The Unapologetic Mathematician

Mathematics for the interested outsider

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)-g(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 | 11 Comments

   

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