Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $x_n \to x$ weakly. My question is: does it hold that $\|x_n\|\to \|x\|$?

I haven't been able to work out the answer and I'd appreciate help with it but here are my thoughts:

Given the inverse $\Delta$-inequality: $|\|x_n\|-\|x\|| \le \|x_n -x\|$ it's clear that if they converge strongly then $\|x_n\|\to \|x\|$.

If the norm $\|.\|$ was continuous then it would hold but I suspect the norm is not continuous in the weak topology (although unfortunately I cannot seem to argue why this is so. Any hints appreciated).

My goal therefore is to show that $\|x_n \| \not \to \|x\|$ by finding an example of a Banach space $X$ and a sequence $x_n$ with $x_n \to x$ weakly but not $\|x_n\|\to \|x\|$.

share|improve this question
2  
No. Consider the sequence of standard unit vectors in $\ell_2$. –  David Mitra Feb 16 at 12:57
add comment

2 Answers 2

Example. Let $X=L^2[0,2\pi]$ and $f_n(x)=\sin nx$. Then $f_n\to 0$ weakly, due to Riemann-Lebesgue Lemma, but $\|f_n\|=\sqrt{\pi}$.

On the other hand, if $\|f_n\|\to 0$, then $f_n\to 0$, strongly.

Note. If the case of Hilbert spaces, if $f_n\to f$ weakly, then $$ (f_n-f,f_n-f)=(f_n,f_n)+(f,f)-(f_n,f)-(f,f_n). $$ Clearly $(f_n,f),\, (f,f_n)\to (f,f)$ but $(f_n,f_n)$ does not in general converge to $(f,f)$. This means that $$ \limsup_{n\to\infty}(f_n-f,f_n-f)=\limsup_{n\to\infty}(f_n,f_n)-(f,f). $$ Thus $\limsup_{n\to\infty}(f_n,f_n)\ge (f,f)$, and the convergence $f_n\to f$ is strong iff
$\limsup_{n\to\infty}(f_n,f_n)= (f,f)$.

share|improve this answer
add comment

The answer is "no". In $c_0$ and $\ell_p$, $1<p<\infty$, the sequence of the standard unit vectors provides a counterexample.

One can find counterexamples in a large class of spaces:

Let $X$ be a Banach space that lacks the Schur property. ($X$ is said to have the Schur property if every weakly convergent sequence in $X$ is norm convergent. Note infinite dimensional reflexive spaces lack the Schur property since their closed unit balls are weakly sequentially compact but not norm compact.)

Let $(x_n)$ be a sequence in $X$ that converges weakly to $x$ but which does not converge in norm. Then the sequence $(x_n-x)$ converges weakly to $0$. Since $(x_n-x)$ does not converge to $0$ in norm, there is an $\alpha>0$ and a subsequence $(x_{n_k}-x)$ such that $\Vert x_{n_k}-x\Vert>\alpha$ for all $k$.


One class, more general than the class of reflexive spaces, of infinite dimensional Banach spaces that lack the Schur property are those that do not contain $\ell_1$:

Let $X$ be an infinite dimensional Banach space that does not contain $\ell_1$. Let $(x_n)$ be an $\epsilon$-separated sequence from $S(X)$, the closed unit sphere of $X$, for some $\epsilon>0$. By Rosenthal's $\ell_1$-theorem, $(x_n)$ has a weakly Cauchy subsequence. Call this subsequence (still) $(x_n)$. Then the sequence $(x_{2n}-x_n)$ converges weakly to $0$ but not in norm (since it isn't norm Cauchy).


(One can show that if $X$ is an infinite dimensional Banach space, then $S(X)$ is weakly sequentially dense in $B(X)$ if and only if $X$ does not have the Schur property.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.