= Solution
Let $p'$ be the <Holder conjugate exponent> of $p$, with $1/\infty=0$. The <indicator function> $\mathbf1_{(-h,h)}$ has <Lp norm> $(2h)^{1/p'}$ in $L^{p'}$, including the value one when $p'=\infty$. The <Holder inequality> therefore gives
$$
|T_hf|\leq h^{-1/p'}\|f\|_p\|\mathbf1_{(-h,h)}\|_{p'}
=2^{1/p'}\|f\|_p.
$$
The <Lebesgue integral> is unchanged by equality <almost everywhere>, so $T_h$ is well-defined on the <Lp space>; its <linearity> and the bound put it in the <continuous dual space> $X^*$.
For $p<\infty$, the <function> $f_h=(2h)^{-1/p}\mathbf1_{(-h,h)}$ has <Lp norm> one and satisfies
$$
T_hf_h=h^{-1/p'}(2h)^{1-1/p}=2^{1/p'}.
$$
For $p=\infty$, use $f_h=\mathbf1_{(-h,h)}$, whose <L-infinity norm> is one and whose image is $2$. Thus all endpoints are included in
$$
\boxed{T_h\in X^*,\qquad\|T_h\|=2^{1/p'}\quad(0<h<1).}
$$
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