1617-257/Classnotes for Friday October 14

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Dror's notes above / Students' notes below

Symmetry of Second Partial Derivatives

(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)

Theorem

Let be an open subset of , and let be of class . Then, for all , we have that .

Proof

Step 1

We begin by defining a function that will help us in our proof. Let be an arbitrary point. We then define the function as follows:

.

Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of :

And similarly for , but with the two limits taken in the opposite order.

Step 2

Now, let's make use of this function. Let us consider the square , where and are so small that . We will show that there exist points such that:

The proofs of the two equalities are symmetric, and thus we only explicitly prove the first one. We will prove this through a double application of the mean value theorem.

Consider the function , defined such that . Then . By hypothesis, exists at all points of , so we can differentiate with respect to the first variable of on the interval . By the mean value theorem, this means that there exists a point such that .

Now we want to apply the mean value theorem one more time. Consider the function , defined such that . By hypothesis, exists at all points of , so we can differentiate with respect to the second variable of on the interval . So, by the mean value theorem, there exists a point such that . Thus:

Step 3

Now we can prove the theorem. Let , and let be so small that . By what we have just shown, , for some .

is the length of the sides of the rectangle , so as , . By hypothesis, is continuous, and so as , . Thus:

.

We can use the same argument, and the second equality from step 2, to show that:

.

Therefore, by the uniqueness of limits, the two quantities must be equal.


Handwritten Lecture Notes in PDF

MAT257 - Lecture14 (Oct 14)