# In a normed vector space X: $x_n \to x$ weakly iff $d(x_n) \to d(x) ~\forall d \in D$, $D$ dense in $X^*$

Good day, I have the following task:

Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in the normed vector space $(X, || \cdot || )$ and let $x \in X$. Show that the following are equivalent:

(i) $x_n$ converges weakly to $x$

(ii) the sequence $(||x_n||)_{n \in \mathbb{N}}$ is bounded and there exists a dense subset $D \subset X^*$ such that $d(x_n) \to d(x)$ for all $~d \in D$.

I know from the lecture:

• $X^* = B(X,\mathbb{R})=\{T:X \to \mathbb{R} ~| ~T ~\text{linear} \wedge \exists M < \infty ~\forall x \in X: ||T(x)|| \leq M ||x|| \}$
• let $X$ be a normed space then $\sigma(X,X^*)$ is called the weak topology and it is the weakest topology on $X$ such that every $f \in X^*$ is continuos
• $x_n \to x$ weakly $\Leftrightarrow$ $f(x_n) \to f(x)$ for all $~f \in X^*$

The last equivalency looks similar to what I have to prove just by using a dense subset of $X^*$. Dense means that $\bar D = X^*$ or equivalently for every $f \in X^*$ I can find a sequence $(d_n)$ in D such that $d_n \to f$.

Therefore: $$x_n \to x ~\text{weakly}~ \Leftrightarrow f(x_n) \to f(x) ~\forall~f \in X^* \Leftrightarrow \lim\limits_{m \rightarrow \infty}{d_m}(x_n) \to \lim\limits_{m \rightarrow \infty}{d_m}(x) ~\forall~ (d_m) \subset D ~\text{s.t.} ~d_m \to f$$

Thanks a lot, Marvin

(ii) $\implies$ (i): Fix $f\in X^*$. Let $k=\sup\{\|x_n\|\}$. For any $m$ \begin{align} \|f(x_n)-f(x)\|&\leq\|f(x_n)-d_m(x_n)\|+\|d_m(x_n)-d_m(x)\|+\|d_m(x)-f(x)\|\\ \ \\ &\leq \|x_n\|\,\|f-d_m\|+\|d(x_n)-d\|+\|x\|\,\|f-d_m\|\\ \ \\ &\leq k\,\|f-d_m\|+\|d(x_n)-d\|+\|x\|\,\|f-d_m\|. \end{align} Then $$\limsup_{n\to\infty}\|f(x_n)-f(x)\|\leq (k+\|x\|)\,\|f-d_m\|.$$ As $\|d-f_m\|\to0$, we get $\limsup_{n\to\infty}\|f(x_n)-f(x)\|=0$, from where $$\lim_{n\to\infty}\|f(x_n)-f(x)\|=0.$$

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(i) $\implies$ (ii): We can take $D=X^*$. The question is how to show that $\{x_n\}$ is bounded. For any $f\in X^*$, we know that the sequence of numbers $\{f(x_n)\}$ converges to $f(x)$, so it is bounded. Thus, for any $f\in X^*$, $$\sup_n |f(x_n)|<\infty.$$ Now we apply the Uniform Boundedness Principle, with the $x_n$ as the functionals (that is, we think of $x_n\in X^{**}$). The conclusion is that $$\sup_n\|x_n\|<\infty.$$ The only thing is that these are the norms of the $x_n$ in the double dual; but these agree with the norms in $X$.

• Thanks a lot, didn't think of Banach-Steinhaus :) Good structure in your proof – Fritz Dec 3 '15 at 20:41
• Glad I could help! – Martin Argerami Dec 3 '15 at 21:55

If $p\notin D$, for all $n \in \mathbb{N}$ there exists a $d_n\in D$ so that $||p-d||≤\frac{1}{n}$ (in supnorm), via triangle inequality you get

$$||p(x)-p(x_m)||≤||p(x)-d_n(x)|| + ||d_n(x)-d_n(x_m)|| +||p(x_n)-d_n(x_m)||≤\frac{||x_m||+||x||}{n}+||d_n(x)-d_n(x_m)||$$

If $||x_m||$ is bounded the right hand side converges to $\frac{||x||+\ \text{bound}}{n}$,since $n$ is arbitrary this means $p(x_m)$ converges to $p(x)$.

For the other direction the only difficult step is to show that weak convergence implies boundedness of the series. To do this we note every element of $X$ can be considered an element of the bidual $X^{**}$ and that $X^*$ is a Banach space. Since $x_m(d) = d(x_m) \to d(x) = x(d)$ we have $\sup_{m \in \mathbb{N}}|x_m(d)|$ is finite for all $d \in X^{*}$. Now we can apply the Banach-Steinhaus theorem to get that there exists an upper bound on $\{||x_m||\}_{m\in\mathbb{N}}$ and as such $||x_m||$ is bounded.

• Thanks a lot for your help – Fritz Dec 3 '15 at 20:41