Lemma 24.4.10987A41833787

Proof. For $p=0$ the inequality holds because $F(M,T)>0$. Otherwise set $a=\max p>0$, $-b=\min p<0$ and $r=a+b$. Integrating $-p'<1$ over a forward interval $I$ from a maximum to the next minimum gives $r<|I|\leq T$; also $r<2M$, since $|p|<M$. The function $G(s)=(a-s)(s+b)$ is nonnegative on the range of $p$. Since $p$ has mean zero and $-p'<1$, integration over $I$ gives

As $ab\leq r^2/4$, the average of $p^2$ is at most $r^2/4-r^3/(6T)$. This expression increases strictly for $0<r<T$, and its value at $r=\min \{ 2M,T\} $ is $F(M,T)$. Since $r<\min \{ 2M,T\} $, the strict inequality follows.

For sharpness, consider the following $T$-periodic profiles $p_0$. When $T\geq 2M$, let $p_0$ be constant equal to $M$ and to $-M$ on two intervals of length $(T-2M)/2$ each, joined by a descent of slope $-1$ and length $2M$, and close the period by an upward jump. When $T\leq 2M$, let $p_0(s)=T/2-s$ for $0\leq s<T$, with its upward jump at the end of the period. In both cases $p_0$ has mean zero, $|p_0|\leq M$, $p_0'\geq -1$ in the sense of distributions, and the average of $p_0^2$ is $F(M,T)$; both constructions apply when $T=2M$. Convolution with a nonnegative periodic approximate identity preserves the mean and the bound on the distributional derivative, and multiplication by $1-\varepsilon $ makes both inequalities strict. As the convolution scale and $\varepsilon $ tend to zero, the resulting smooth functions converge to $p_0$ in $L^2$, which proves that the supremum is $F(M,T)$. $\square$