Weak convergence of norms of sequence

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 $\|x_n\|\to \|x\|$.

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No. Consider the sequence of standard unit vectors in $\ell_2$. –  David Mitra Feb 16 at 12:57

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
$\limsup_{n\to\infty}(f_n,f_n)= (f,f)$.

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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.)

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