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Let $A$ be open in $\mathbb{R}^2$; let $f:A\rightarrow\mathbb{R}$ be of class $C^2$. Let $Q$ be a rectangle contained in $A$.

(a) Use Fubini's theorem and the fundamental theorem of calculus to show that $$\int_QD_2D_1f=\int_QD_1D_2f.$$

(b) Give a proof that $D_2D_1f(\textbf{x})=D_1D_2f(\textbf{x})$ for each $\textbf{x}\in A$.

I'm quite confused about this exercise. Shouldn't (b) come first before (a)? If we know $D_2D_1f(\textbf{x})=D_1D_2f(\textbf{x})$ for each $\textbf{x}\in A$, then certainly it also holds for each $\textbf{x}\in Q$, and the two integrals in part (a) must be equal.

Also, in part (a), I don't see how to apply the two theorems here. The fundamental theorem of calculus is for one variable. Fubini needs the assumption that the integral over $Q$ exists in the first place, and then states that you can integrate over direction $y$ (or find the lower/upper integrals) and then integrate over direction $x$.

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Mika H.
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  • $(b)$ is sufficient for $(a)$ but not necessary. you can do them in the order provided. – obataku Aug 30 '13 at 21:22
  • If you prove $(b)$ first, $(a)$ is an easy consequence of that. But $(a)$ is easier to prove than $(b)$ (and can help proving $(b)$). – Daniel Fischer Aug 30 '13 at 21:23

1 Answers1

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For part (a), write out the rectangle explicitly as say $[a,b]\times[c,d]$ then use Fubini's theorem to calculate the integrals explicitly, choosing to do the $x_1$ integration first for one integral and $x_2$ first for another. You should get an answer in terms of f evaluated at the corners of the rectangle. Then notice the answers are the same.

James S. Cook
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Matt Rigby
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  • Thanks, I get the idea. However, to use Fubini's theorem, I think we need to show first that $D_2D_1f$ is integrable over $Q$. How can we show that? – Mika H. Aug 30 '13 at 23:24
  • Continuity! There you go. :) – Ted Shifrin Aug 30 '13 at 23:39
  • @TedShifrin I realize the Fundamental Theorem of Calculus guarantees the existence of an antiderivative for a continuous function in one dimensional. However, here the integral is over $Q$, which is two dimensional. How does continuity guarantee integrability over $Q$ in this case? – Mika H. Aug 30 '13 at 23:42
  • There is a theorem that says functions are integrable if and only if its set of points where it is discontinuous has measure $0$. Continuous functions are discontinuous nowhere so it is integrable. – Pratyush Sarkar Aug 30 '13 at 23:44
  • Continuity on Q of the second derivatives implies that $D_2D_1f$ is bounded on Q by some constant C, so by the positive-valued function version of Fubini, the integral of |D2D1f| is less than $C * Area (Q)$. So $D_2D_1 f$ is integrable, and the integrable function-version of Fubini works. – Matt Rigby Aug 30 '13 at 23:46
  • Thanks, PratyushSarkar and MattRigby. Now onto part (b)... – Mika H. Aug 31 '13 at 00:38
  • Without quoting any hard theorems, you should be able to prove a continuous function is integrable on a rectangle by knowing that a continuous function on a compact set is uniformly continuous. – Ted Shifrin Aug 31 '13 at 00:43
  • @TedShifrin Right, for any $\epsilon$ there exists $\delta$ such that $|x-y|<\delta\rightarrow |f(x)-f(y)|<\epsilon$. Choose a partition so that the diameter of each rectangle is $<\delta$. The the difference between the lower and upper integrals is at most $\epsilon\cdot(\text{Volume of }Q)$ – Mika H. Aug 31 '13 at 00:49
  • For part b - it can be shown that if $g$ is a continuous function on $A$, $x=(x_1,x_2)$ is in $A$ and $Q_\epsilon$ is the rectangle $[x_1 - \epsilon, x_1+ \epsilon] \times [x_2 - \epsilon, x_2 + \epsilon] $, then the limit as $\epsilon$ tends to $0$ of the integral of $g$ over $Q_\epsilon$ is $g(x)$. Then apply this with $g = D_1D_2 f - D_2D_1 f$. – Matt Rigby Aug 31 '13 at 00:50
  • @MattRigby You mean the limit of the integral over $Q_\epsilon$, divided by the volume of $Q_\epsilon$? – Mika H. Aug 31 '13 at 00:53
  • Yes - was just about to edit that but it won't let me – Matt Rigby Aug 31 '13 at 01:00