# Contexts For Bessel's Inequality?

Bessel's inequality appears to be about orthonormal sequences.

But (in the context of inner product spaces), I've thought of this inequality as being a demonstration that the hypotenuse of triangles is always at least as long as its cosine projection (and only equal when the angle is zero and the cosine is 1).

Are the two uses related? Are they different versions of the same thing or two totally different things? And is there a "third" context that I've missed?

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I'll be going in a circle while trying to answer the question:

First of all, given an orthonormal basis $\{e_n\}_{n\in \mathbb{N}}$ of a (separable infinite-dimensional) Hilbert space, Parseval's identity—the equality case of Bessel's inequality—implies that $x = \sum_{n \in \mathbb{N}} \langle x, e_n\rangle e_n$, where the sum converges in norm.

The equality $$\sum_{n\in\mathbb{N}} |\langle x, e_n\rangle|^{2} = \|x\|^2$$ can be seen as an infinite-dimensional Pythagorean theorem.

Now if $\{f_n\}_{n \in \mathbb{N}}$ is an orthonormal system, the vector $Px = \sum_{n\in\mathbb{N}} \langle x,f_n\rangle f_n$ lies in the closed linear span of $\{f_n\}_{n}$ and its squared length is $\|Px\|^2 = \sum_{n\in\mathbb{N}} |\langle x,f_n\rangle|^2$ by Parseval. Now Bessel's inequality tells us that $\|Px\| \leq \|x\|$ and thus the map $x \mapsto Px$ is continuous of norm $1$ since $P$ is obviously linear. It satisfies $P^2 = P$ and it is easy to check that $P^{\ast} = P$, so we see that $P$ is an orthogonal projection and by definition its range is the closed span of $\{f_n\}$. Obviously, $x = Px + (1-P)x$ and $P^{\ast} = P$ gives us that $$\langle Px, (1-P)x \rangle = \langle x, (P-P^2)x \rangle = 0,$$ so $Px \perp (1-P)x$. Now in the (at most) two-dimensional subspace spanned by $Px$ and $(1-P)x$ we have a right-angled triangle with corners $0,Px,x$. By expanding the scalar product, we see

$$\|x\|^2 = \langle Px + (1-P)x, Px + (1-P)x \rangle = \|Px\|^2 + \|(1-P)x\|^2$$ which is of course nothing but Pythagoras.

Now this equality together with $\|Px\|^2 = \sum_{n\in\mathbb{N}} |\langle x,f_n\rangle|^2$ gives us back Bessel's inequality

$$\sum_{n\in\mathbb{N}} |\langle x,f_n\rangle|^2 \leq \|x\|^2$$

and we see that the defect is measured by $(1-P)x$, the part of $x$ orthogonal to the span of the $\{f_n\}$.

Finally, the angle $\phi$ between $x$ and $Px$ is defined by $$\cos{\phi} = \frac{\langle x, Px \rangle}{\|x\|\,\|Px\|},$$ where the fraction is in $[-1,1]$ by Cauchy-Schwarz, and that gives us the interpretation of $Px$ as the "cosine projection" of the hypotenuse $x$ of our triangle with corners $0,Px,x$.

I'm not aware of a third thing that you missed (except if you're willing to count Parseval). I hope this answers your question, I for one found these considerations illuminating when I made them for the first time.

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"I for one found these considerations illuminating when I made them for the first time." I've a similar experience trying to explain simpler topics to people who are less advanced than I am, which is why I don't mind asking people to do these things. (And I do count Pareval as a "third" context.) –  Tom Au Jun 28 '11 at 19:47
@Tom: Ok, very good, so we've got three things. Do you think this answer addressed your question or do you want me to expand on something? –  t.b. Jun 28 '11 at 19:53
It addressed my questions, and I have accepted your answer. –  Tom Au Jun 28 '11 at 20:24

Bessel's inequality governs inner product spaces, by showing that the hypotenuse of triangles is always at least as long as its cosine projection (and only equal when the angle is zero and the cosine is 1) in such spaces.

Many inner product spaces are defined over orthnormal sequences. So the two uses are connected.

Parseval's Identity is a "special" (equality) case of Bessel's inequality that holds when the orthonormal sequence in question has a basis.

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