3
$\begingroup$

Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like $$\lim_{\varepsilon\to0^+} \int_{-\infty}^\infty dk f(k,\varepsilon) \overset{?}{=} \int_{-\infty}^\infty dk\lim_{\varepsilon\to0^+} f(k,\varepsilon).$$ Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)

$\endgroup$
5
$\begingroup$

The statement of the dominated convergence theorem (DCT) is as follows:

"Discrete" DCT. Suppose $\{f_n\}_{n=1}^\infty$ is a sequence of (measurable) functions such that $|f_n| \le g$ for some integrable function $g$ and all $n$, and $\lim_{n\to\infty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $\int |f-f_n| \to 0$. In particular, $\lim_{n\to\infty}\int f_n = \int f$ (by the triangle inequality). This can be written as $$ \lim_{n\to\infty}\int f_n = \int \lim_{n\to\infty} f_n.$$

(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)

As you note, the statements of these theorems involve sequences of functions, i.e., a $1$-discrete-parameter family of functions $\{f_n\}_{n=1}^\infty$. To apply these theorems to a $1$-continuous-parameter family of functions, say $\{f_\epsilon\}_{0<\epsilon<\epsilon_0}$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:

Proposition. If $f$ is a function, then $$\lim_{\epsilon\to0^+}f(\epsilon) = L \iff \lim_{n\to\infty}f(a_n) = L\quad \text{for $\mathbf{all}$ sequences $a_n\to 0^+$.}$$

With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):

"Continuous" DCT. Suppose $\{f_\epsilon\}_{0<\epsilon<\epsilon_0}$ is a $1$-continuous-parameter family of (measurable) functions such that $|f_\epsilon| \le g$ for some integrable function $g$ and all $0<\epsilon<\epsilon_0$, and $\lim_{\epsilon\to0^+}f_\epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $\lim_{\epsilon\to 0^+}\int f_\epsilon = \int f$. This can be written as $$ \lim_{\epsilon\to0^+}\int f_\epsilon = \int \lim_{\epsilon\to0^+} f_\epsilon.$$

The way we use the continuous DCT in practice is by picking an arbitrary sequence $\pmb{a_n\to 0^+}$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $a_n$, using only the assumption that $a_n\to 0^+$ and properties of the family $\{f_\epsilon\}$ that are known to us.

$\endgroup$
  • $\begingroup$ Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,\varepsilon)| \leq g(k), \forall k \in\mathbb{R}$ and all $\varepsilon$ between $0$ and some positive $\varepsilon_0$. Then I check if $f(k,\varepsilon) \to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right? $\endgroup$ – PhysSE is Cancer Mar 30 at 23:45
  • $\begingroup$ @IvanV.: Yes, that's correct! $\endgroup$ – Alex Ortiz Mar 31 at 0:23
  • $\begingroup$ Alright, thank you, much appreciated! $\endgroup$ – PhysSE is Cancer Mar 31 at 2:00
4
$\begingroup$

Let's look at it in a sample case. We want to prove by DCT that $$\lim_{\varepsilon\to0^+} \int_0^\infty e^{-y/\varepsilon}\,dy=0$$

This is the case if and only if for all sequences $\varepsilon_n\to 0^+$ it holds $$\lim_{n\to\infty}\int_0^\infty e^{-y/\varepsilon_n}\,dy=0$$

And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $e^{-x}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.