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The book "Introductory Functional Analysis with Applications" (Kreyszig) presents the following definitions.

A sequence $(x_n)$ in a normed space $X$ is said to be strongly convergent if there is an $x\in X$ such that $\lim\|x_n-x\|=0$. (page 256)

A sequence $(x_n)$ in a normed space $X$ is said to be weakly convergent if there is an $x\in X$ such that $\lim f(x_n)=f(x)$ for every $f\in X'$. (page 257)

Then is proved that strong convergence implies weak convergence but the converse is not generally true (unless that $X$ is finite-dimensional). To prove it, he gives an example in a Hilbert space and uses the Riesz Representation Theorem and the Bessel inequality (page 259).

I would like an exemple that weak convergence does not implies strong convergence in a normed space $X$ that is not a Hilbert space. Is there one?

Thanks.

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Consider $\ell^p$, the space of real sequences for which $\sum_j|x_j|^p$ is finite with natural norm. If $p\neq 2$, this space is not a Hilbert space. If $1\lt p\lt \infty$, take $e_{j}=(0,\dots,0,1,0,\dots)$, where the $1$ is at the $j$-th position. This sequence converges weakly to $0$ in $\ell^p$ but not strongly. In order to see the weak convergence, we can use the fact that $f$ can be represented as $f\left(\left(x_n\right)_{n\geqslant 1}\right)=\sum_{n\geqslant 1}f_n x_n$ where $\left(f_n\right)_{n\geqslant1}\in \ell^{p'}$ and $p'$ is the conjugate exponent of $p$, that is,$1/p+1/p'=1$. Then $f\left(e_n\right)=f_n$ and $\left\lvert f_n\right\rvert^{p'}\to 0$ because of the convergence of $\sum_{n\geqslant 1}\left\lvert f_n\right\rvert^{p'}$.

The case $p=1$ is different. If weak convergence and convergence in norm in $(X,\lVert\cdot\rVert)$ are equivalent, we say that $(X,\lVert\cdot\rVert)$ has the Schur property.

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  • $\begingroup$ What's the natural norm? $\endgroup$ – Pedro Nov 29 '13 at 16:43
  • $\begingroup$ The norm $\lVert x\rVert_p:=\left(\sum_{j=1}^\infty |x_j|^p\right)^{1/p}$. $\endgroup$ – Davide Giraudo Nov 29 '13 at 16:44
  • $\begingroup$ Your sequence $(e_n)_{n\in\mathbb{N}}$ doesn't converges strongly to $0$ because $\|e_n-0\|=(\sum_j|e_j|^p)^{1/p}=1$ for all $n\in\mathbb{N}.$ How can we see that $|f(e_n)|\overset{n\to\infty}\longrightarrow 0$ for all $f$ in the dual of $\ell^p$? $\endgroup$ – Pedro Nov 29 '13 at 19:05
  • $\begingroup$ It's indeed not straightforward. If we know that the dual of $\ell^p$ for $1\lt p\lt \infty$ is $\ell^q$ with $p^{-1}+q^{-1}=1$ it's okay. $\endgroup$ – Davide Giraudo Nov 29 '13 at 19:06
  • $\begingroup$ Even with the knowledge that the dual of $l^p$ is $l^q$ with $1/p + 1/q = 1$, I'm not sure how to prove $|f(e_n)| \to 0$ for all $f$ in the dual of $l^p$. Would you mind outlining how to prove it? $\endgroup$ – user2139 May 27 '18 at 5:20
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Another easy non-Hilbert example: Take $X = C_0(\mathbb{R})$ with the uniform norm. $X^*$ is the Radon measures on $\mathbb{R}$. The sequence $\{ \chi_{[n, n + 1]} \}_n$ goes to $0$ weakly but is not Cauchy in norm.

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