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I want to prove that in a normed linear space $H$, that if $x_n$ is weak convergent to $x$, and $\lim_{n\to\infty} \|x_n\| = \|x\|$ then:

$$\lim_{n\to\infty}\|x_n-x\|=0$$

Can I please have a hint? Also does $\langle x,x\rangle^{\frac12}=\|x\|$ or something?

The impression I have is, essentially $\lim_{n\to\infty}\|x_n\|-\|x\|=\lim_{n\to\infty}\|x_n - x\|$ due to the weak convergence, otherwise in the general case:

$$\|x\|-\|y\|=\|x - y\|$$ could be true, but only if they are in opposite directions.

My definition of weak convergence is $\lim_{n\to\infty} \langle x_n,y\rangle =\langle x,y\rangle$, so I suspect that this is a hilbert space.

Functional Analysis
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    Really important: is this a Hilbert space? – bartgol Aug 13 '15 at 15:45
  • Hmmm I suspect so based on weak convergence, but it isn't specified – Functional Analysis Aug 13 '15 at 15:46
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    There is generally no relationship between $\langle x,y \rangle^{1/2}$ and $|y-x|$ – Ben Grossmann Aug 13 '15 at 15:46
  • @Omnomnomnom See edit, sorry – Functional Analysis Aug 13 '15 at 15:47
  • Weak convergence makes perfect sense in Banach spaces. But if you are in a Hilbert space, we can prove the claim quite easily. – bartgol Aug 13 '15 at 15:47
  • @bartgol I edited in my definition of weak convergence at the end. Since it is an inner product, I suspect it is an inner-product space, but without completeness I don't know – Functional Analysis Aug 13 '15 at 15:50
  • @Bartgol: Does weak convergence together with $\left|x_{n}\right|\rightarrow\left|x\right|$ imply strong convergence in any arbitrary Banach space $X$? If $X$ is reflexive, then it would seem true by simple modification of the Hilbert space proof. – Matt Rosenzweig Aug 13 '15 at 16:02
  • @MattRosenzweig Consider $u_n(x)=1+\sin(nx)$ for $x\in(0,2\pi)$. Then $u_n\to u\equiv 1$ weakly in $L^1(0,2\pi)$ and $|u_n|=|u|$ for all $n$, but $u_n$ does not converge to $u$ in $L^1$ norm. –  Aug 13 '15 at 16:22
  • @ByronSchmuland: Thanks. I believe I may have mispoke too about the radon-riesz property holding in a reflexive Banach space. – Matt Rosenzweig Aug 13 '15 at 16:32
  • @MattRosenzweig I think it would for the reflexive case, using the canonical isometric isomorphism from $X$ to $X^{**}$ and some corollary of the Hanh-Banach theorem... For the general Banach space case, I have the feeling it would fail, but I haven't thought deeply about it... – bartgol Aug 13 '15 at 19:15
  • @Bartgol: See the comments to Hamza's answer below. – Matt Rosenzweig Aug 13 '15 at 19:21

2 Answers2

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This property is called "(H)" property and the space is called a "Radon Riesz" space.

As example of this space we can say every Hilbert space as the proof given by @bartgol and $L^p$ for $1<p<\infty$, and schur space (in particular $\ell^1$).

Hamza
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  • This link? https://en.wikipedia.org/wiki/Radon%E2%80%93Riesz_property – Functional Analysis Aug 13 '15 at 16:23
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    Does look extremely relevant, thanks! – Functional Analysis Aug 13 '15 at 16:26
  • @Hamza Thanks for this. I knew there was a name for this, but I'd forgotten it! –  Aug 13 '15 at 16:27
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    Is there a simple counterexample of a reflexive Banach space that does not have the Radon-Riesz property? – Matt Rosenzweig Aug 13 '15 at 16:33
  • such example can be found in SMITH and TURETT paper : "ROTUNDITY IN LEBESGUE-BOCHNER FUNCTION SPACES" page 113. http://www.ams.org/journals/tran/1980-257-01/S0002-9947-1980-0549157-4/S0002-9947-1980-0549157-4.pdf – Hamza Aug 13 '15 at 18:11
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    Thank you for this nice paper. Just for reference: The counterexample is the following: Take $X = (\mathbb{R^2}, |\cdot|_{\ell^1})$. Then, $L^2(0,1;X)$ is a counterexample. – gerw Aug 13 '15 at 18:17
  • I was not aware of Radon-Riesz spaces. Thanks for the link! Do you think/know if it is possible to say that every reflexive space is a Radon-Riesz space (I'm thinking exploiting some corrollary of the Hanh-Banach theorem)? – bartgol Aug 13 '15 at 19:13
  • @bartgol, the above counter-example is reflexive. However, it seems that uniformly convex spaces should fall into this class. – Tomasz Kania Aug 13 '15 at 22:51
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Ok, assuming you are in a Hilbert space (otherwise I think we must use Hanh-Banach), you can simply consider the functional

$$ F(x_n) = \langle x_n,x \rangle $$ and see what happens when $n\to\infty$.

Then what happens to $\|x-x_n\|^2=\langle x-x_n,x-x_n\rangle$?

Edit: yes, in any Hilbert space, by definition, $\|x\|^2 = \langle x,x \rangle$.

Ben Grossmann
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bartgol
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