Bessels inequality proof

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Studying a proof of Bessels' inequality. something confused me here is the proof:

Lemma 1: Let $H$ be an inner product space if $\{ e_{1}, e_{2} ... , e_{n} \}$ is an orthonormal set then for all $h \in H$

$\sum^{n}_{k = 1} <e_{k},h>^2 \leq ||h||^{2}$

Proof let $g = \sum^{\infty}_{k = 1} <e_{k}, h>e_{k}$ (first question, why do we let g equal this? is it just '' a good idea''?

Then

$||g||^{2} = <g,g>\\ =<\sum^{\infty}_{k = 1} <e_{k}, h>e_{k}, \sum^{\infty}_{k = 1} <e_{k}, h>e_{k}>\\ = \sum^{n}_{k = 1} <e_{k}, h>||e_{k}||^{2}\\ = \sum^{n}_{k = 1} <e_{k}, h>$

Now the above calculation, why do we suddenly get that we sum over n instead of $\infty$

$\forall h \in H$\begin{aligned} 0 &\leq||h - g||^{2} = ||h||^{2} - 2 <h,g> + ||g||^{2}\\ &= ||h||^{2} - 2<h,\sum^{\infty}_{k = 1} <e_{k}, h>e_{k}> + ||g||^{2}\\ &= ||h||^{2} -2 \sum^{\infty}_{k = 1} <e_{k}, h>^{2} + ||g||^{2}\\ &= ||h||^{2} - 2||g||^{2} + ||g||^{2}\\ &= ||h||^{2} - ||g||^{2}\\ &\implies ||g||^{2} \leq ||h||^{2}\\ &\implies \sum^{n}_{k=1} <e_{k}, h>^{2} \leq ||h||^{2} \end{aligned}

Again why do we get sum from $\infty$ suddenly becomes $n$

and how can this proof the inequality I mean there's no absolute value around LHS

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1 Answer

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You should have defined $g$ in the first place as a finite sum:$$g = \sum_{k=1}^n \langle h,e_k \rangle e_k.$$Note the order in the inner product. Next your calculation of $\|g\|^2$ is incorrect. It should be$$\|g\|^2 = \sum_{k=1}^n |\langle h,e_k \rangle|^2.$$This follows easily from the properties of the inner product and the fact the $e_k$ are orthonormal.

You can evaluate the inner product of $h$ and $g$ by$$\langle h,g \rangle = \langle h,\sum_{k=1}^n \langle h,e_k \rangle e_k \rangle = \sum_{k=1}^n \overline{\langle h,e_k\rangle} \langle h,e_k \rangle = \sum_{k=1}^n |\langle h,e_k \rangle|^2 = \|g\|^2.$$

Instead of considering $\|h-g\|$ just apply Cauchy-Schwarz to find$$\|g\|^2 = \langle h,g \rangle \le \|h\| \|g\|$$ which yields the inequality you want.

p.s. use $\langle$ and $\rangle$ instead of $<$ and $>$ for the delimiters on an inner product.

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