1617-257/Classnotes for Friday October 14
| 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 [math]\displaystyle{ A }[/math] be an open subset of [math]\displaystyle{ \mathbb{R}^2 }[/math], and let [math]\displaystyle{ f: A \rightarrow \mathbb{R} }[/math] be of class [math]\displaystyle{ C^2 }[/math]. Then, for all [math]\displaystyle{ c \in A }[/math], we have that [math]\displaystyle{ D_1 D_2 f(c) = D_2 D_1 f(c) }[/math].
Proof
Step 1
We begin by defining a function that will help us in our proof. Let [math]\displaystyle{ (a, b) \in A \subseteq \mathbb{R}^2 }[/math] be an arbitrary point. We then define the function [math]\displaystyle{ \lambda }[/math] as follows:
[math]\displaystyle{ \lambda(h, k) = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b) }[/math].
Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of [math]\displaystyle{ \lambda }[/math]:
[math]\displaystyle{ \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} }[/math]
And similarly for [math]\displaystyle{ D_2 D_1 f(a, b) }[/math], 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 [math]\displaystyle{ Q = [a, a + h] \times [b, b + k] }[/math], where [math]\displaystyle{ h }[/math] and [math]\displaystyle{ k }[/math] are so small that [math]\displaystyle{ Q \subseteq A }[/math]. We will show that there exist points [math]\displaystyle{ p, q \in Q }[/math] such that:
[math]\displaystyle{ \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} }[/math]
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 [math]\displaystyle{ \phi }[/math], defined such that [math]\displaystyle{ \phi(s) = f(s, b + k) - f(s, b) }[/math]. Then [math]\displaystyle{ \phi(a + h) - \phi(a) = \lambda(h, k) }[/math]. By hypothesis, [math]\displaystyle{ D_1 f }[/math] exists at all points of [math]\displaystyle{ A }[/math], so we can differentiate [math]\displaystyle{ \phi }[/math] with respect to the first variable of [math]\displaystyle{ f }[/math] on the interval [math]\displaystyle{ [a, a + h] }[/math]. By the mean value theorem, this means that there exists a point [math]\displaystyle{ s_0 \in [a, a + h] }[/math] such that [math]\displaystyle{ \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 }[/math].
Now we want to apply the mean value theorem one more time. Consider the function [math]\displaystyle{ \gamma }[/math], defined such that [math]\displaystyle{ \gamma(t) = D_1 f(s_0, t) }[/math]. By hypothesis, [math]\displaystyle{ D_2 D_1 f }[/math] exists at all points of [math]\displaystyle{ A }[/math], so we can differentiate [math]\displaystyle{ \gamma }[/math] with respect to the second variable of [math]\displaystyle{ f }[/math] on the interval [math]\displaystyle{ [b, b + k] }[/math]. So, by the mean value theorem, there exists a point [math]\displaystyle{ t_0 \in [b, b + k] }[/math] such that [math]\displaystyle{ \gamma(b + k) - \gamma(b) = \gamma^\prime(t_0) \cdot k }[/math]. Thus:
[math]\displaystyle{ \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} }[/math]
Step 3
Now we can prove the theorem. Let [math]\displaystyle{ c = (a, b) \in A }[/math], and let [math]\displaystyle{ t \gt 0 }[/math] be so small that [math]\displaystyle{ Q_t = [a, a + t] \times [b, b + t] \subseteq A }[/math]. By what we have just shown, [math]\displaystyle{ \lambda(t, t) = D_2 D_1 f(p_t) \cdot t^2 }[/math], for some [math]\displaystyle{ p_t \in Q_t }[/math].
[math]\displaystyle{ t }[/math] is the length of the sides of the rectangle [math]\displaystyle{ Q_t }[/math], so as [math]\displaystyle{ t \rightarrow 0 }[/math], [math]\displaystyle{ p_t \rightarrow c }[/math]. By hypothesis, [math]\displaystyle{ D_2 D_1 f }[/math] is continuous, and so as [math]\displaystyle{ p_t \rightarrow c }[/math], [math]\displaystyle{ D_2 D_1 f(p_t) \rightarrow D_2 D_1 f(c) }[/math]. Thus:
[math]\displaystyle{ \lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_2 D_1 f(c) }[/math].
We can use the same argument, and the second equality from step 2, to show that:
[math]\displaystyle{ \lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_1 D_2 f(c) }[/math].
Therefore, by the uniqueness of limits, the two quantities must be equal. [math]\displaystyle{ \square }[/math]
