Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 22H (Analysis of Functions)

Let $X$ have basis $e_1,\dots ,e_d$, and write $x={\sum }_j{a}_j{e}_j$. On the compact coordinate sphere ${\sum }_j|a_j|=1$, the continuous positive functionhas a positive minimum $m$. Homogeneity and the triangle inequality therefore give constants $m,M>0$ such thatIn particular, each coordinate functional ${\varphi }_j(x)=a_j$ belongs to the continuous dual space $X^{\ast }$.

Every member of $X^{\ast }$ is norm-continuous, so the weak topology is coarser than the norm topology. Conversely, given $x_0\in X$ and $\epsilon >0$, the basic weak neighbourhoodsatisfies $U\subseteq \{x:\Vert x-x_0\Vert <\epsilon \}$. Thus every norm-open set is weakly open, proving that the two topologies coincide. This is the finite-dimensional weak and norm topologies coincide theorem.

The Uniform boundedness principle states that if $E$ is a Banach space and a family of bounded linear operators $S_{\alpha }:E\to Y$ is pointwise bounded, then ${\sup }_{\alpha }\Vert S_{\alpha }\Vert <\infty$.

Suppose $x_n\rightharpoonup x$ in the normed vector space $X$. For each $n$, define the evaluation functionalThe completeness of the dual space says that $X^{\ast }$ is Banach even when $X$ is not. For each $f\in X^{\ast }$, the scalar sequence $J_n(f)=f(x_n)$ converges to $f(x)$ and is therefore bounded. Uniform boundedness givesBy the canonical embedding into the bidual and the Hahn-Banach theorem, $\Vert J_n\Vert =\Vert x_n\Vert$. Hencewhich proves that a weakly convergent sequence is bounded.

Suppose $a^k\rightharpoonup a$ in ${\ell }^1$ but $a^k$ does not converge to $a$ in the norm topology. There are $\epsilon >0$ and a subsequence $(a^{m_k})$ such thatSet $x^k=a^{m_k}-a$. Thenwhich proves the first assertion.

We now use a gliding hump argument. Put $b_0=0$. Having chosen $b_{n-1}$ and $k_{n-1}$, coordinatewise convergence $x_i^k\to 0$ lets us choose $k_n>k_{n-1}$ so thatFor this fixed element of ${\ell }^1$, choose $b_n>b_{n-1}$ so far out thatDefine one sequence $y$ byThe blocks partition the positive integers and $|y_i|\le 1$, so $y\in {\ell }^{\infty }$. On the $n$th assigned block, the signs agree; outside it, use $|y_i|\le 1$. The duality of l1 and l infinity givesBut $x^{k_n}\rightharpoonup 0$ requires $⟨x^{k_n},y⟩\to 0$ for this fixed $y\in ({\ell }^1{)}^{\ast }={\ell }^{\infty }$, a contradiction. Therefore every weakly convergent sequence in ${\ell }^1$ converges in norm. This is the Schur property of l1.

No. In the Lebesgue space $L^1([0,2\pi ])$, letFor every $g\in L^{\infty }([0,2\pi ])$, finiteness of the interval gives $g\in L^1([0,2\pi ])$, and the Riemann-Lebesgue lemma yieldsThe duality of Lp spaces therefore shows that $f_n\rightharpoonup 0$ in $L^1$. On the other hand,for every $n$. Thus weak convergence does not imply norm convergence in this $L^1$ space; the Weakly null sine sequence in L1 is the required counterexample.