} \). In class we proved that \( \|M_g\| \le m \), so I’ll prove the converse. Also, I apologize for the strange formatting of the line breaks, but I don’t know how to insert linebreaks in MathJax, and MathJax doesn’t detect that there’s a border so it doesn’t automatically add the break for me.
Let \( E_\epsilon = \left\{ {x \colon x \in [0,2 \pi], g(x) \ge m- \epsilon } \right\} \). As \(m \) is the supremum, this set is nonempty, so the following function is nonzero on a set of nonzero measure. Define \[ f(x) = \left\{ \begin{array} {lr} 1 \colon x \in E_\epsilon \\ 0 \colon x \in (E_\epsilon)^c \end{array} \right\} \]
\[ \|M_g f\|^2 = \frac{1}{2 \pi} \int_0^{2 \pi} | M_g f(x) |^2 dx = \frac{1}{2 \pi} \left ( \int_{E_\epsilon} | g(x) f(x) |^2 dx + \int_{(E_\epsilon)^c} | g(x) f(x) |^2 dx \right ) \]
\[= \frac{1}{2 \pi} \int_{E_\epsilon} | g(x) f(x) |^2 dx \ge (m- \epsilon)^2 \frac{1}{2 \pi} \int_{E_\epsilon} | f(x) |^2 dx = (m- \epsilon)^2 \|f\|^2 \]
Hence \( \|M_g\| \ge \frac{\|M_g f \|} {\|f\|} \ge m-\epsilon \), and as epsilon was arbitrary, \( \|M_g\| \ge m \).