An alternative proof for $D_2 D_1 f(x) = D_1 D_2 f(x)$ when $f \in C^2$

Question

In the book of Analysis On Manifolds by Munkres, at page 103, question 4-b, it is asked to show that

$$D_2 D_1 f(x) = D_1 D_2 f(x)$$ for all $x \in A$, where $A \subseteq \mathbb{R}^2 $ is open and $f\in C^2(A)$.

Proof:

Let $Q$ be a rectangle in $A$, so we do know that $$\int_Q D_2 D_1 f(x) = \int_Q D_1 D_2 f(x),$$ hence consider $$\int_Q |D_2 D_1 f(x) - D_1 D_2 f(x)| = $$ Since the integral above exists, and the integrand is non-negative and the integral is zero, by a theorem, the integrand $|D_2 D_1 f(x) - D_1 D_2 f(x)|$ vanishes except on a a set of measure zero.

Now we claim that, the integrand $|D_2 D_1 f(x) - D_1 D_2 f(x)|$ vanishes everywhere on $Q$.

Assume that is it not the case; then $\exists a \in Q$ s.t $|D_2 D_1 f(a) - D_1 D_2 f(a)| \not = 0$, so by a lemma, there exists a neighbourhood $U$ of $a$ s.t that the integrand is still non-zero, but that contradicts with the fact that the integrand vanishes on $Q$ except on a set of measure zero, since a open ball does not have a measure zero.

So, $|D_2 D_1 f(x) - D_1 D_2 f(x)| = 0$ $\forall x \in Q$ implies $$D_2 D_1 f(x) = D_1 D_2 f(x) \quad \forall x \in Q.$$

Question:

Is there any flaw in my proof ? or is there anything that is not clear ? or do you have any suggestion ?

Answer 1

Since $f \in C^2(A)$, the function $g =D_2D_1f - D_1D_2f$ is continuous.

Choose any point $a \in A$. If $g(a) > 0$, then there exists a rectangle $Q$ such that $g(x) > 0 $ for all $x \in Q$ and $\int_Qg > 0$, contradicting

$$\int_Q D_2D_1f = \int_QD_1D_2 f.$$

Make a similar argument if $g(a) < 0$, proving that $g(a) = 0$ for all $a \in A$.

Answer 2

There is indeed a big flaw in your proof:

$$\int f =0$$ doesn’t imply that $$\int \vert f \vert=0$$ which is basically what you’re using.