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

For a normed space $X$, let $X^{\ast }$ be its continuous dual space and $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$ its bidual space. Its canonical embedding into the bidual isThe definition of a reflexive space in this setting isThe Hahn-Banach theorem makes $J_X$ a linear map that is an isometry. For any normed space $Z$, its continuous dual space is a Banach space: a Cauchy sequence $({\ell }_n)$ in the operator norm has scalar limits $\ell (z)={\lim }_n{\ell }_n(z)$, and the uniform Cauchy bound proves both boundedness of $\ell$ and $\Vert {\ell }_n-\ell \Vert \to 0$. Apply this with $Z=X^{\ast }$ to see that $X^{\ast \ast }$ is complete. If $J_X$ is onto, the isometry transfers this completeness to $X$. Therefore

The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.

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

Put $M=2^{1/p^{\prime }}$, the common operator norm from part (b). Given $\epsilon >0$, choose $N$ with $\Vert f-f_N{\Vert }_p<\epsilon /(3M)$. Since $f_N\in Y$, the scalar limit of $T_h{f}_N$ exists. The Cauchy criterion for a function limit gives $\delta >0$ such thatUsing the triangle inequality and the uniform operator norm bound,This is the requested estimate. The completeness of $\mathbb{R}$ makes the scalar Cauchy criterion for a function limit sufficient, hence ${\lim }_{h\to 0^{+}}{T}_{h}f$ exists and $f\in Y$.

For $u,v\in Y$ and real $a,b$, the linearity of $T_h$ gives $T_h(au+bv)=a{T}_{h}u+b{T}_{h}v$, which also has a limit. Thus $Y$ is a linear subspace. The preceding argument shows it contains every norm convergence limit of its sequences, so the sequential criterion for closed sets in a metric space gives

Use $D=C([-1,1])$, restricted to $(-1,1)$ and regarded as a linear subspace of $L^p((-1,1))$. The density of continuous functions in Lp on an interval says that $D$ is a dense subspace whenever $1\le p<\infty$.

For $g\in D$, the supremum norm givesIn particular $D\subseteq Y$, with zero limit. Part (c), together with $D$ being a dense subspace, implies $Y=X$. To identify the limit throughout $X$, for $f\in X$ and $g\in D$ useFirst choose $g$ in the dense subspace to make the first term arbitrarily small, then let $h\to 0^{+}$. This is convergence on a dense subspace under a uniform operator bound, and provesThe zero linear functional belongs to the continuous dual space $X^{\ast }$.

When $p=\infty$, $p^{\prime }=1$ and $T_{h}f=h^{-1}{\int }_{-h}^{h}f(x)dx$. Take the bounded measurable functionIt represents an element of the Lp space $L^{\infty }((-1,1))$, with L-infinity norm at most one. Sincewhere the lower endpoint contributes zero, its symmetric Lebesgue integral givesFor $h_n=e^{-2\pi n}$ and ${\tilde{h}}_n=e^{-(2\pi n+\pi )}$, both sequences tend to zero through $(0,1)$, but $T_{h_n}f=-1$ and $T_{{\tilde{h}}_n}f=1$. Therefore