= Solution
When $p=\infty$, $p'=1$ and $T_hf=h^{-1}\int_{-h}^h f(x)\,dx$. Take the bounded <measurable function>
$$
f(x)=\begin{cases}\sin(\log|x|),&x\ne0,\\0,&x=0.\end{cases}
$$
It represents an element of the <Lp space> $L^\infty((-1,1))$, with <L-infinity norm> at most one. Since
$$
\int_0^h\sin(\log x)\,dx
=\frac h2\bigl(\sin(\log h)-\cos(\log h)\bigr),
$$
where the lower endpoint contributes zero, its symmetric <Lebesgue integral> gives
$$
T_hf=\sin(\log h)-\cos(\log h).
$$
For $h_n=e^{-2\pi n}$ and $\widetilde h_n=e^{-(2\pi n+\pi)}$, both <sequences> tend to zero through $(0,1)$, but $T_{h_n}f=-1$ and $T_{\widetilde h_n}f=1$. Therefore
$$
\boxed{f\notin Y,\qquad Y\subsetneq L^\infty((-1,1)).}
$$
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