Let
$$
w=e^{-\Phi},
\qquad
m_R=\min_{[-R,R]}w>0,
\qquad
E_R(h)=\int_{-R}^R|h'|^2w.
$$
The <fundamental theorem of calculus> and the <Cauchy-Schwarz inequality> give
$$
|h(x)-h(y)|
\leq|x-y|^{1/2}\left(\int_{-R}^R|h'|^2\right)^{1/2}
\leq m_R^{-1/2}|x-y|^{1/2}E_R(h)^{1/2}.
$$
Thus the one-half <Hölder seminorm> is at most $m_R^{-1/2}E_R(h)^{1/2}$.

Writing $Z_R=\int_{-R}^Rw>0$ and using the zero weighted mean, for each $x$ we have
$$
h(x)=\frac1{Z_R}\int_{-R}^R(h(x)-h(y))w(y)\,dy.
$$
Since $|x-y|\leq2R$, the preceding estimate yields
$$
|h(x)|\leq(2R)^{1/2}m_R^{-1/2}E_R(h)^{1/2}.
$$
Consequently the required inequality holds, for example, with
$$
\boxed{C_R=(1+\sqrt{2R})m_R^{-1/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 23F / b / Solution

{c}
{title2=$[f]_{C^{0,\alpha}}$}

The Hölder seminorm of exponent $0<\alpha\leq1$ is
$$
[f]_{C^{0,\alpha}}=\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}.
$$
It measures <Hölder continuity> without including the <supremum norm> of the function itself.

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