2
$\begingroup$

I'm currently learning my lecture and try to have a very deep understanding of definitions, theorems and proves.

Hello everyone,

The theorem which I'm stuck on the proof is:

Let $H$ be an Hilbert space and $\{e_n\}_{n\in \mathbb{N}}$ be a complete orthonormal sequence. We have that for all $x\in H$, $x=\sum_{n=1}^{\infty}\langle x,e_n\rangle x$

And I looked at the proof, I understand clearly (I think) until the end:

We have $\sum_{n=1}^{\infty}\langle x,e_n\rangle e_n$ is convergent (proved previously). So we can let $y=x-\sum_{n=1}^{\infty}\langle x,e_n\rangle e_n$ and note that again using the continuity of the inner product we have that for all $j\in\mathbb{N}$:

$\langle y,e_j\rangle=\langle x,e_j\rangle-\langle \sum_{n=1}^{\infty}\langle x,e_n\rangle e_n,e_j\rangle=\langle x,e_j\rangle-\langle x,e_j\rangle=0$

So if we have that if the only $z\in H$ for which $\langle z,e_j\rangle =0$ for all $j\in \mathbb{N}$ then we have $x=\sum_{n=1}^{\infty}\langle x,e_n\rangle x$

I really don't understand the last sentence. I don't know how the $x$ came to replace $e_n$ in the Fourier series.

Can you help me to "visualize" or well understand this please ?

Thank you

$\endgroup$
1
$\begingroup$

The final sentence should say:

"Since we have shown that $\langle y , e_j \rangle = 0$ for all $j$, it follows that $y = 0$. Hence $x = \sum_j \langle x, e_j \rangle e_j$."

Indeed, one could define a complete orthonormal basis to be an orthonormal set $\{ e_j \}$ such that any $z \in H$ such that $\langle z , e_j \rangle = 0$ for all $j \in \mathbb N$ is the zero element. Since you have just shown that your $y$ obeys this condition, you can deduce that $y = 0$.

[Or perhaps you define a complete orthonormal basis to be an orthonormal set $\{ e_j \}$ such that the closure of the set of finite linear combinations of the $e_j$'s is the entire Hilbert space. But then, the orthogonal complement of this closure would be the zero space, and if $\langle z , e_j \rangle = 0$ for all $j \in \mathbb N$, then $z$ clearly lies in this orthogonal complement (by continuity of the inner product) and hence $z = 0$.]

$\endgroup$
  • $\begingroup$ Thank you a lot! So that was just a mistake in the lectures ? Because it made me doubt since we have a $x$ instead of $e_n$ twice: in the proof and in the theorem $\endgroup$ – Holog Mar 27 '17 at 23:00
  • $\begingroup$ I think your lecturer was trying to say, "Since we proved in the previous lecture that any $z \in H$ such that $\langle z , e_j \rangle = 0$ for all $j$ must be the zero element, and since we have just checked that $y$ obeys this property, we can conclude that $y = 0$, hence $x = \sum_i \langle x, e_j \rangle e_j$". $\endgroup$ – Kenny Wong Mar 27 '17 at 23:02
  • $\begingroup$ Oh you mean that typo! Yes, it should be $x = \sum_i \langle x, e_j \rangle e_j$. $\endgroup$ – Kenny Wong Mar 27 '17 at 23:03
  • $\begingroup$ Yes the typo, I waste so much time for a mistake. Anyway, now I'm not going to forget this theorem :) $\endgroup$ – Holog Mar 27 '17 at 23:08
  • $\begingroup$ And I will never forget that nice fact about complete orthonormal bases! So some good has come of this for both of us. :) $\endgroup$ – Kenny Wong Mar 27 '17 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.