# The Unapologetic Mathematician

## Clairaut’s Theorem

Now for the most common sufficient condition ensuring that mixed partial derivatives commute. If $f$ is a function of $n\geq2$ variables, we can for the moment hold the values of all but two of them constant. We’ll only consider two variables at a time, which will simplify our notation. For the moment, then, we write $f(x,y)$. We will also assume that $f$ is real-valued, and deal with vector values one component at a time.

I assert that if the partial derivatives $D_xf$ and $D_yf$ are continuous in a neighborhood of the point $(a,b)$, and if the mixed second partial derivative $D_{y,x}f$ exists and is continuous there, then the other mixed partial derivative $D_{x,y}f$ exists at $(a,b)$, and we have the equality

$\displaystyle\left[D_{x,y}f\right](a,b)=\left[D_{y,x}f\right](a,b)$

By definition, within the neighborhood in the statement of the theorem the partial derivative $\frac{\partial f}{\partial y}$ is given by the limit

$\displaystyle\left[D_yf\right](x,y)=\lim\limits_{k\to0}\frac{f(x,y+k)-f(x,y)}{k}$

So the numerator of the difference quotient defining the desired mixed partial derivative is

\displaystyle\begin{aligned}\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)=&\lim\limits_{k\to0}\frac{f(a+h,b+k)-f(a+h,b)}{k}\\-&\lim\limits_{k\to0}\frac{f(a,b+k)-f(a,b)}{k}\end{aligned}

For a fixed $k$, we define the function

$\displaystyle g_k(t)=f(a+t,b+k)-f(a+t,b)$

We compute the derivative of $g_k$ as

$\displaystyle g_k'(t)=\left[D_xf\right](a+t,b+k)-\left[D_xf\right](a+t,b)$

so we can apply the mean value theorem to write

$\displaystyle\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)=\lim\limits_{k\to0}\frac{g_k(h)-g_k(0)}{k}=\lim\limits_{k\to0}\frac{hg_k'(\bar{h})}{k}$

for some $\bar{h}$ between ${0}$ and $h$. We use the above expression for $g_k'$ to write the difference quotient

$\displaystyle\frac{\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)}{h}=\lim\limits_{k\to0}\frac{\left[D_xf\right](a+\bar{h},b+k)-\left[D_xf\right](a+\bar{h},b)}{k}$

In a similar trick to the one above, we can see that $\left[D_xf\right](a+\bar{h},b+s)$ is differentiable as a function of $s$ with derivative $\left[D_{y,x}f\right](a+\bar{h},b+s)$. And so the mean value theorem tells us that we can write our difference quotient as

$\displaystyle\frac{\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)}{h}=\lim\limits_{k\to0}\left[D_{y,x}f\right](a+\bar{h},\bar{y})$

for some $\bar{y}$ between $b$ and $b+k$.

And so we come to try taking the limit

$\displaystyle\lim\limits_{h\to0}\lim\limits_{k\to0}\left[D_{y,x}f\right](a+\bar{h},\bar{y})=\left[D_{y,x}f\right](a,b)$

If $\bar{h}$ didn’t depend in its definition on $k$, this would be easy. First we could let $k$ go to zero, which would make $\bar{y}$ go to $b$, and then letting $h$ go to zero would make $\bar{h}$ go to zero as well. But it’s not going to be quite so easy, and limits in two variables like this usually call for some delicacy.

Given an $\epsilon>0$, there (by the assumption of continuity) is some $\delta>0$ so that

$\displaystyle\lvert\left[D_{y,x}f\right](x,y)-\left[D_{y,x}f\right](a,b)\rvert<\frac{\epsilon}{2}$

for $(x,y)$ within a radius $\delta$ of $(a,b)$. As long as we keep $\lvert h\rvert$ and $\lvert k\rvert$ below $\frac{\delta}{2}$, the point $(a+\bar{h},\bar{y})$ will be within this radius. So we can keep $h$ fixed at some small enough value, and find that $\lvert k\rvert<\frac{\delta}{2}$ implies the inequality

$\displaystyle\lvert\left[D_{y,x}f\right](a+\bar{h},\bar{y})-\left[D_{y,x}f\right](a,b)\rvert<\frac{\epsilon}{2}$

Now we can take the limit as $k$ goes to zero. As we do so, the inequality here may become an equality, but since we kept it below $\frac{\epsilon}{2}$, we still have some wiggle room. So, if $\lvert h\rvert<\frac{\delta}{2}$, we have the inequality

$\displaystyle\left\lvert\lim\limits_{k\to0}\left[D_{y,x}f\right](a+\bar{h},\bar{y})-\left[D_{y,x}f\right](a,b)\right\rvert\leq\frac{\epsilon}{2}<\epsilon$

which gives us the limit we need.

Of course we could instead assume that the second mixed partial derivative exists and is continuous near $(a,b)$, and conclude that the first one exists and is equal to the second.

October 15, 2009 Posted by | Analysis, Calculus | 14 Comments