# Interchanging limits and differentiation/integration

1. Nov 4, 2008

### magic_castle32

I am wondering as to what are the (regularity) conditions which would permit the interchanging of limits and differentiation (for a sequence of functions)? What about the interchanging of differentiation and integration?

2. Nov 4, 2008

### jostpuur

Suppose $$\Omega\subset\mathbb{R}^n$$ is an open set, $$Y\subset\mathbb{R}^m$$ is a measurable set, and $$f:\Omega\times Y\to \mathbb{R}$$ some function so that $$y\mapsto f(x,y)$$ is integrable with all $$x\in\Omega$$, and so that $$x\mapsto f(x,y)$$ has the partial derivative $$\partial_i f(x,y)$$ with all $$y\in Y$$. Now it makes sense to ask if the equation

$$\partial_i \int\limits_Y dy\; f(x,y) = \int\limits_{Y} dy\; \partial_i f(x,y)$$

is correct. If we write down the definition of the derivative, we see that the equation is equivalent with

$$\lim_{\epsilon\to 0} \int\limits_Y dy\; \frac{f(x+\epsilon e_i, y) - f(x,y)}{\epsilon} = \int\limits_Y dy\; \lim_{\epsilon\to 0} \frac{f(x+\epsilon e_i, y) - f(x,y)}{\epsilon}.$$

So actually the question is that can you change the order of the limit and the integration.

This is a basic problem in the integration theory: Given a sequence of integrable functions $$f_n:X\to\mathbb{R}$$, that converge pointwisely to some function $$f:X\to\mathbb{R}$$, on some measure space, is the equation

$$\lim_{n\to\infty} \int\limits_X dx\; f_n(x) = \int\limits_X dx\; f(x)$$

true? The answer is that it is not always true, but the Lebesgue's dominated convergence theorem, http://en.wikipedia.org/wiki/Dominated_convergence_theorem, describes a sufficient and useful condition that guarantees that the limit and integration can be commuted. The condition is that a dominating function $$g:X\to\mathbb{R}$$ must exist so that

$$|f_n(x)| \leq g(x),\quad\quad \forall x\in X,\quad \forall n\in\mathbb{N}$$

and

$$\int\limits_X dx\; g(x) < \infty.$$

In similar spirit, a following condition gives an answer to the original problem: If there exists a function $$h:Y\to\mathbb{R}$$ so that $$|\partial_i f(x,y)| \leq h(y)$$ for all $$x\in\Omega$$ and $$y\in Y$$, and so that

$$\int\limits_Y dy\; h(y) < \infty,$$

then the differentiation and integration can be commuted.

The proof uses the dominated convergence, and the mean value theorem which guarantees the existence of $$\xi_{x,y,\epsilon}\in\Omega$$ such that

$$\frac{f(x + \epsilon e_i, y) - f(x,y)}{\epsilon} = \partial_i f(\xi_{x,y,\epsilon}, y).$$

So if you want to commute differentiation and integration rigorously, the ready formula for it is that you must prove the existence of this dominating function.

Last edited: Nov 4, 2008
3. Nov 4, 2008

### jostpuur

btw. I didn't answer your first question, because I don't know answer to it. It looks easier at least... perhaps somebody else says something to it?

4. Nov 4, 2008

### jostpuur

Considering the fact that the Riemann integral of a (Riemann integrable) function is always the same as the Lebesgue integral, the comment in Wikipedia:

Is not very correct IMO. Besides, according to my experience (which is not very much of experience yet, but anyway...), the dominated convergence is useful especially when you want to actually prove some concrete formulas.

Also, part of the power of the theorem lies in the fact that it holds for all measures, not only for the Lebesgue measure. So it can be used to prove, for example, formulas like this:

$$\lim_{k\to \infty} \sum_{n=1}^{\infty} a_{n,k} = \sum_{n=1}^{\infty} \big(\lim_{k\to\infty} a_{n,k}\big)$$

Since the infinite summation is an integration of certain kind.