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I'm starting to study Hilbert Spaces for the very first time in my life and I had difficulty to understand one very simple proof:

Let $\{x_n:n=1,2,...\}$ be a sequence of vectors in the space; the sequence is said to converge to an element $x$ (of the space) $x_n\rightarrow x$ if $\lim_{n\rightarrow\infty}||x_n-x||=0$.

And then it's also stated that if $x_n\rightarrow x$ then $\langle x_n,z\rangle\rightarrow \langle x,z\rangle$.

I'm a little confused, first the definition of convergence implies the use of the norm, so I guess that I should take absolute values instead for the inner product.

The book mentions the Schwarz Inequality as the tool needed to prove the statement, but honestly I don't see how could this be used in this context. Any hint will be greatly appreciated.

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    $\begingroup$ I corrected the names of Hilbert and of Schwarz. $\endgroup$ – Giuseppe Negro Nov 6 '18 at 14:27
  • $\begingroup$ What does the Schwarz inequality say? $\endgroup$ – saulspatz Nov 6 '18 at 14:28
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Hint:

$$\langle x_n, z\rangle = \langle x_n - x + x, z\rangle = \langle x_n-x, z\rangle + \langle x, z\rangle$$

Now, use the Cauchy-Scwhartz inequality on $$\langle x_n-x, z\rangle.$$


Note To refresh your memory, the inequality is $$\langle a,b\rangle \leq \|a\|\cdot\|b\|$$

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  • $\begingroup$ Oh now I see it! Thanks, as soon as the site allow me I'll accept your answer. $\endgroup$ – RScrlli Nov 6 '18 at 14:33
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In case you didn't notice I would remind you, result is direct consequence of the fact that linear functional $\phi(x) = \langle x, z \rangle$ defined for $x \in H$ is continuous. This of course follows from continuity of inner product which is also proved by help of Cauchy-Schwartz inequality. So if $\lim_{n \to \infty}x_n = x$ and since $\phi$ is continuous, then $\lim_{n \to \infty} \phi(x_n) = \phi(x)$.

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  • $\begingroup$ Indeed, but the fact is that the author showed the continuity of the norm operator later in the Chapter, now that I've read this it makes completely sense. Thank you $\endgroup$ – RScrlli Nov 6 '18 at 16:16

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