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 $A$ 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 $\lambda$:

${\begin{aligned}D_{1}D_{2}f(a,b)&={\frac {\partial }{\partial x}}{\frac {\partial f}{\partial y}}(a,b)\\&=\lim _{h\to 0}({\frac {1}{h}})[{\frac {\partial f}{\partial y}}(a+h,b)]-[{\frac {\partial f}{\partial y}}(a,b)]\\&=\lim _{h\to 0}({\frac {1}{h}})[\lim _{k\to 0}({\frac {1}{k}})f(a+h,b+k)-f(a+h,b)]-[\lim _{k\to 0}({\frac {1}{k}})f(a,b+k)-f(a,b)]\\&=\lim _{h\to 0}({\frac {1}{h}})[\lim _{k\to 0}({\frac {1}{k}})f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)]\\&=\lim _{h\to 0}\lim _{k\to 0}({\frac {1}{hk}})[f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)]\\&=\lim _{h\to 0}\lim _{k\to 0}({\frac {1}{hk}})\lambda (h,k)\end{aligned}}$

And similarly for $D_{2}D_{1}f(a,b)$, 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:

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

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 $\phi$, 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:

${\begin{aligned}\lambda (h,k)&=[D_{1}f(s_{0},b+k)-D_{1}f(s_{0},b)]\cdot h\\&=[\gamma (b+k)-\gamma (b)]\cdot h\\&=\gamma ^{\prime }(t_{0})\cdot hk\\&=D_{2}D_{1}f(s_{0},t_{0})\cdot hk\end{aligned}}$

Step 3

Now we can prove the theorem. Let $c=(a,b)\in A$, 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)