Bessels inequality proof

Question

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 $h \in H$

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

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

Then

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

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 \langle h,g\rangle + ||g||^{2}\\ &= ||h||^{2} - 2\langle h,\sum^{\infty}_{k = 1} \langle e_{k}, h\rangle e_{k}\rangle + ||g||^{2}\\ &= ||h||^{2} -2 \sum^{\infty}_{k = 1} \langle e_{k}, h\rangle^{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} \langle e_{k}, h\rangle^{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

Answer 1

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.

Answer 2

The post is old, but I will say something because I have the same doubts 4 years latter.

The answer to the question 'Why define g like that?' comes from the Gram-Schmidt process.

Suppose we are in inner product space and we have a set of $\{x_1,...,x_n\}$ vectors, which is not orthonormal, and we want a new set, of orthonormal vectors.

Of course we can, arbitrarily define the first orthonormal vector as $e_1=\frac{x_k}{||x_k||}$, I will choose

$$e_1=\frac{x_1}{||x_1||}$$ The next step is define the second vector.

The plan is as follows: we will project $x_2$ onto $x_1$ and define $z_1$ as $z_1 = x_2 - proj_{x_2}$. Can you see that $z_1$ is orthogonal to $x_1$?

$$e_2:=\frac{z_1}{||z_1||}=$$

But what exactly is $z_1$?

Note that, (see image) the condition is $\langle x_2-\alpha x_1,e_1 \rangle=0$, with $\alpha e_1 = proj_{x_2}$, which gives us $\alpha = \langle x_2,e_1 \rangle$. Therefore

$$e_1=x_2-\langle x_2,e_1 \rangle e_1$$

I will omit the rest of the calculations and directly tell you that, $$z_n=x_n-\sum_{k=1}^{n-1}\langle x_n,e_k \rangle e_k$$ and $$e_n=\frac{v_n}{||v_n||}$$

Note que $z_n \perp e_k$.

More generally, given $x \in X$ where $X$ is inner product space, and $Y_n=\{e_1,...,e_n\}$, with $x$ not necessarily in $Y_n$. We can define $y$ in $Y_n$ as $$y=\sum_{k=1}^{n}\langle x,e_k\rangle e_k$$ with $n$ fixed, e define $z=x-y$, in that case, $z \perp y$

This is where the definition of $g$ arises.