missing a condition for convergence or not

Question

In preparation of a test I have some problems solving the following problem: Let $A:=\{a\in \ell^{2}: \phantom{x} \|a\|_{2}\leq 1\}$, $(a_{n})_{n} \in A, a \in A$. Prove that for all $b \in \ell^{2}$ the following two statements are equivalent

\begin{align*} (i) & \phantom{x} \langle a_{n},b \rangle\rightarrow \langle a,b \rangle \\ (ii) & \phantom{x} a_{n}\rightarrow a \\ \end{align*}

as $n\rightarrow\infty$ are equivalent.

Also the following metric is given with $a,b\in A$

$$d(a,b)=\sum_{n=1}^{\infty}(1/2)^{n}|a_{n}-b_{n}|$$

My work:

(2) $\Rightarrow$ (1):

Let $(a_{n})_{n}$ and $a \in A$ with $A:=\{a\in \ell^{2}| \phantom{x} \|a\|_{2}\leq 1\}$. Suppose that $a_{n}\rightarrow a$ as $n\rightarrow\infty$. We note that the norm-maping $A\longrightarrow\mathbb{R}_{+}$, $a\mapsto \|a\|$ is continuous, i.e.,

$$a_{n}\rightarrow a \phantom{x}\mathrm{in}\phantom{x}A\phantom{x}\Rightarrow \phantom{x} \|a_{n}\|\rightarrow \|a\| \phantom{x}\mathrm{in}\phantom{x}\mathbb{R}.$$

So we have $\|a_{n}-a\|\rightarrow 0$ as $n\rightarrow\infty$. Take an arbitrary $b\in\ell^{2}$ and get

\begin{align*} |\langle a_{n},b \rangle -\langle a,b\rangle| &=|\langle a_{n}-a,b \rangle|\phantom{xxxx}\mathrm{(sesquilinearity)} \\ &\leq \|a_{n}-a\|\|b\| \phantom{xxxx}\mathrm{(Cauchy-Schwarz\phantom{x} inequality)} \\ &\rightarrow 0, \end{align*}

since $\|a_{n}-a\|\rightarrow 0$ as $n\rightarrow\infty$. As $b$ was chosen arbitrarely we have

$$\langle a_{n},b\rangle\rightarrow \langle a,b\rangle\phantom{x} \forall b \in \ell^{2}.$$

Question 1: Have I done this part correctly?

Now (1) $\Rightarrow$ (2):

Suppose that $\langle a_{n},b\rangle\rightarrow \langle a,b\rangle$ for all $b\in\ell^{2}$. We have using sesquilinearity, symmetry and the fact that $z+\overline{z}=2$Re(z) for every complex number $z\in\mathbb{C}$

\begin{align*} \|a_{n}-a\|^{2} &= \langle a_{n}-a,a_{n}-a \rangle \\ &= \langle a_{n}, a_{n} \rangle - \langle a_{n}, a \rangle - \langle a,a_{n} \rangle + \langle a, \rangle \\ &= \|a_{n}\|^{2}-\langle a_{n},a \rangle- \overline{\langle a_{n},a \rangle} +\|a\|^{2} \\ &= \|a_{n}\|^{2}-2Re\langle a_{n}, a \rangle +\|a\|^{2} \\ \end{align*}

Question 2: Now I would like to use $\|a_{n}\|\rightarrow\|a\|$ as $n\rightarrow\infty$ but I don't believe I can conclude that from (ii) or am I wrong? If so how do I prove this? Or is $\|a_{n}\|\rightarrow\|a\|$ as $n\rightarrow\infty$ a necessary assumption?

Question 3: Can I get it from the metric $d(a,b)=\sum_{n=1}^{\infty}(1/2)^{n}|a_{n}-b_{n}|$ with $a,b\in A$. Would it than be possible to conclude $\|a_{n}\|\rightarrow\|a\|$ as $n\rightarrow\infty$?

The rest of the proof I would do it like this:

Since $\|a_{n}\|\rightarrow\|a\|$ as $n\rightarrow\infty$ we can write

\begin{align*} \|a_{n}-a\|^{2} &= \|a_{n}\|^{2}-2Re\langle a_{n},a \rangle +\|a\|^{2} \\ &\rightarrow \|a\|^{2}-2Re\langle a,a \rangle +\|a\|^{2} \\ &= 2\|a\|^{2}-2\|a\|^{2} \\ &=0. \end{align*}

So we have taking the square root $\|a_{n}-a\|\rightarrow 0$, thus we have $a_{n}\rightarrow a$ as $n\rightarrow\infty$.

Answer 1

With the metric

$$d(a,b) = \sum_{n=1}^\infty \frac{1}{2^n}\lvert a_n - b_n\rvert,$$

the first part is not right, since convergence in $d$ is much weaker than convergence in the norm.

Since $x \in A \Rightarrow \lVert x\rVert_2 \leqslant 1 \Rightarrow \lvert a_n \rvert \leqslant 1$, so $\lvert a_n - b_n\rvert \leqslant 2$ for all $a,b \in A$ and $n \in\mathbb{Z}^+$, convergence in $d$ is componentwise convergence, i.e.

$$\lim_{k\to\infty} d(a^{(k)},a) = 0 \iff \bigl(\forall n\bigr)\bigl(\lvert a^{(k)}_n - a_n\rvert \to 0\bigr).$$

The one direction is clear, fixing $n$, we have

$$d(a^{(k)},a) \to 0 \iff 2^n d(a^{(k)},a) \to 0 \Rightarrow \lvert a^{(k)}_n - a_n\rvert \to 0.$$

For the other direction, given $\varepsilon > 0$, choose $N$ large enough that $\sum\limits_{n=N}^\infty 2^{-n}\cdot 2 < \varepsilon/2$. Then, if $a^{(k)} \to a$ componentwise, you can choose $k_0$ large enough that for $k \geqslant k_0$ you have $\lvert a^{(k)}_n - a_n\rvert < \varepsilon/2$ for all $n \leqslant N$. Then

$$\begin{align} d(a^{(k)},a) &= \sum_{n=1}^\infty 2^{-n}\lvert a^{(k)}_n - a_n\rvert\\ &= \sum_{n=1}^N 2^{-n}\lvert a^{(k)}_n - a_n\rvert + \sum_{n=N+1}^\infty 2^{-n}\lvert a^{(k)}_n-a_n\rvert\\ &\leqslant \sum_{n=1}^N 2^{-n} \frac{\varepsilon}{2} + \sum_{n=N+1}^\infty 2^{-n}\cdot 2\\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \end{align}$$

for $k \geqslant k_0$, so $\lim\limits_{k\to\infty} d(a^{(k)},a) = 0$.

Now the implication $(ii) \Rightarrow (i)$ is the easier one, just note that

$$\langle a, e_n\rangle = a_n$$

where $e_n$ is the standard unit vector with $1$ in the $n$-th coordinate and $0$ elsewhere. So if $\langle a^{(k)},b\rangle \to \langle a,b\rangle$ for all $b \in \ell^2$, in particular you have $\langle a^{(k)}, e_n\rangle = a^{(k)}_n \to a_n = \langle a,e_n\rangle$ for all $n$. By the above, that implies that $d(a^{(k)},a) \to 0$.

For the implication $(i) \Rightarrow (ii)$, I suppose we are not yet in a position to argue with the fact that $A$ is weakly compact, so we need an elementary argument. Let $b \in \ell^2$ arbitrary. For all $\varepsilon > 0$, there is an $N \in \mathbb{Z}^+$ with

$$\sum_{n=N+1}^\infty \lvert b_n\rvert^2 < \frac{\varepsilon^2}{9}.$$

Let $b_N$ denote the approximation to $b$ whose first $N$ components are the corresponding components of $b$, and whose later components are $0$, so $b_N = \sum\limits_{n=1}^N \langle b,e_n\rangle e_n$. Then $\lVert b-b_N\rVert < \varepsilon/3$. By the componentwise convergence of $a^{(k)}$ to $a$, there is a $k_0$ such that for all $k \geqslant k_0$ and $n \leqslant N$ we have

$$\lvert a^{(k)}_n - a_n\rvert < \frac{\varepsilon}{3N(1+\lvert b_n\rvert)}.$$

Then, for $k \geqslant k_0$ we have

$$\lvert \langle a^{(k)}, b\rangle -\langle a,b\rangle\rvert \leqslant \underbrace{\lvert \langle a^{(k)}, b-b_N\rangle \rvert}_{\leqslant \lVert a^{(k)}\rVert\cdot \lVert b-b_N\rVert < \varepsilon/3} + \underbrace{\lvert \langle a^{(k)} -a , b_N\rangle\rvert}_{< \varepsilon/3} + \underbrace{\lvert \langle a,b_N - b\rangle\rvert}_{\leqslant \lVert a\rVert\cdot \lVert b-b_N\rVert < \varepsilon/3} < \varepsilon$$

and hence $\langle a^{(k)},b\rangle \to \langle a,b\rangle$.