Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 22F (Analysis of Functions)

For $1\le p<\infty$, the Lebesgue space $L^p(X)$ consists of equivalence classes modulo equality almost everywhere of measurable functions satisfyingThe space $L^{\infty }(X)$ consists of essentially bounded measurable functions modulo equality almost everywhere, with norm

Let $1\le p<q<\infty$ and put $r=q/p>1$. By Hölder's inequality,Taking $p$th roots givesFor $q=\infty$, directly integrating $|f{|}^p\le \Vert f{\Vert }_{\infty }^p$ gives the same endpoint estimate. Hence $L^q(X)\subseteq L^p(X)$ when $\mu (X)<\infty$, as in Lp inclusion on a finite measure space.

Let $q$ be the Holder conjugate exponent of $p$. On ${\mathbb{R}}^n$ with Lebesgue measure, the duality of Lp spaces isunder the pairing ${\Lambda }_g(f)=\int fg$. Thus $(L^1{)}^{\ast }=L^{\infty }$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.

A Banach space is reflexive when its canonical map into its bidual is surjective. The spaces $L^p({\mathbb{R}}^n)$ are reflexive exactly forFor these exponents, applying $L^p$ duality twice identifies the bidual with $L^p$. The endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive; for example, disjoint unit cubes give closed copies of ${\ell }^1$ and ${\ell }^{\infty }$, respectively.

Integration gives an isometric embedding $L^1\hookrightarrow (L^{\infty }{)}^{\ast }$, but it is not onto. On the subspacedefine $\Lambda (c+h)=c$, the limit at infinity. Since $|\Lambda (v)|\le \Vert v{\Vert }_{\infty }$, the Hahn-Banach theorem extends $\Lambda$ to all of $L^{\infty }$. If this extension were integration against some $g\in L^1$, its vanishing on every compactly supported bounded function would force $g=0$ almost everywhere, but $\Lambda (1)=1$. ThereforeThis is the Dual of L infinity is larger than L1.

First suppose $1\le p<\infty$. Choose a subsequence so thatBy Tonelli theorem,Hence the sum inside the integral is finite almost everywhere, which forces $|f_{n_k}(x)-f(x)|\to 0$ almost everywhere.

If $p=\infty$, choose representatives and remove the countable union of the null sets on whichOutside that null set, norm convergence bounds the pointwise difference and the entire sequence converges to $f$. Thus in every case an almost-everywhere convergent subsequence exists, as stated in Almost-everywhere convergent subsequence from Lp convergence.

Regard the set-theoretic inclusionas a linear map between Banach spaces. Its graph is closed. Indeed, if $f_n\to f$ in $L^q$ and $f_n\to g$ in $L^p$, part (i) gives an almost-everywhere convergent subsequence for the first convergence and then a further subsequence converging almost everywhere for the second; hence $f=g$ almost everywhere. The closed graph theorem therefore gives

Let $E_R=X\cap B(0,R)$. It has finite measure, and $\mu (E_R)\uparrow \mu (X)$. Applying the bound to $1_{E_R}$ gives, for finite $q$,so $\mu (E_R{)}^{1/p-1/q}\le C$. For $q=\infty$, it gives $\mu (E_R{)}^{1/p}\le C$. In either case the measures $\mu (E_R)$ are uniformly bounded. Continuity from below now yieldsThis proves the converse summarized by Lq inclusion implies finite measure on a Euclidean Borel subset.