Under what minimal conditions are the following interchange of operations valid (including a question of existence, if not given explicitly)? \begin{align*} \lim \int f_n&=\int \lim f_n \\ \lim_{x\to a} \lim_{y \to b} f(x,y)&=\lim_{y\to b}\lim_{x \to a} f(x,y) \\ \frac{d}{dx}\lim f_n&=\lim \frac{d}{dx}f_n \\ \lim_{x \to c} \int f(x,y)dy&=\int \lim_{x \to c} f(x,y)dy \\ \frac{d}{dx} \int f(x,t) dt &= \int \frac{d}{dx} f(x,t)dt \end{align*}

This is probably my weakest area in analysis. The only loose idea I have in my head for a general program surrounds uniform convergence, but I wouldn't even know how to apply that notion to, e.g. the fourth question.

I have books, like Royden, Rudin, etc., that provide a laundry list of conditions, but I generally need visualization and intuitive rationale for theorems - and I just don't see it here. I once had the advice to always think of these in basic terms, like sequences and series, but even then, it's hard to picture what's going on geometrically.

I have similar issues with $l_p$ vs $L_p$ spaces - norms for $l_p$ are easy to visualize, but norms for $L_p$ have no geometric meaning for me.

Thanks for helping me get over this hump - I know this is an important area for understanding analysis deeply.

I cannot remember the tag for non-specific questions and would appreciate anyone adding that tag if she or he knows.

  • $\begingroup$ I'm not in a position to answer your question, but it might help if you also list what you do know and what you have seen, so that other people can better address your concerns. For instance, it seems you are familiar with the various convergence theorems (monotone, bounded, dominated)? $\endgroup$
    – angryavian
    Commented Jun 22, 2014 at 2:37
  • $\begingroup$ Angryavian, yes, I am familiar with them - but I don't have them conceptually "memorized," as I do, say, with the definition of countability or with Riemann integration. Simply put, my trouble is that I could not describe the various "interchange theorems" from a lay point of view, and so I must look them up every time I use them. This leads to conceptual difficulties e.g. with Monotone and Dominated Convergence, identifying advantages to complex functions that are uniformly convergent on compact sets, etc. $\endgroup$
    – Darrin
    Commented Jun 22, 2014 at 3:05
  • $\begingroup$ I believe it's hard to identify what minimal means. One could also think of sequences converging weakly in a certain sense. If the domain of integration is bounded one obtains the first identity for such converging sequences. (But (at least for continuous functions) weak convergence in $C([a,b])$ is equivalent to the convergence of integrals) $\endgroup$ Commented Jun 24, 2014 at 6:21
  • 2
    $\begingroup$ The third condition doesn't need to hold even if $f_n$ converges uniformly to $f$ and $f_n'$ and $f'$ exists. $\endgroup$ Commented Jun 24, 2014 at 17:02
  • $\begingroup$ The reason I've deleted my answer is that, in retrospect, it didn't even conform to my own standards. And not much would be gained by another try to improve it. Apologies. $\endgroup$ Commented Jul 1, 2014 at 17:38

1 Answer 1


A lot of insight comes from having counterexamples in your head. For example, take the 1st, which is essentially the same as the 4th: I can never remember which way Fatou's lemma goes until I do some simple examples like:

(Counter)example: Let $f_n(x) = 1/n$ for all $x$. Clearly $\int_{-\infty}^{\infty} f_n(x)dx = \infty$ for all $n$, so its limit is also $\infty$. But $f_n(x)$ converges point wise to $0$, so the integral of the limit is $0$. So we now remember that Fatou's lemma for non-negative functions $f_n(x)$ is: \begin{eqnarray*} \liminf_{n\rightarrow\infty} \int f_n \geq \int \liminf_{n\rightarrow\infty} f_n \end{eqnarray*}

The Lebesgue dominated convergence theorem is the usual thing to apply if you want to get equality results. That is actually proven directly from Fatou: Suppose $f_n(x)$ are non-negative, converge pointwise to $f(x)$, and there is an L1-integrable $y(x)$ such that $f_n(x) \leq y(x)$ for all $x$ and $n$. Then $y(x)-f_n(x)$ is non-negative, so by Fatou: \begin{eqnarray*} \liminf_{n\rightarrow\infty} \int (y-f_n) \geq \int (y-f) \end{eqnarray*} Since $\int y$ is finite we can subtract it from both sides to get: \begin{eqnarray*} \liminf_{n\rightarrow\infty} \int -f_n \geq \int -f \end{eqnarray*} Multiplying the above by $-1$ gives: \begin{eqnarray*} \limsup_{n\rightarrow\infty} \int f_n \leq \int f \end{eqnarray*} On the other hand, by regular Fatou we know for the non-negative function $f_n(x)$: \begin{eqnarray*} \liminf_{n\rightarrow\infty} \int f_n \geq \int f \end{eqnarray*} The previous two inequalities together imply that the $\limsup$ and $\liminf$ are the same, so: \begin{eqnarray*} \lim_{n\rightarrow\infty} \int f_n = \int f \end{eqnarray*}

Then the general Lebesgue result for functions $f_n(x)$ that are possibly negative uses a bounding function $y(x)$ that satisfies $|f_n(x)|\leq y(x)$ for all $x$, and is proven by breaking $f_n(x)$ into positive and negative parts.

The last one is also directly related to this issue of pushing limits thru integrals. Assume $f(x,t)$ is L1-integrable (over $t$) for all $x$. So we can talk about $\int f(x+h,t)dt - \int f(x,t)dt$ without worrying about pesky cases of $\infty - \infty$. Then, from the definition of derivative we wish to look at: \begin{eqnarray*} \lim_{h\rightarrow 0}\int \frac{f(x+h,t)-f(x,t)}{h}dt \end{eqnarray*} So we can define $g_h(t) = \frac{f(x+h,t)-f(x,t)}{h}$. If we assume $\lim_{h\rightarrow0}g_h(t) = \frac{\partial}{\partial x}f(x,t)$, then we can use the Lebesgue Dominated Convergence theorem to show $\lim_{h\rightarrow 0} \int g_h(t)dt = \int \frac{\partial}{\partial x}f(x,t)dt$. This is justified if we can find an L1-integrable bounding function $y(t)$ that satisfies $|g_h(t)| \leq y(t)$ for all $t$ and $h$. Such bounding functions can often be found by exploiting certain properties of $f(x,t)$, such as Lipschitz properties and/or finite support over $t$.

The second property is interesting and had me fooled for a bit! I found a link to a good counterexample here: Under what conditions can I interchange the order of limits for a function of two variable?

The simplest (not the strongest) sufficient condition is that $f(x,y)$ be continuous at all points in an open neighborhood of $(a,b)$ (including $(a,b)$ itself). You can prove that and it leads to good "geometric" intuition.

A more general sufficient condition for $\lim_{x\rightarrow a} [\lim_{y\rightarrow b} f(x,y)]$ and $\lim_{y\rightarrow b} [\lim_{x\rightarrow a} f(x,y)]$ to exist and be equal is if all of the following three conditions hold:
(i) $f(x,y)$ is continuous at $(a,b)$.
(ii) There is a $c>0$ such that $\lim_{x\rightarrow a} f(x,y)$ exists as a real number whenever $0<|y-b|<c$.
(iii) There is a $d>0$ such that $\lim_{y\rightarrow b} f(x,y)$ exists as a real number whenever $0<|x-a|<d$.

Peter Franek's comment perhaps leads to the most "geometrical" intuition (for the 3rd problem). You can consider functions that have their actual value uniformly pushed down to 0, but they still oscillate wildly.

  • $\begingroup$ I added a more general sufficient condition for interchanging limits. To clarify the first sufficient condition, by "neighborhood of $(a,b)$," I meant an open ball containing the point $(a,b)$, which of course includes $(a,b)$ itself. $\endgroup$
    – Michael
    Commented Jul 1, 2014 at 16:04

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