# 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 ${\displaystyle A}$ be an open subset of ${\displaystyle \mathbb {R} ^{2}}$, and let ${\displaystyle f:A\rightarrow \mathbb {R} }$ be of class ${\displaystyle C^{2}}$. Then, for all ${\displaystyle c\in A}$, we have that ${\displaystyle D_{1}D_{2}f(c)=D_{2}D_{1}f(c)}$.

### Proof

#### Step 1

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

${\displaystyle \lambda (h,k)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)}$.

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

{\displaystyle {\begin{aligned}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{aligned}}}

And similarly for ${\displaystyle D_{2}D_{1}f(a,b)}$, 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 ${\displaystyle Q=[a,a+h]\times [b,b+k]}$, where ${\displaystyle h}$ and ${\displaystyle k}$ are so small that ${\displaystyle Q\subseteq A}$. We will show that there exist points ${\displaystyle p,q\in Q}$ such that:

{\displaystyle {\begin{aligned}&\lambda (h,k)=D_{2}D_{1}f(p)\cdot hk\\&\lambda (h,k)=D_{1}D_{2}f(q)\cdot hk\end{aligned}}}

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 ${\displaystyle \phi }$, defined such that ${\displaystyle \phi (s)=f(s,b+k)-f(s,b)}$. Then ${\displaystyle \phi (a+h)-\phi (a)=\lambda (h,k)}$. By hypothesis, ${\displaystyle D_{1}f}$ exists at all points of ${\displaystyle A}$, so we can differentiate ${\displaystyle \phi }$ with respect to the first variable of ${\displaystyle f}$ on the interval ${\displaystyle [a,a+h]}$. By the mean value theorem, this means that there exists a point ${\displaystyle s_{0}\in [a,a+h]}$ such that ${\displaystyle \lambda (h,k)=\phi (a+h)-\phi (a)=\phi ^{\prime }(s_{0})\cdot h=[D_{1}f(s_{0},b+k)-D_{1}f(s_{0},b)]\cdot h}$.

Now we want to apply the mean value theorem one more time. Consider the function ${\displaystyle \gamma }$, defined such that ${\displaystyle \gamma (t)=D_{1}f(s_{0},t)}$. By hypothesis, ${\displaystyle D_{2}D_{1}f}$ exists at all points of ${\displaystyle A}$, so we can differentiate ${\displaystyle \gamma }$ with respect to the second variable of ${\displaystyle f}$ on the interval ${\displaystyle [b,b+k]}$. So, by the mean value theorem, there exists a point ${\displaystyle t_{0}\in [b,b+k]}$ such that ${\displaystyle \gamma (b+k)-\gamma (b)=\gamma ^{\prime }(t_{0})\cdot k}$. Thus:

{\displaystyle {\begin{aligned}\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{aligned}}}

#### Step 3

Now we can prove the theorem. Let ${\displaystyle c=(a,b)\in A}$, and let ${\displaystyle t>0}$ be so small that ${\displaystyle Q_{t}=[a,a+t]\times [b,b+t]\subseteq A}$. By what we have just shown, ${\displaystyle \lambda (t,t)=D_{2}D_{1}f(p_{t})\cdot t^{2}}$, for some ${\displaystyle p_{t}\in Q_{t}}$.

${\displaystyle t}$ is the length of the sides of the rectangle ${\displaystyle Q_{t}}$, so as ${\displaystyle t\rightarrow 0}$, ${\displaystyle p_{t}\rightarrow c}$. By hypothesis, ${\displaystyle D_{2}D_{1}f}$ is continuous, and so as ${\displaystyle p_{t}\rightarrow c}$, ${\displaystyle D_{2}D_{1}f(p_{t})\rightarrow D_{2}D_{1}f(c)}$. Thus:

${\displaystyle \lim _{t\to 0}{\frac {\lambda (t,t)}{t^{2}}}=D_{2}D_{1}f(c)}$.

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

${\displaystyle \lim _{t\to 0}{\frac {\lambda (t,t)}{t^{2}}}=D_{1}D_{2}f(c)}$.

Therefore, by the uniqueness of limits, the two quantities must be equal. ${\displaystyle \square }$