1617-257/Classnotes for Friday October 14

From Drorbn
Jump to: navigation, search
Dror's notes above / Students' notes below

Contents

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 A be an open subset of \mathbb{R}^2, and let f: A \rightarrow \mathbb{R} be of class C^2. Then, for all c \in A, we have that D_1 D_2 f(c) = D_2 D_1 f(c).

Proof

Step 1

We begin by defining a function that will help us in our proof. Let (a, b) \in A \subseteq \mathbb{R}^2 be an arbitrary point. We then define the function \lambda as follows:

\lambda(h, k) = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b).

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


\begin{align}
D_1 D_2 f(a, b) & = \frac{\partial}{\partial x} \frac{\partial f}{\partial y} (a, b) \\
& = \lim_{h \to 0} (\frac{1}{h}) [\frac{\partial f}{\partial y} (a + h, b)] - [\frac{\partial f}{\partial y} (a, b)] \\
& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b)] - [\lim_{k \to 0} (\frac{1}{k}) f(a, b + k) - f(a, b)] \\
& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) [f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) \lambda(h, k)
\end{align}

And similarly for D_2 D_1 f(a, b), 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 Q = [a, a + h] \times [b, b + k], where h and k are so small that Q \subseteq A. We will show that there exist points p, q \in Q such that:


\begin{align}
& \lambda(h, k) = D_2 D_1 f(p) \cdot hk \\
& \lambda(h, k) = D_1 D_2 f(q) \cdot hk
\end{align}

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 \phi, defined such that \phi(s) = f(s, b + k) - f(s, b). Then \phi(a + h) - \phi(a) = \lambda(h, k). By hypothesis, D_1 f exists at all points of A, so we can differentiate \phi with respect to the first variable of f on the interval [a, a + h]. By the mean value theorem, this means that there exists a point s_0 \in [a, a + h] such that \lambda(h, k) = \phi(a + h) - \phi(a) = \phi^\prime(s_0) \cdot h = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h.

Now we want to apply the mean value theorem one more time. Consider the function \gamma, defined such that \gamma(t) = D_1 f(s_0, t). By hypothesis, D_2 D_1 f exists at all points of A, so we can differentiate \gamma with respect to the second variable of f on the interval [b, b + k]. So, by the mean value theorem, there exists a point t_0 \in [b, b + k] such that \gamma(b + k) - \gamma(b) = \gamma^\prime(t_0) \cdot k. Thus:


\begin{align}
\lambda(h, k) & = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h \\
& = [\gamma(b + k) - \gamma(b)] \cdot h \\
& = \gamma^\prime(t_0) \cdot hk \\
& = D_2 D_1 f(s_0, t_0) \cdot hk
\end{align}

Step 3

Now we can prove the theorem. Let c = (a, b) \in A, and let t > 0 be so small that Q_t = [a, a + t] \times [b, b + t] \subseteq A. By what we have just shown, \lambda(t, t) = D_2 D_1 f(p_t) \cdot t^2, for some p_t \in Q_t.

t is the length of the sides of the rectangle Q_t, so as t \rightarrow 0, p_t \rightarrow c. By hypothesis, D_2 D_1 f is continuous, and so as p_t \rightarrow c, D_2 D_1 f(p_t) \rightarrow D_2 D_1 f(c). Thus:

\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_2 D_1 f(c).

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

\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_1 D_2 f(c).

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


Handwritten Lecture Notes in PDF

MAT257 - Lecture14 (Oct 14)