Proving that cross partials in R^2 are equal if

[GridironCPJ]

There is a theorem stating the following:

Let f be defined in a neighborhood of (x0, y0) in R^2. Suppose f has partial derivatives f_1, f_2, f_12, and f_21 in this neighborhood and that the cross partials f_12 and f_21 are continuous at (x0, y0). Then the cross partials f_12 and f_21 are equal at (x0, y0).

This is a theorem proven in TBB's Elementary Real Analysis. It is stated that there are other conditions that could be met instead. I will reqrite the theorem with this other condition:

Let f be defined in a neighborhood of (x0, y0) in R^2. Suppose f has partial derivatives f_1, f_2, f_12, and f_21 in this neighborhood and that the partials f_1 and f_2 are differentiable at (x0, y0). Then the cross partials f_12 and f_21 are equal at (x0, y0).

How would you prove this or start a proof of this? No, this is not a homework problem, this is a theorem I'm curious about myself for my own satisfaction. Also, if my notation is not clear, f_1 is a partial derivative with respect to x, f_2 is the partial derivative with respect to y.

 

[lugita15]

Well, the result you want to prove follows trivially from the theorem in your book, along with the theorem that differentiability implies continuity. So I assume the latter is what you want to prove, right? See here.

 

[GridironCPJ]

Well, the result you want to prove follows trivially from the theorem in your book, along with the theorem that differentiability implies continuity. So I assume the latter is what you want to prove, right? See here.

Yes, differentiability implies continuity, but we are given f_1 and f_2 as being differentiable at (x0, y0), so this tells us that f_1 and f_2 are continuous there, but this does not say anything about the continuity of f_12 and f_21, which is the criteria in the first theorem I gave (the one proven in TBB).

 

[gopher_p]

You might try using the fact that derivatives have the intermediate value property.

Or just look at the proof of the original statement and see where and how the continuity of $f_{12}$ and $f_{21}$ is used. You might see how that condition is a bit more than necessary.

 

[blue_raver22]

The theorem you have stated is known as Clairaut's theorem, and it is a fundamental result in multivariable calculus. To prove this theorem, we can use the definition of partial derivatives and the mean value theorem.

First, let's define the notation used in the theorem. The partial derivatives f_1 and f_2 represent the rates of change of the function f with respect to the variables x and y, respectively. The cross partials f_12 and f_21 represent the rates of change of the partial derivatives f_1 and f_2 with respect to the other variable. In other words, f_12 is the rate of change of f_1 with respect to y, and f_21 is the rate of change of f_2 with respect to x.

Now, to prove the theorem, we will use the mean value theorem for functions of one variable. This theorem states that if a function is differentiable on an interval, then there exists a point within that interval where the function's derivative is equal to the slope of the secant line connecting the endpoints of the interval.

In our case, we will apply the mean value theorem to the partial derivatives f_1 and f_2. Since we know that f_1 and f_2 are differentiable at (x0, y0), we can apply the mean value theorem to each of them independently. This gives us two points, (x0, y1) and (x1, y0), where the partial derivatives are equal to the slopes of the secant lines connecting (x0, y0) to (x0, y1) and (x0, y0) to (x1, y0), respectively.

Now, let's consider the difference between these two points in terms of the cross partials f_12 and f_21. We have:

f_12(x0, y1) - f_12(x0, y0) = f_1(x0, y1) - f_1(x0, y0)
f_21(x1, y0) - f_21(x0, y0) = f_2(x1, y0) - f_2(x0, y0)

Since the partial derivatives f_1 and f_2 are differentiable at (x0, y0), we can apply the mean value theorem again to each of them to obtain two points, (x0