- What does it mean to say that $x_n ⇀ x$ in $X$, i.e., a sequence $x_n$ converges to $x$ weakly in $X$ ?
- Prove that any weakly converging sequence in $X$ is bounded.
- Let $x_n ⇀ 0$ in $X$. Show that one can select a subsequence $(x_{n_k})$ with the following property \[ ⟨x_{n_1}+x_{n_2}+…+x_{n_k}, x_{n_{k+1}}⟩ ⩽ 1 \] for all $k ∈ ℕ$. Prove by induction that \[ \|x_{n_1}+x_{n_2}+…+x_{n_k}\| ⩽(2+M) k \] for all $k ∈ ℕ$, where $M=\sup_n\|x_n\|^2$. Deduce from the above inequality that \[ y_m=\frac{1}{m} \sum_{k=1}^m x_{n_k} → 0 \] as $m → ∞$.

- $x_n ⇀ x$ if and only if $⟨ x_n, y⟩ →⟨ x, y⟩$ for all $y ∈ X$.
- We let $π_n^*(y)=⟨ x_n, y⟩$ for $y ∈ X$. We have $\|π_n^*\|_{X^*}=\|x_n\|$ and the sequence $π_n^*(y)$ is bounded for each $y ∈ X$. The result follows from the Uniform Boundedness Principle.
- We let $n_1=1$, sequence $⟨ x_{n_1}, x_n⟩$ converges to zero as $n → ∞$ and thus we can find $n_2>n_1$ such that $⟨ x_{n_1}, x_{n_2}⟩ ⩽ 1$. Sequence $⟨ x_{n_1}+x_{n_2}, x_n⟩ → 0$ and thus we can find $n_3>n_2$ such that $⟨ x_{n_1}+x_{n_2}, x_{n_3}⟩ ⩽ 1$ and so on.

By (ii), there is positive $M$ such $\sup_n\|x_n\|^2 ⩽ M$. Let us show by induction the following identity \[ \|x_{n_1}+x_{n_2}+…+x_{n_k}\|^2 ⩽(2+M) k \] for all $k ∈ ℕ$. For $k=1$ it is true. Assume that the inequality above holds and let us show that it true for $k+1$. Indeed, we have \begin{aligned} \|x_{n_1}+x_{n_2}+…+x_{n_{k+1}}\|^2 &⩽\|x_{n_1}+x_{n_2}+…+x_{n_k}\|^2+2⟨ x_{n_1}+x_{n_2}+…+x_{n_k}, x_{n_{k+1}}⟩+\|x_{n_{k+1}}\|^2\\&⩽(2+M) k+2+M=(2+M)(k+1) . \end{aligned} The final result follows from the estimate \[ \|y_m\|^2=\frac1{m^2}\|x_{n_1}+x_{n_2}+…+x_{n_m}\|^2⩽ \frac{2+M}{m}→0\text {. } \]

- It is said that a subset $K$ of $X$ is
*sequentially weakly closed*if the limit of any weakly converging sequence $x_n ∈ K$ belongs to $K$. Prove that a convex subset $K$ of $X$ is sequentially weakly closed if and only if $K$ is closed. - Assume that $X$ is infinite dimensional. Show that for any $x$ such that $\|x\| ⩽ 1$ there is a sequence $(x_n)$ on the unit sphere, i.e., $\|x_n\|=1$, such that $x_n ⇀ x$.

- Obviously, sequentially weakly closed set is closed. To prove the opposite statement, assume that $x_j ∈ K$ and $x_j ⇀ x$. By the previous question, there exists a subsequence $x_{j_k}$ such that \[ y_m=\frac{1}{m} \sum_{k=1}^m x_{j_k} → x \] as $m → ∞$. By convexity of $K, y_m ∈ K$ for any $m$. Since $K$ is closed, $x ∈ K$ and thus $K$ is sequentially weakly closed.
- We let $Y=\operatorname{Span}(x)$. Then $X=Y ⊕ Y^{⟂}$. Let $\{e_n^⟂\}_{n=1}^{∞}$ be an orthonormal sequence in $Y^{⟂}$. By Bessel's inequality, $e_n^{⟂} ⇀ 0$ in $Y^{⟂}$ and thus in $X$. We let $x_n=λ e_n^{⟂}+x$ with $λ=\sqrt{1-\|x\|^2}$. By construction, $\|x_n\|=1$ and $x_n ⇀ x$.

- $F$ is continuous on $K$;
- $F$ is convex on $K$, i.e., $F(λ x+(1-λ) y) ⩽ λ F(x)+(1-λ) F(y)$ for any $x$ and $y$ from $K$ and for any $0<λ<1$;
- if $K$ is unbounded then $\|x_n\| → ∞$ implies $F(x_n) → ∞$.

We let $d=\inf_{x ∈ K} F(x)$. Let $x_n ∈ K$ be a minimising sequence, i.e, $F(x_n) → d$. By the property (iii), $\sup\|x_n\|<∞$. In a Hilbert space, a bounded set is sequentially weakly compact and there is a subsequence, denoted again by $x_n$, that converges weakly to $x_0$ in $X$. By the result of question (a), we can find a subsequence $x_{n_k}$ such that
\[
y_m=\frac{1}{m} \sum_{k=1}^m x_{n_k} → x_0 .
\]
By convexity of $K, y_m ∈ K$ and thus $x_0 ∈ K$. By convexity of $F$,
\[
d ⩽ F(y_m) ⩽ \frac{1}{m} \sum_{k=1}^m F(x_{n_k}) → d .
\]
So, we have $F(y_m) → d$ as $m → ∞$. By continuity of $F$, we find $F(y_0)=d$.