- Let $X$ be a Banach space. What does it mean to say that a subset $A$ of $X$ is weakly sequentially compact? State a necessary and sufficient condition for the weak sequential compactness of the closed unit ball of $X$.
- Give an example of a bounded sequence in $L^1(0,1)$, the space of real-valued integrable functions on $(0,1)$, which does not have any weakly convergent subsequence. Justify your answer.

[You may use without proof that the dual of $L^1(0,1)$ is $L^{∞}(0,1)$, and the fact that if an integrable function $h$ satisfies $∫_E h d x=0$ for all measurable subsets $E$ of some given interval $[α, β]$, then $h=0$ a.e. in $[α, β]$.]

- $A$ is weakly sequentially compact if every sequence in $A$ has a subsequence that converges weakly to an element $x$ of $A$. The closed unit ball of $X$ is weakly sequentially compact if and only if $X$ is reflexive.
- We show that the sequence $f_n=n χ_{[0,1 / n]}$ is bounded but contains no weakly convergent subsequence in $L^1(ℝ)$. First, we compute \[ \|f_n\|=∫_0^{1 / n} n d x=1 \] hence $(f_n)$ is bounded in $L^1(0,1)$. We show that $f_n$ has no weakly convergent subsequence by a contradiction argument. Suppose that a subsequence $(f_{n_k})$ converges weakly to some $f$, i.e. \[ \lim_{k → ∞} ∫_0^1 f_{n_k} g d x=∫_0^1 f g d x \text { for all } g ∈ L^{∞}(0,1) . \] Now note that, for any measurable subset $E$ which is contained in a compact subset $[α, β]$ of $(0,1)$, the supports of $f_{n_k}$ are eventually disjoint from $E$. Hence, taking $g=χ_E$ yields $∫_E f d x=0$. This implies that $f=0$ a.e. in $[α, β]$. Since $[α, β]$ is arbitrary, $f=0$ a.e. in $(0,1)$. We now take $g=1$ as a test function to obtain \[ 1=\lim_{k → ∞} ∫_{ℝ} f_{n_k} d x=∫_{ℝ} f d x=0 \]

- State the principle of uniform boundedness.

For the rest of this question, $H$ is a real Hilbert space with inner product $⟨⋅, ⋅⟩$ and norm $\|⋅\|$ and $(x_n)$ and $(z_n)$ are two sequences in $H$. You may use without proof the Cauchy-Schwarz inequality and the Riesz representation theorem. - Suppose that $(⟨y, x_n⟩)$ is convergent for each $y ∈ H$. Show that $(x_n)$ is weakly convergent.
- Prove that if $(⟨y+\frac{1}{\|x_n\|^2+1} x_n, x_n⟩)$ is bounded from below for each $y ∈ H$, then $(x_n)$ has a weakly convergent subsequence. If we have instead that $(⟨y+x_n, x_n⟩)$ is bounded from below, is it true that $(x_n)$ has a weakly convergent subsequence?

- Principle of uniform boundedness states that if $X$ is a Banach space and $Y$ is a normed space, and if $ℱ$ is a family of bounded linear operators from $X$ to $Y$ such that $\sup_{T∈ℱ}\|T_n x\|_Y<∞$ for all $x ∈ X$, then $\sup_{T∈ℱ}\|T_n\|_{ℬ(X,Y)}<∞$.
- For $y ∈ H$, let $T_n(y)=⟨y, x_n⟩$. By the property of the inner product, $T_n$ is linear. The boundedness of $T_n$ is a consequence of the Cauchy-Schwarz inequality: $\|T_n\|_* ⩽\|x_n\|$. Hence $T_n ∈ H^*$.

By hypothesis, the family $\{T_n\}$ is pointwise bounded. Since $H$ is complete, the principle of uniform boundedness implies that $(T_n)$ is bounded.

By the Riesz representation theorem, $\|T_n\|_*=\|x_n\|$. (Alternatively, student can argue as follows: If $x_n=0$, this is clear. Suppose that $x_n ≠ 0$. We have $\|x_n\|^2=T_n(x_n) ⩽\|T_n\|_*\|x_n\|$, which implies that $\|x_n\| ⩽\|T_n\|_*$. Together with the inequality $\|T_n\|_* ⩽\|x_n\|$ which was proved earlier, this gives the claim.)

The last two assertions show that $(x_n)$ is bounded, say $\|x_n\| ⩽ M$. Consider now $ℓ(y)=\lim ⟨y, x_n⟩$. It is clear that $ℓ$ is linear. Also, by the Cauchy-Schwarz inequality, $|ℓ(y)| ⩽ M\|y\|$ and so $ℓ$ is bounded. By the Riesz representation theorem, we have that $ℓ(y)=⟨y, x⟩$ for some $x$. It is now readily seen that $(x_n)$ converges weakly to $x$. - The statement is true. Suppose for each $y ∈ H$ that \[ ⟨y+\frac{1}{\|x_n\|^2+1} x_n, x_n⟩ ⩾-M(y) \text { for all } n \text {. } \] Then, \[ ⟨y, x_n⟩ ⩾-M(y)-\frac{\|x_n\|^2}{\|x_n\|^2+1} ⩾-M(y)-1 . \] This implies that \[ ⟨-y, x_n⟩ ⩾-M(-y)-1 \text { and so }⟨y, x_n⟩ ⩽ M(-y)+1 . \] It follows that $(⟨y, x_n⟩)$ is bounded for each $y ∈ H$. As in the previous part, we then have that $(x_n)$ is bounded. Note that $H$ is reflexive thanks to the Riesz representation theorem. Hence bounded subsets of $H$ are weakly sequentially compact. In particular, $(x_n)$ has a weakly convergent subsequence. For the last part, one has by Cauchy-Schwarz' inequality that $⟨y+x_n, x_n⟩=\|x_n\|^2+⟨y, x_n⟩ ⩾\|x_n\|^2-\|y\|\|x_n\| ⩾-\frac{1}{4}\|y\|^2$. So the statement $(⟨y+x_n, x_n⟩)$ is bounded from below does not give any information about $(x_n)$.