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I'm trying to understand the proof of Parseval's Theorem I have in my notes, but I'm failing to comprehend and follow it fully. In particular, the conclusion which is drawn at the end.

Parseval's Theorem: Suppose that $\{f_k\}_{k=1}^\infty$ is an orthonormal sequence in an inner product space $V$. Then $\{f_k\}_{k=1}^\infty$ is a complete orthonormal sequence if and only if for every $f\in V$, $\sum_{k=1}^\infty|\langle f,f_k\rangle|^2=\|f\|^2$.

Recall that $\{f_k\}_{k=1}^\infty$ is a complete orthonormal sequence if $\forall f\in V\,\exists\, c_k\in\mathbb C:f=\sum_{k=1}^\infty c_kf_k$ where $c_k=\langle f,f_k\rangle.$ We also recall Bessel's Inequality which tells us that if $\{f_k\}_{k=1}^\infty$ is an orthonormal sequence in an inner product space $V$. Then for every $f\in V$ the series of nonnegative numbers $\sum_{k=1}^\infty|\langle f,f_k\rangle|^2$ converges and $\sum_{k=1}^\infty|\langle f,f_k\rangle|^2\le\|f\|^2$. I think that I need to make use of Bessel's Inequality in some places.

We start by considering:

$$\|f-\sum_{k=1}^n\langle f,f_k\rangle f_k\|^2=\langle f-\sum_{k=1}^n\langle f,f_k\rangle f_k,f-\sum_{k=1}^n\langle f,f_k\rangle f_k\rangle$$

$$=\|f\|^2-\sum_{k=1}^n|\langle f,f_k\rangle|^2$$

Where we use the continuity of the inner product to take out several sums and combine the final sum. Combining the final sum means that it will cancel with one of the factors obtained on expanding the inner product.

Now, $\|f\|^2-\sum_{k=1}^n|\langle f,f_k\rangle|^2\ge0$, by the definition of the norm (or by Bessel's inequality) so that,

$$\iff\sum_{k=1}^n|\langle f,f_k\rangle|^2\le\|f\|^2$$

This is the point from which I don't fully follow the proof I have. My notes go on to say that:

"Hence, $\lim_{n\to\infty}\sum_{k=1}^n|\langle f,f_k\rangle|^2=\|f\|^2\iff\lim_{n\to\infty}\sum_{k=1}^n\langle f,f_k\rangle f_k=f$".

Now, Bessel's inequality tells us that the series of nonnegative numbers $\sum_{k=1}^\infty|\langle f,f_k\rangle|^2$ converges, so that it's limit of partial sums $\sum_{k=1}^n|\langle f,f_k\rangle|^2$ exists. But what, exactly, allows you draw equality here? (Here, being the expression on the left hand side of the equivalence.) Of course it's what you want to follow, indeed it is what you are trying to show, but am I using to say that the limit of partial sums can no longer be less that $\|f\|^2$? What, also, let's you draw equivalence to the statement on the right hand?

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Consider the identity ($ N > M $):

$$ \| \sum_{k=1}^{N} (f_k,f)f_k - \sum_{k=1}^{M} (f_k,f)f_k \|^2 = \sum_{k=M+1}^N |(f_k,f)|^2 \qquad (1) $$ which holds true due to Pythagoras' theorem. By the inequality you have shown and this identity we immediately get that the sequence $$ \left( \sum_{k=1}^N (f_k,f)f_k \right)_{N} $$ is Cauchy and hence convergent, say to some $\overline{f}$. It is then easy to verify that:

$$ (f-\overline{f},f_k)=0 \qquad \forall k \in \mathbb{N} $$ so that $$ f=\overline{f}=\lim_{N} \sum_{k=1}^N (f_k,f)f_k \qquad (2) $$ In the end, consider another identity following from Pythagoras' (and the definition of orthonormal basis), namely for every $N \in \mathbb{N}$:

$$ \| f \|^2 = \| f - \sum_{k=1}^N (f,f_k)f_k \| ^ 2 + \sum_{k=1}^N |(f,f_k) |^2 $$ Keeping in mind $(2)$ and taking the limit gives the desired result.

The equivalence statement you mentioned lies inside the last equality I wrote. Hope this helps!

Added note (detailed passages to get $(1)$): \begin{align*} \| \sum_{k=1}^{N} (f_k,f)f_k - \sum_{k=1}^{M} (f_k,f)f_k \|^2 &= \| \sum_{k=M+1}^N (f_k,f)f_k \|^2 \\ &= \sum_{k=M+1}^N \| (f_k,f)f_k \|^2 \\ &= \sum_{k=M+1}^N | (f_k,f) |^2 \| f_k \|^2 \\ &= \sum_{k=M+1}^N | (f_k,f) |^2 \end{align*} where Pythagoras' theorem (generalized to finite sums) was used to justify the second equality. Also, do remember that $(f,f_k)$ terms are just real numbers.

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  • $\begingroup$ It seems that you should make use of the triangle inequality for sums both at the start (to take the square of the norm inside the sum from $k=M+1$ to $N$) and at the end (adding zero to $\|f\|_2^2$ and then splitting this up). In these cases, how is it you draw equality and not less than or equal to? $\endgroup$ – Jeremy Jeffrey James Apr 29 '18 at 7:29
  • $\begingroup$ Also, do you need to be careful in using things like the triangle inequality for the square of the norm? Why is it not that $| f \|^2 = (\| f - \sum_{k=1}^N (f,f_k)f_k \| + \|\sum_{k=1}^N (f,f_k)f_k \|)^2$? Are you just treating $\|\cdot\|^2$ as a "new" norm in it's own right? $\endgroup$ – Jeremy Jeffrey James Apr 29 '18 at 7:33
  • $\begingroup$ No, I'm not using triangle inequality anywhere. I'm using the fact that, if $(x,y)=0$, then $ \| x+ y \|^2 = \| x \|^2 + \| y \|^2 $ (this can be generalized to finite sums of course), which I call Pythagora's theorem. Checking orthogonality only requires lineariry of scalar product! $\endgroup$ – GaC Apr 29 '18 at 11:55
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    $\begingroup$ Write $ \| f \|^2 = \| f - \sum_{k=1}^N (f,f_k)f_k + \sum_{k=1}^N (f,f_k)f_k \|^2 $, use Pythagoras' theorem with $ x = f - \sum_{k=1}^N (f,f_k)f_k $ and $y = \sum_{k=1}^N (f,f_k)f_k$ (checking orthogonality is an easy calculation). Then use again the theorem on the sum $ \| \sum_{k=1}^N (f,f_k)f_k \|^2 $ along with $\| f_k \|=1 $ and $ \| (f,f_k) \|= |(f,f_k)| $. $\endgroup$ – GaC Apr 29 '18 at 12:16
  • $\begingroup$ I will write it out again with what you have said regarding Pythagoras' theorem, but I think I have it now. Thanks again for clarifying! $\endgroup$ – Jeremy Jeffrey James Apr 29 '18 at 18:28

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