# 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 $A$ 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. $\square$