Sharp Poincare bound for the uniform law on an interval

Let $m=\mathbb{E}[g(U)]$ and $h(x)=g(a+Lx)-m$ for $0\le x\le 1$. Define $H(t)=h(2t)$ for $0\le t\le 1\mathrm{/}2$ and $H(t)=h(2-2t)$ for $1\mathrm{/}2\le t\le 1$, then extend $H$ periodically to $\mathbb{T}$. The two absolutely continuous pieces match at $t=1\mathrm{/}2$, and $H(0)=H(1)$. A finite paste of absolutely continuous pieces with matching endpoint values is absolutely continuous: split every interval crossing the junction into its at most two pieces and apply [F1] on each piece with half the error tolerance. Thus $H$ is absolutely continuous on the period interval. Its mean is zero, $\mathrm{\parallel }H{\mathrm{\parallel }}_{L^2(\mathbb{T})}^2=\mathrm{Var}(g(U))$, and affine changes of variables give $\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(\mathbb{T})}^2=4L^2\mathbb{E}\mathrm{\mid }{g}^{\mathrm{\prime }}(U){\mathrm{\mid }}^2$.

For each integer $n$, apply integration by parts [F2] to $H(t)e^{-2\pi int}$ on $[0,1]$, taking real and imaginary parts. The boundary term vanishes since $H(0)=H(1)$ and $e^{-2\pi in}=1$, giving $\widehat{H^{\mathrm{\prime }}}(n)=2\pi in\widehat{H}(n)$. Parseval [F4] and the zero mean in step 1.1 yield $\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2={\sum }_{n\mathrm{\ne }0}4{\pi }^2{n}^2\mathrm{\mid }\widehat{H}(n){\mathrm{\mid }}^2\ge 4{\pi }^2{\sum }_{n\mathrm{\ne }0}\mathrm{\mid }\widehat{H}(n){\mathrm{\mid }}^2=4{\pi }^2\mathrm{\parallel }H{\mathrm{\parallel }}_2^2\mathrm{.}$ Substituting the two norm identities from step 1.1 and dividing by $4{\pi }^2$ proves the claimed bound. Equality holds for $g(x)=\cos (\pi (x-a)\mathrm{/}L)$: its mean is zero, its variance is $1\mathrm{/}2$, and its derivative second moment is ${\pi }^2\mathrm{/}(2L^2)$. Hence the constant $L^2\mathrm{/}{\pi }^2$ is sharp.