Question about weak and strong convergence

Question

I have the following exercise that concerns me:

Let $X$ be a normed space and $X^*$ its dual space. Suppose $\text{s-lim}\ l_n = l$ and $x_n \to x$. Then $l_n(x_n) \to l(x)$. Does this still hold, if $\text{s-lim}\ l_n = l$ and $x_n \rightharpoonup x$?

We use the following definitions for strong and weak convergence: if $l(x_n) \to l(x)$, we say that $x_n$ converges weakly to $x$ and we write $x_n \rightharpoonup x$. Furthermore, we have $\text{s-lim} \ l_n = l$ iff $l_n(x) \to l(x)$.

My answer to the first question: we look at

$$\lVert l_n(x_n) - l(x) \rVert \leq \lVert l_n(x_n) - l(x_n)\rVert + \lVert l(x_n) - l(x)\Vert \leq \lVert l_n(x_n) - l(x_n)\rVert + \lVert l \rVert \lVert x_n - x \rVert \to 0$$

where we have used the strong convergence $\text{s-lim}\ l_n = l$ for the first and the norm convergence $x_n \to x$ for the second summand.

Now, based on my approach and the definitions I have given, the second question also has to be true. But I am a little bit unsure about that and I think someone really has to pay attention to the details when it comes to dealing with weak and strong convergence. Could anyone maybe help me with that?

$l_n(x) \to l(x)$ for all $x$ does not give you $l_n(x_n) \to l(x_n)$. You cannot make $x$ dependent on $n$. — Kavi Rama Murthy, Jan 05 '24 at 23:09

Martin Argerami · Accepted Answer · 2024-01-06T05:16:12.473

1

The estimate you need is similar to what you did. You have \begin{align} |l_n(x_n)-l(x)| &\leq |l_n(x_n)-l_n(x)|+|l_n(x)-l(x)|\\[0.2cm] &\leq\|l_n\|\,\|x_n-x\|+|l_n(x)-l(x)|. \end{align} Now, because a weak$^*$-convergent sequence is bounded, there exists $c>0$ with $\|l_n\|\leq c$. This gives us \begin{align} |l_n(x_n)-l(x)| &\leq c\,\|x_n-x\|+|l_n(x)-l(x)|. \end{align} Then \begin{align} \limsup_n|l_n(x_n)-l(x)| &\leq \limsup_n c\,\|x-x_n\|+|l(x_n)-l(x)|=0. \end{align} This proves that the limit exists and $\lim_n|l_n(x_n)-l(x)|=0$.

When $x_n\rightharpoonup x$, the result can fail. For instance let $X=c_0$, and consider the canonical basis $e_n$. Then $e_n\rightharpoonup 0$. Consider also $e_n\in \ell^1=c_0^*$. This gives us the functionals $l_n(x)=x_n$. Given any $x\in c_0$, $$ l_n(x)=x_n\to0, $$ so $l_n\to0$ in the weak$^*$-topology. But $l_n(e_n)=1$ for all $n$.

edited Jan 06 '24 at 05:16

answered Jan 06 '24 at 03:09

Martin Argerami

217,281

I think the expression $$l_n(x_n)-l(x)=l_n(x_n-x)+l_n(x)-l(x)$$ gives the convergence faster, with no need to introduce $x_m.$ – Ryszard Szwarc Jan 06 '24 at 05:10
Yes indeed! $ \ $ – Martin Argerami Jan 06 '24 at 05:11
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Moreover a slightly simpler example is provided by any orthonormal sequence $e_n$ in a Hilbert space. It tends to $0$ weakly by the Bessel inequality and $\langle e_n,e_n\rangle=1.$ Sorry for being pain ... – Ryszard Szwarc Jan 06 '24 at 05:21
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No worries. But it's the same example, isn't it. Just in $\ell^2$-$\ell^2$ instead of $c_0$-$\ell^1$. – Martin Argerami Jan 06 '24 at 05:24
The same, except that the underlying spaces are different. I always consider inner product spaces simpler than the normed ones. – Ryszard Szwarc Jan 06 '24 at 07:23

Question about weak and strong convergence

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