# Convergence of the inner product in Hilbert Space

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.

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

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 $$\langle a,b\rangle \leq \|a\|\cdot\|b\|$$

• Oh now I see it! Thanks, as soon as the site allow me I'll accept your answer. – RScrlli Nov 6 '18 at 14:33

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 $$\lim_{n \to \infty} \phi(x_n) = \phi(x)$$.

• 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 – RScrlli Nov 6 '18 at 16:16