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In this theorem, available here on page 52, we prove that differentiation does not discriminate between the order of partial differentiation. In his proof, he proves a 'second-order' mean-value theorem for $f$. In particular, let $Q = [a, a+h] \times [b,b+k]$ be a rectangle contained in an open set $A \subseteq \mathbb{R}^m$. Define $\lambda(h,k) = f(a,b)- f(a+h, b) - f(a, b+k) + f(a+h, b+k)$. He shows that $\lambda(h,k) = D_2 D_1 f(q) \cdot hk$ and $\lambda(h,k) = D_1 D_2 f(p)\cdot hk$ for some points $p, q$ of $Q$. However, this geometric argument is confusing me slightly.

The question I'm having is where did $\lambda(h,k)$ come from? As far as I can tell he assigns positive and negative values by the corners of the rectangle Q and says that $\lambda$ is the sum of the values of $f$ at the four vertices of $Q$. But, why do we need this rectangle $Q$ in the proof?

Perhaps, I'm asking someone to explain to me the steps in this proof a little clearer than Munkres writes.

dansidorkin
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  • Intuitively, you can think of "inverting" the process of differentiating by seeing how $f$ changes when you first change $x$ and then change $y$ versus the other way — first change $y$ and then change $x$. These correspond, respectively, to doing $\big(f(x+h,y)-f(x,y)\big) + \big(f(x+y,y+k)-f(x+h,y)\big)$ versus $\big(f(x,y+k)-f(x,y)\big) + \big(f(x+h,y+k)-f(x,y+k)\big)$. The lower-right path from the bottom left to the upper right versus the upper-left path. – Ted Shifrin Oct 18 '22 at 20:38
  • See here for a intuitive phrasing of this proof: https://math.stackexchange.com/questions/4318702/intuition-behind-schwarz-theorem-partial-derivatives/4318716#4318716 Essentially, we use commutativity of discrete difference quotients, and then take the limit as the step size $h \to 0$. – Mason Oct 19 '22 at 06:54

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