# Bessel's inequality proof

• Jun 2nd 2013, 11:36 AM
Stormey
Bessel's inequality proof
Hi guys.
I need to prove Bessel's inequality for a finite set of orthonormal vectors $\left \{ e_1, e_2,...e_n \right \}$
I looked around the internet and found this on Wikipedia:

"Bessel's inequality follows from the identity:
$0\leq \left \| x-\sum_{i=1}^{n}e_i \right \|^2=\left \| x \right \|^2-2\sum_{i=1}^{n}||^2+\sum_{i=1}^{n}|| ^2=\left \| x \right \|^2-\sum_{i=1}^{n}||^2$
"

but something just does not make sense to me.
how did they make the first transition from $\left \| x-\sum_{i=1}^{n}e_i \right \|^2$ to $\left \| x \right \|^2-2\sum_{i=1}^{n}||^2+\sum_{i=1}^{n}|| ^2$?
you can't just use the binomial expanding formula here, can you?

Is there another, more simple proof for this inequality?

• Jun 2nd 2013, 01:55 PM
Plato
Re: Bessel's inequality proof
Quote:

Originally Posted by Stormey
Hi guys.
I need to prove Bessel's inequality for a finite set of orthonormal vectors $\left \{ e_1, e_2,...e_n \right \}$

This is one I have used.

Let $\alpha_j=.$

\begin{align*} 0 &\le\left\langle {x - \sum\limits_{j = 1}^n {{\alpha _j}{e_j}} ,x - \sum\limits_{i = 1}^n {{\alpha _i}{e_i}} } \right\rangle \\ &\le{\left\| x \right\|^2} + \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {{\alpha _j}\overline {{\alpha _i}} \left\langle {{e_j},{e_i}} \right\rangle } } - \sum\limits_{i = 1}^n {\overline {{\alpha _i}} \left\langle {x,{e_i}} \right\rangle } - \sum\limits_{j = 1}^n {{\alpha _j}\left\langle {{e_j},x} \right\rangle } \\&\le {\left\| x \right\|^2} + \sum\limits_{j = 1}^n {{{\left| {{\alpha _j}} \right|}^2}} - \sum\limits_{i = 1}^n {{{\left| {{\alpha _i}} \right|}^2}} - \sum\limits_{j = 1}^n {{{\left| {{\alpha _j}} \right|}^2}}\\&\le {\left\| x \right\|^2} - \sum\limits_{i = 1}^n {{{\left| {{\alpha _i}} \right|}^2}} \end{align*}
• Jun 2nd 2013, 02:29 PM
Stormey
Re: Bessel's inequality proof
Great. thank you.
that's more clear.
• Jun 2nd 2013, 07:21 PM
johng
Re: Bessel's inequality proof
Hi,
The proof that you indicate is not hard to understand. It mainly uses distribution of inner product (dot product) over addition.

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