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This is a follow-up question to this question. I read a more elegant looking proof in [1] Proposition 8.44, p208. So it goes like this.

Proposition If $x_n \rightharpoonup x$ then

$$\left\Vert x \right\Vert \leq \liminf_{n \to \infty}\left\Vert x_n \right\Vert $$.

Proof. Using the weak convergence of $(x_n)$ and th Cauchy-Schwarz inequality, we find that $$\left\Vert x \right\Vert^2 = \langle x, x \rangle =\lim_{n \to \infty}\langle x, x_n \rangle \leq \left\Vert x \right\Vert\liminf_{n\to \infty}\left\Vert x_n \right\Vert.$$

What I don't understand is that how after the Cauchy-Schwarz the lim becomes liminf and still the inequality still holds. I would think that replacing the limit with liminf should have a less than or equal. But ofcourse I know the result is correct, infact if we use the weak lower semicontinuity of the norm then we are good. As weak semicontinuity is defined a bit after the proof in the book.

So is there a way to do it without invoking sequential weak lower semicontinuity?

[1] J. K. Hunter and B. Nachtergaele, Applied Analysis. World Scientific Publishing Company, 2001.

mmcrjx
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  • because there is no reason for $\lim_{n\to \infty }|x_n|$ to exist. However, $\liminf_{n\to \infty }|x_n|$ always exist. – Surb May 16 '23 at 08:35
  • But, how is it true that $\lim_{n\to \infty} \left \Vert x_n \right \Vert \leq \liminf_{n \to \infty} \left \Vert x_n \right \Vert$ Shouldn't it be $\geq$? – mmcrjx May 16 '23 at 08:38
  • $\le$ comes from Cauchy Schwarz inequality. – geetha290krm May 16 '23 at 08:41
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    Yes Cauchy-Schwarz gets me here -> $\lim_{n\to \infty} \langle x ,x_n \rangle \leq \left \Vert x \right \Vert \lim_{n \to \infty} \left \Vert x_n \right \Vert$. Then how do I get to the liminf? – mmcrjx May 16 '23 at 08:45
  • In fact, at this stage if I use $\lim_{n \to \infty} \left \Vert x_n \right\Vert \leq \liminf_{n \to \infty} \left \Vert x_n \right\Vert$, then doesn't it become a circular argument? I am using the same thing that I want o prove. – mmcrjx May 16 '23 at 08:53
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    Note that your "more elegant looking proof" is exactly the same one, but in the particular case of Hilbert spaces, where the $\phi$ of this answer is equal to $\langle\frac x{|x|},\cdot\rangle.$ That answer and its comments could have sufficed to enlighten you. – Anne Bauval May 16 '23 at 09:05
  • Yes, you are right, this is a more general property of a Banach space. In fact, the use case I have for this is in a space which is not Hilbert. Thank you! – mmcrjx May 16 '23 at 09:10

1 Answers1

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As mentioned in the comment, since the strong limit of $x_n$ might not exists, one cannot write $\lim_{n\to \infty} \|x_n\|$. Instead, the Cauchy Schwarz gives

$$ \langle x, x_n\rangle \le \|x\| \cdot \|x_n\|,$$ now take liminf on both sides:

$$\liminf_{n\to\infty} \langle x, x_n\rangle \le \|x\| \cdot \liminf_{n\to \infty} \|x_n\|,$$ since $\langle x, x_n\rangle$ does has a limit, so

$$\liminf _{n\to\infty}\langle x, x_n\rangle = \lim_{n\to\infty} \langle x, x_n\rangle,$$ which is what you want.

Arctic Char
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