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Suppose I have a function $x:\mathbb{R}\rightarrow\mathbb{R}$ such that is square-integrable:


Suppose also that $x(t)$ contains no higher frequencies than $B$ Hz. Then I can use the Nyquist theorem to recover $x(t)$ from its samples as long as these samples are spaced $T_s=\frac{1}{2B}$ seconds apart. I am trying to prove that the sequence $\{x(kT_s)\}_{k=-\infty}^\infty$ of samples of $x(t)$ is square-summable. Here is my approach:

$$\begin{eqnarray} \infty&>&\int_{-\infty}^\infty|x(t)|^2dt\\ &=&\int_{-\infty}^\infty\left(\sum_{k=-\infty}^\infty x(kT_s)\operatorname{sinc}(t/T_s-k)\right)\left(\sum_{l=-\infty}^\infty x(lT_s)\operatorname{sinc}(t/T_s-l)\right)dt\\ &=&\sum_{k=-\infty}^\infty \sum_{l=-\infty}^\infty x(kT_s)x(lT_s)\int_{-\infty}^\infty\operatorname{sinc}(t/T_s-k)\operatorname{sinc}(t/T_s-l)dt\\ &=&T_s^2\sum_{k=-\infty}^\infty |x(kT_s)|^2 \end{eqnarray}$$

However, I am unsure about the third step. Can I interchange the sums with integral there? Does Fubini's Theorem apply here?

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up vote 1 down vote accepted

Since the Integral of the sums converges, Then Fubini's theorem guarantees that the sum of the integral will also converge to the same answer.

A similar question was answered here.

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Each of the sums converges to $x(t)$, which is finite (I left that out of the question, but $|x(t)|\leq 1$). Is this what you meant by "each of the sums converging"? Also, what do you mean by "an appropriate measure"? Sorry if this is a dumb question, but I am shaky on the measure theory... – M.B.M. Jan 12 '13 at 17:21
@M.B.M. - for your first question, yes. By "an appropriate measure" I just meant that sums are just integral with a certain measure. Since Fubini's theorem let's you change the order of integration, you can use the measure that turns integrals to sums to get the result that you can change the order of sums and integration as well. – nbubis Jan 12 '13 at 17:32
Ahh, so one uses the measure to essentially make sums "behave like" integrals for the purposes of the Fubini's theorem. Back to my first question though, where does the requirement for the sums to converge come from? I don't see it in the statement of the Fubini's theorem on Wikipedia. Is it somehow implicit? Thank you for your help, and sorry for asking what are probably very basic questions. I just want to figure this out so that I can use in the future... – M.B.M. Jan 12 '13 at 18:27
It does appear on Wikipedia.. But in your case the condition is met so I wouldn't worry. – nbubis Jan 12 '13 at 18:41

Partial answer: define suitable parameters $t$ in a suitable metrics spaces $T$ and $Y$ in theorem.

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : Y \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $Y$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $Y$ over the base $\mathcal{B}_{T}$ to a function $F : Y \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(y)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{Y}}(\lim_{\mathcal{B}_{T}}F_t(y))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{Y}}F_t(y))$ exist and the equality and holds $$ \lim_{\mathcal{B}_{Y}}(\lim_{\mathcal{B}_{T}}F_t(y))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{Y}}F_t(y)). $$

This theorem can be found in books of Zorich (Mathematical Analysis II p. 381). In this case, if the integral $\displaystyle\lim_{t\to\infty}\int_{-t}^t|x(t)|^2dt$ or the sequence $$ \lim_{y\to\infty}\left(\sum_{k=-y}^y x(kT_s)\operatorname{sinc}(t/T_s-k)\right)\left(\sum_{l=-y}^y x(lT_s)\operatorname{sinc}(t/T_s-l)\right) $$ converge uniformly com respect to $y$ or $t$ in this order, then set $$ F_t(y)= \int_{-t}^t\left(\sum_{k=-y}^y x(kT_s)\operatorname{sinc}(t/T_s-k)\right)\left(\sum_{l=-y}^y x(lT_s)\operatorname{sinc}(t/T_s-l)\right)dt $$ and use the Theorem.

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