# College Math Teaching

## August 18, 2014

### Interchanging infinite sums with integrals

Filed under: Uncategorized — collegemathteaching @ 1:03 pm

This part of this series of lectures got me to thinking about this topic; the relevant part starts at 13 minutes:

(side note: I am enjoying these and hope to finish all 30!)

Of interest here: if $u(x,t)$ describes the heat in a one dimensional circle of metal and the radius is one, one can do the analysis of the heat equation $u_t = a u_{xx}$ with initial condition $u(x,0) = f(x)$ and assuming that $u$ has a Fourier expansion (complex coefficients) one can obtain: $u(x,t) = \Sigma_k \hat{f_k}e^{-4ak \pi^2t}e^{2k \pi i x}$ which can be rewritten as $\Sigma_k (\int^1_0 f(w) e^{-2k \pi i w} dw) e^{-4ak \pi^2t}e^{2k \pi i x} = \Sigma_k (\int^1_0 f(w) e^{-2k \pi i (x-w)} ) e^{-4ak \pi^2t} dw$. Note: $k$ ranges from $-\infty$ to $\infty$ and by $\Sigma^{\infty}_{k=-\infty} c_k$ we mean $lim_{n \rightarrow \infty} \Sigma^{n}_{k=-n}c_k$ and note that $c_k, c_{-k}$ are complex conjugates for all $k$.

Now IF we could interchange the summation sign and the integral sign we’d have: $\int^1_0 \Sigma_k f(w) e^{-2k \pi i (x-w)} e^{-4ak \pi^2t} dw$. Now let $g(x,t) = e^{-2k \pi i (x)} e^{-4ak \pi^2t}$ then we could say that $u(x,t) = \int^1_0 g(x-w,t) f(w) dw$ which is a convolution product; $g(x,t)$ is the heat kernel which is a nice form. But about that interchange: when can we do it?

First note that by $\Sigma_k f_k$ we mean the limit of the sequence of partial sums: $\phi_n = \Sigma_{k = -n}^{n} f_k$ and if $\int^1_0 lim_n \phi_n = lim_n \int^1_0 \phi_n$ then the interchange is valid. NOTE: I am following the custom of not using the “differential” $dx$ and of letting it be understood that $lim_n$ means $lim_{n \rightarrow \infty}$.

The “I don’t want “TL;DR” answer” version
If you are comfortable with Lebesgue integration, then the Dominated Convergence Theorem is the standard: if $\phi_n$ are all measurable functions and $lim_n \phi_n = \phi$ (pointwise) and there exists an integrable function $g$ where $g \ge |\phi_n|$ for all $n$, then $\int^1_0 lim_n \phi_n = lim_n \int^1_0 \phi_n = \int^1_0 \phi$.

Now if the terms “Lebesgue integration” and “measurable” has you scratching your head, you can either learn a bit about it or, if you are in the “tl;dr” mode I’ll make some (hopefully) “practitioner friendly” remarks.

First of all, all Riemann integrable functions are Lebesgue integrable (provided we are NOT talking about improper integrals) and a “measurable function” is one in which the inverse image of a “measurable set” is a “measurable set”. Now “measurable set”: these include sets that single point sets, open intervals, closed intervals, countable intersections of such, countable unions of such and complements of such unions and intersections. Unfortunately there are measurable sets that aren’t formed in this manner, and there are such things as non-measurable sets. See here for the definition of “measurable set”.

Upshot: the sort of functions that appear in Fourier Series are measurable so you probably don’t have to worry. So there is probably no harm in assuming that the $\phi_n$ are Riemann integrable functions.

Pointwise convergence: this means for all $x$ in the domain of interest (here, $x \in [0,1]$), $lim \phi_n(x) = \phi(x)$.

Of course, when we are talking about the Fourier series for a given function $f$, there are conditions that must be met to get that the the series converges to a function that is “almost $f$; the video assumes that $f$ is $L_2$ which means that $\int^1_0 (f)^2$ exists. The mathematics of convergence of a Fourier series is rich; for this note we will assume that the Fourier series in question converges.

Now for the conclusion: assuming that the $\phi_n$ converge pointwise to some $\phi$ then $lim_n \int^1_0 \phi_n = \int^1_0 lim_n \phi_n$ but we need to use the Lebesgue integral to guarantee this equality..in general. This is why:

For example, suppose we enumerate the rational numbers by $q_{1},q_{2},...q_{k}...$ and define $f_{1}(x)=\left\{\begin{array}{c}1,x\neq q_{1} \\ 0,x=q_{1}\end{array}\right.$ and then inductively define $f_{k}(x)=\left\{\begin{array}{c}1,x\notin \{q_{1},q_{2},..q_{k}\} \\ 0,x\in \{q_{1},q_{2},..q_{k}\}\end{array}\right.$. Then $f_{k}\rightarrow f=\left\{\begin{array}{c}1,x\notin \{q_{1},q_{2},..q_{k}....\} \\ 0,x\in \{q_{1},q_{2},..q_{k},...\}\end{array}\right.$ and for each $k$, $\int_{0}^{1}f_{k}(x)dx=1$ but $f$, the limit function, is not Riemann integrable. It is Lebesgue integrable though, and the integral remains 1.

But, given the types of series that the practitioner will be working with (typically: only a finite number of maximums and minimums on a given interval and a finite number of jump discontinuities), one will probably not encounter such pathological behavior with the functions. I give this example to explain why the Dominated Convergence Theorem uses Lebesgue integrals.

Wait a minute you might say, didn’t I read something about “uniform convergence” of functions that lead to the limiting behavior that we need? Well, yes, and I’ll explain that here:

we say that $\phi_n \rightarrow \phi$ uniformly if for any $\epsilon > 0$ there exists $N$ such that for all $n > N$, $|\phi_n(x) - \phi(x)| < 0$ for ALL $x$ in the interval of interest. Then, it a routine exercise in Riemann integration to see that if $\phi_n \rightarrow \phi$ uniformly then $\int^1_0 \phi_n \rightarrow \int^1_0 \phi$. The down side is that we rarely have uniform convergence when we are talking about Fourier series terms. Here is why: it is known that if $\phi_n \rightarrow \phi$ uniformly and that if all of the $\phi_n$ are continuous, then the limit function $\phi$ is continuous as well. However when one obtains the Fourier series for a function with jump discontinuities (say, for a pulse wave) one sees that the terms (and hence the sequence of partial sums) of the Fourier series are continuous but what the Fourier series converges to is not continuous; hence the convergence of the series is NOT uniform.