# Why every x in an Hilbert space with a complete orthonormal sequence can be written as this Fourier series

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

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 $z = 0$.]

• 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 – Holog Mar 27 '17 at 23:00
• 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 $x = \sum_i \langle x, e_j \rangle e_j$". – Kenny Wong Mar 27 '17 at 23:02
• Oh you mean that typo! Yes, it should be $x = \sum_i \langle x, e_j \rangle e_j$. – Kenny Wong Mar 27 '17 at 23:03
• Yes the typo, I waste so much time for a mistake. Anyway, now I'm not going to forget this theorem :) – Holog Mar 27 '17 at 23:08
• And I will never forget that nice fact about complete orthonormal bases! So some good has come of this for both of us. :) – Kenny Wong Mar 27 '17 at 23:09