Is this a valid proof of Young's theorem?
Mia Lopez 
I tried to prove Young's theorem (symmetry of mixed partial derivatives) myself, but my proof seems considerably easier than the one I could find in my textbook. Can anybody point out where I went wrong?
Let $f$ be a $C^2$ function, so that the first two derivatives of $f$ are continues and that $f$ is two times differentiable in a point $(a,b)$. We wish to show that $\frac{\partial^2 f}{\partial x \partial y}(a,b)=\frac{\partial^2 f}{\partial y \partial x}(a,b)$.
Define $$\Delta(h,k)=f(a+h,b+k)-f(a,b+k)-f(a+h,b)+f(a,b)$$
Notice that $$\lim_{h\rightarrow 0} \frac{\Delta(h,k)}{h}=\lim_{h \rightarrow 0}(\frac{f(a+h,b+k)-f(a,b+k)}{h}-\frac{f(a+h,b)-f(a,b)}{h})=$$ $$\frac{\partial f(a,b+k)}{\partial x}-\frac{\partial f(a,b)}{\partial x}$$ and $$\lim_{k \rightarrow 0} \frac{1}{k}\cdot \left (\lim_{h\rightarrow 0} \frac{\Delta(h,k)}{h} \right )=\lim_{k \rightarrow 0} \frac{\frac{\partial f(a,b+k)}{\partial x}-\frac{\partial f(a,b)}{\partial x}}{k}=\frac{\partial}{\partial y}(\frac{\partial f(a,b)}{\partial x})=\frac{\partial^2 f(a,b)}{\partial y \partial x}$$
Likewise, taking the limits in opposite order gives
$$\lim_{h \rightarrow 0}(\lim_{k \rightarrow 0}\Delta(h,k))=\frac{\partial^2 f(a,b)}{\partial x \partial y}$$
I then conclude that:
$$\lim_{(h,k) \rightarrow (0,0)} \frac{\Delta(h,k)}{hk}=\frac{\partial^2 f(a,b)}{\partial y \partial x}=\frac{\partial^2 f(a,b)}{\partial x \partial y}$$
as $f$ is twice differentiable at $(a,b)$, so it shouldn't matter how we approach $(a,b)$. Is this argument correct?
$\endgroup$ 21 Answer
$\begingroup$Your approach starts in the right direction. But, the interchange of the order of limits is in general not correct. Instead, we will use the Mean-Value Theorem to facilitate the development.
First, note that as it is defined the limits of $\frac{\Delta (h,k)}{hk}$ are
$$\lim_{h\to 0}\lim_{k\to 0}\frac{\Delta(h,k)}{hk}=\frac{\partial }{\partial x}\frac{\partial f}{\partial y}(a,b)$$
and
$$\lim_{k\to 0}\lim_{h\to 0}\frac{\Delta(h,k)}{hk}=\frac{\partial }{\partial y}\frac{\partial f}{\partial x}(a,b)$$
Next, we exploit the Mean-Value Theorem and write
$$\begin{align} \frac{\Delta(h,k)}{hk}&=\frac{\frac{\partial f}{\partial x}(a+\theta h,b+k)\,h-\frac{\partial f}{\partial x}(a+\theta h,b)\,h}{hk}\\\\ &=\frac1k \left(\frac{\partial f}{\partial x}(a+\theta h,b+k)-\frac{\partial f}{\partial x}(a+\theta h,b)\right)\\\\ &=\frac{\partial }{\partial y}\frac{\partial f}{\partial x}(a+\theta h,b+\eta k) \tag 1 \end{align}$$
where $0<\theta<1$ and $0<\eta<1$.
Analogously, we can exploit the Mean-Value Theorem and write
$$\begin{align} \frac{\Delta(h,k)}{hk}&=\frac{\frac{\partial f}{\partial y}(a+ h,b+\nu k)\,k-\frac{\partial f}{\partial y}(a,b+\nu k)\,k}{hk}\\\\ &=\frac1h \left(\frac{\partial f}{\partial y}(a+ h,b+\nu k)-\frac{\partial f}{\partial y}(a,b+\nu k)\right)\\\\ &=\frac{\partial }{\partial x}\frac{\partial f}{\partial y}(a+\phi h,b+\nu k) \tag 2 \end{align}$$
where $0<\nu<1$ and $0<\phi<1$.
Using $(1)$ and $(2)$ reveals that for each $x$ and $y$, and $h$ and $k$, there exist four numbers $\theta$, $\eta$, $\nu$, and $\phi$ (all between $0$ and $1$) such that
$$\frac{\partial }{\partial y}\frac{\partial f}{\partial x}(a+\theta h,b+\eta k)=\frac{\partial }{\partial x}\frac{\partial f}{\partial y}(a+\phi h,b+\nu k) \tag 3$$
Under the assumption that both mixed partial derivatives of $f$ exist and are continuous at $(a,b)$, then we can take the limit as $(h,k)\to (0,0)$ of both sides of $(3)$ and obtain the coveted equality of mixed partials.
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