Bessel's inequality proof

Hi guys.

I need to prove Bessel's inequality for a finite set of orthonormal vectors $\displaystyle \left \{ e_1, e_2,...e_n \right \}$

I looked around the internet and found this on Wikipedia:

"Bessel's inequality follows from the identity:

$\displaystyle 0\leq \left \| x-\sum_{i=1}^{n}<x,e_i>e_i \right \|^2=\left \| x \right \|^2-2\sum_{i=1}^{n}|<x,e_i>|^2+\sum_{i=1}^{n}|<x,e_i>| ^2=\left \| x \right \|^2-\sum_{i=1}^{n}|<x,e_i>|^2$

"

but something just does not make sense to me.

how did they make the first transition from $\displaystyle \left \| x-\sum_{i=1}^{n}<x,e_i>e_i \right \|^2$ to $\displaystyle \left \| x \right \|^2-2\sum_{i=1}^{n}|<x,e_i>|^2+\sum_{i=1}^{n}|<x,e_i>| ^2$?

you can't just use the binomial expanding formula here, can you?

Is there another, more simple proof for this inequality?

Thanks in advanced!

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 $\displaystyle \left \{ e_1, e_2,...e_n \right \}$

This is one I have used.

Let $\displaystyle \alpha_j=<x,e_j>.$

$\displaystyle \begin{align*} 0 &\le\left\langle {x - \sum\limits_{j = 1}^n {{\alpha _j}{e_j}} ,x - \sum\limits_{i = 1}^n {{\alpha _i}{e_i}} } \right\rangle \\ &\le{\left\| x \right\|^2} + \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {{\alpha _j}\overline {{\alpha _i}} \left\langle {{e_j},{e_i}} \right\rangle } } - \sum\limits_{i = 1}^n {\overline {{\alpha _i}} \left\langle {x,{e_i}} \right\rangle } - \sum\limits_{j = 1}^n {{\alpha _j}\left\langle {{e_j},x} \right\rangle } \\&\le {\left\| x \right\|^2} + \sum\limits_{j = 1}^n {{{\left| {{\alpha _j}} \right|}^2}} - \sum\limits_{i = 1}^n {{{\left| {{\alpha _i}} \right|}^2}} - \sum\limits_{j = 1}^n {{{\left| {{\alpha _j}} \right|}^2}}\\&\le {\left\| x \right\|^2} - \sum\limits_{i = 1}^n {{{\left| {{\alpha _i}} \right|}^2}} \end{align*}$

Re: Bessel's inequality proof

Great. thank you.

that's more clear.

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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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