Past exam of the functional analysis I course of the University of Oxford / 2023 / 2 / b / Solution

Let $p^{\prime }$ be the Holder conjugate exponent of $p$, with $1/\infty =0$. The indicator function $1_{(-h,h)}$ has Lp norm $(2h{)}^{1/p^{\prime }}$ in $L^{p^{\prime }}$, including the value one when $p^{\prime }=\infty$. The Holder inequality therefore givesThe 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 $X^{\ast }$.

For $p<\infty$, the function $f_h=(2h{)}^{-1/p}{1}_{(-h,h)}$ has Lp norm one and satisfiesFor $p=\infty$, use $f_h=1_{(-h,h)}$, whose L-infinity norm is one and whose image is $2$. Thus all endpoints are included in