Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 22I (Linear Analysis)

The Baire category theorem says that a complete metric space cannot be a countable union of nowhere dense sets, equivalently that every countable intersection of open dense sets is dense. To prove it, let $G_1,G_2,\dots$ be open dense subsets and let $V$ be nonempty and open. Choose nested closed ballsstarting inside $V\cap G_1$. The centres are a Cauchy sequence; completeness gives a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense.

Choose a countable sequence $q_j\uparrow p$ with $1\le q_j<p$. Since ${\ell }^q\subset {\ell }^r$ for $q<r$, every ${\ell }^q$ with $q<p$ lies in some ${\ell }^{q_j}$. For integers $m\ge 1$, setPointwise convergence and Fatou lemma for series show that $E_{j,m}$ is closed in ${\ell }^p$. It has empty interior: inside any ${\ell }^p$ ball, add a sufficiently long finite block whose ${\ell }^p$ norm is tiny but whose ${\ell }^{q_j}$ norm is larger than $m$. Therefore every $E_{j,m}$ is nowhere dense. Since ${\ell }^p$ is a Banach space, Baire gives

An explicit element of the difference isIts $p$th powers form a convergent logarithmic series, whereas for every $q<p$ the powers contain $n^{-q/p}$ with exponent below one and their series diverges. Hence $x\in {\ell }^p$ but $x\notin {\ell }^q$ for every $q<p$.

For integers $m,n\ge 1$, let $E_{m,n}$ contain those $f\in C([0,1])$ for which there is some $x\in [0,1]$ such thatwhenever $|y-x|<1/n$. Compactness of $[0,1]$ shows that $E_{m,n}$ is closed in the uniform norm: for a uniformly convergent sequence, choose convergent subsequences of the corresponding witness points and pass to the limit.

Each $E_{m,n}$ has empty interior. Given $f$ and a uniform-error tolerance, first approximate $f$ by a polygonal function, then add a sufficiently small-amplitude, sufficiently high-frequency sawtooth whose slopes dominate both $m$ and every slope of the polygonal approximation. At every point, one can move a distance below $1/n$ along one adjacent linear segment and obtain a difference quotient of magnitude greater than $m$. The perturbed function lies in the prescribed ball but outside $E_{m,n}$.

If $f$ is differentiable at some point, its nearby difference quotients are bounded, so $f\in E_{m,n}$ for some $m,n$. The functions differentiable somewhere therefore form a meagre set. By the Baire category theorem, their complement is nonempty; indeed it is dense. Thus

For $x\in X$, define $e_x\in X^{\ast \ast }$ by $e_x(\varphi )=\varphi (x)$. The map $J:x\mapsto e_x$ is linear, andshows $\Vert Jx\Vert \le \Vert x\Vert$, so $J$ is continuous. The granted equality $\Vert Jx\Vert =\Vert x\Vert$ is the canonical embedding into the bidual.

Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$. Their common continuous dual has two operator norms, say $\Vert \cdot {\Vert }_{\ast }$ and $\Vert \cdot {\Vert }_{\ast }^{\prime }$. Both dual spaces are Banach, even if the second normed space $X$ is not complete. The identity maphas a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The closed graph theorem makes the two dual norms equivalent.

Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such thatThis says the two norms are equivalent norms, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.