# Can dominated convergence justify commuting two limits?

I am evaluating an expression of the form:

$$\lim_{a\to 0^+}\,\,\lim_{b\to 0^+}\,\,\sum_{n=1}^\infty\int_{-\infty}^{+\infty}f_n(x;a,b)\,dx.$$

Suppose I can find dominating functions $$F_n(x)$$ such that

$$\forall n\in\mathbb{N},x\in\mathbb{R},a,b\in[0,1]:|F_n(x)|\geqslant|f_n(x)|$$

and $$\sum_n\int |F_n(x)|dx<\infty$$.

I believe (correct me if I’m wrong) that the dominated converge theorem justifies writing

$$\lim_{a\to 0^+}\,\,\lim_{b\to 0^+}\,\,\sum_{n=1}^\infty\int_{-\infty}^{+\infty}f_n(x;a,b)\,dx = \sum_{n=1}^\infty\int_{-\infty}^{+\infty}\,\,\lim_{a\to 0^+}\,\,\lim_{b\to 0^+}\,\,f_n(x;a,b)\,dx,$$

and that Fubini’s theorem would also justify reversing the order of the sum and integral (again, please correct if wrong).

What I’m curious about (and would greatly simplify my calculation) is, are we justified in reversing the order of the two limits as well? I.e., can we write

$$\lim_{a\to 0^+}\,\,\lim_{b\to 0^+}\,\,\sum_{n=1}^\infty\int_{-\infty}^{+\infty}f_n(x;a,b)\,dx = \sum_{n=1}^\infty\int_{-\infty}^{+\infty}\,\,\lim_{b\to 0^+}\,\,\lim_{a\to 0^+}\,\,f_n(x;a,b)\,dx \,\,?$$

I have seen uniform convergence used as a criterion for reversing the order of limits, and I have also seen cases where dominated convergence appears as a “stronger” condition than uniform convergence, so it seems plausible that my above statement would be valid, but I’m not positive.

I presume you mean $$|F_n(x)| \ge f_n(x;a,b)$$ for all $$a$$ and $$b$$. Then if you know that $$\lim_{a \to 0+} \lim_{b \to 0+} f_n(x;a,b)$$ exists for all $$n$$ and almost every $$x$$, you can write $$\lim_{a \to 0+} \lim_{b \to 0+} \sum \int f_n(x;a,b) \; dx = \sum \int \lim_{a \to 0+} \lim_{b \to 0+} f_n(x;a,b)\; dx$$ and similarly if $$\lim_{b \to 0+} \lim_{a \to 0+} f_n(x;a,b)$$ exists for all $$n$$ and almost every $$x$$, $$\lim_{b \to 0+} \lim_{a \to 0+} \sum \int f_n(x;a,b) \; dx = \sum \int \lim_{b \to 0+} \lim_{a \to 0+} f_n(x;a,b)\; dx$$ But you certainly can't conclude $$\lim_{b \to 0+} \lim_{a \to 0+}$$ and $$\lim_{a \to 0+} \lim_{b \to 0+}$$ for the sum of integrals (or integral of sums) are equal unless you already have $$\lim_{a \to 0+} \lim_{b \to 0+} f_n(x;a,b) =\lim_{b \to 0+} \lim_{a \to 0+} f_n(x;a,b)$$ for all $$n$$ and almost every $$x$$.
• Try $f_n(x;a,b) = e^{-n-x^2} g(a,b)$ where $\lim_{a \to 0+} \lim_{b \to 0+} g(a,b) \ne \lim_{b \to 0+} \lim_{a \to 0+} g(a,b)$, or doesn't exist at all. – Robert Israel Apr 22 at 19:10
Here is an example: let $$f_n(x;a,b) = \frac{a^2}{a^2 + b^2} \cdot \frac{1}{n^2} \cdot \frac 1{1+x^2}.$$ Clearly $$|f_n(x;a,b)| \le \frac{1}{n^2} \cdot \frac 1{1+x^2}$$ and $$\sum_{n=1}^\infty \int_{-\infty}^\infty \frac{1}{n^2} \cdot \frac 1{1+x^2} \, dx = \frac{\pi^3}{6}.$$ However $$\sum_{n=1}^\infty f_n(x;a,b) \, dx = \frac{a^2}{a^2 + b^2} \cdot \frac{\pi^3}{6}.$$ This has limit $$0$$ if you let $$a \to 0^+$$ first; and limit $$\frac{\pi^3}{6}$$ if you let $$b \to 0^+$$ first.