Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 22F (Linear Analysis)

For each $n$, define the bounded linear functionalThe hypothesis is precisely that $x_n$ converges weakly to $x$. In particular, for each fixed $y\in H$, the scalar sequence $T_n(y)$ converges and is therefore bounded. Thus the family $\{T_n\}$ is pointwise bounded.

The Uniform boundedness principle givesBy the Riesz representation theorem, $\Vert T_n\Vert =\Vert x_n{\Vert }_H$. Henceso the sequence is bounded.

Put $z_n=x_n-x$, so $z_n\rightharpoonup 0$. We show that the norm closure of the convex hull of every tailcontains zero. If it did not, the Hahn-Banach separation theorem would give a bounded linear functional $\varphi$, a real number $c>0$, and, after multiplying $\varphi$ by a complex scalar if necessary,for every $z$ in that closed convex hull. In particular $\mathrm{Re}\varphi (z_n)\ge c$ for all $n\ge k$. By the Riesz representation theorem, $\varphi (z)=⟨z,y⟩$ for some $y\in H$, contradicting $z_n\rightharpoonup 0$.

Consequently, for every $k$ there is a finite convex combinationwith $\Vert w_k{\Vert }_H<1/k$. DefineThen every ${\tilde{x}}_k$ is a convex combination of terms of the original sequence andThis is Mazur lemma in the present Hilbert-space setting.

Apply part (b). Since every $x_n$ belongs to the convex set $C$, every convex combination ${\tilde{x}}_k$ also belongs to $C$. Since $C$ is norm closed and ${\tilde{x}}_k\to x$ in norm, it follows that $x\in C$. Thus a norm-closed convex set is weakly closed.

No. Take the Hilbert space $H={\ell }^2$ and its standard orthonormal sequence $(e_n)$. Letthe unit sphere. This set is norm closed but is not convex. By the Bessel inequality, for every $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$; equivalently, this is a weakly null orthonormal sequence. Every $e_n$ lies in $C$, but the weak limit $0$ does not. Hence convexity is essential.