## Clairaut’s Theorem

Now for the most common sufficient condition ensuring that mixed partial derivatives commute. If is a function of 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 . We will also assume that is real-valued, and deal with vector values one component at a time.

I assert that if the partial derivatives and are continuous in a neighborhood of the point , and if the mixed second partial derivative exists and is continuous there, then the other mixed partial derivative exists at , and we have the equality

By definition, within the neighborhood in the statement of the theorem the partial derivative is given by the limit

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

For a fixed , we define the function

We compute the derivative of as

so we can apply the mean value theorem to write

for some between and . We use the above expression for to write the difference quotient

In a similar trick to the one above, we can see that is differentiable as a function of with derivative . And so the mean value theorem tells us that we can write our difference quotient as

for some between and .

And so we come to try taking the limit

If didn’t depend in its definition on , this would be easy. First we could let go to zero, which would make go to , and then letting go to zero would make 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 , there (by the assumption of continuity) is some so that

for within a radius of . As long as we keep and below , the point will be within this radius. So we can keep fixed at some small enough value, and find that implies the inequality

Now we can take the limit as goes to zero. As we do so, the inequality here may become an equality, but since we kept it below , we still have some wiggle room. So, if , we have the inequality

which gives us the limit we need.

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