# Proving equality of mixed second order partial derivatives

1. Oct 18, 2015

### PWiz

Let $f(x,y)$ be a scalar function. Then $$\frac{∂f}{∂x} = \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h} = f_x (x,y)$$ and $$\frac{∂}{∂y} \left (\frac{∂f}{∂x} \right ) = \lim_{k \rightarrow 0} \frac{f_x(x,y+k)-f_x(x,y)}{k} = \lim_{k \rightarrow 0} \left ( \frac{ \displaystyle \lim_{h \rightarrow 0} \frac{f(x+h,y+k)-f(x,y+k)}{h} - \displaystyle \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h}} {k} \right ) = f_{yx} (x,y)$$

Similarly, $$\frac{∂f}{∂y} = \lim_{k \rightarrow 0} \frac{f(x,y+k)-f(x,y)}{k} = f_y (x,y)$$
and $$\frac{∂}{∂x} \left (\frac{∂f}{∂y} \right ) = \lim_{h \rightarrow 0} \frac{f_y(x+h,y)-f_y(x,y)}{h} = \lim_{h \rightarrow 0} \left ( \frac{ \displaystyle \lim_{k \rightarrow 0} \frac{f(x+h,y+k)-f(x+h,y)}{k} - \displaystyle \lim_{k \rightarrow 0} \frac{f(x,y+k)-f(x,y)}{k}} {h} \right ) = f_{xy} (x,y)$$

Now if we use the fact that $$\lim_{\mu \rightarrow \sigma} (g(\mu)) ± \lim_{\mu \rightarrow \sigma} ((h(\mu)) = \lim_{\mu \rightarrow \sigma} (g(\mu) ± h(\mu))$$
then $$f_{yx} (x,y)= \lim_{k \rightarrow 0} \left ( \lim_{h \rightarrow 0} \left ( \frac{f(x+h,y+k)-f(x,y+k) - f(x+h,y) + f(x,y)}{hk} \right ) \right )$$ and $$f_{xy} (x,y) = \lim_{h \rightarrow 0} \left ( \lim_{k \rightarrow 0} \left ( \frac{f(x+h,y+k)-f(x+h,y) - f(x,y+k) + f(x,y)} {hk} \right ) \right )$$

I'm just one short step from establishing the symmetry here (since the expression in both cases is the same), but I'm stuck. How do I justify interchanging the order of the limits? I don't think the answers obtained by applying the limits in a different order should in general commute.

Additionally, it is quite easy to see from above that this proof will be valid only if all the mentioned partial derivatives are defined (namely $f_x(x,y)$,$f_y(x,y)$, $f_{xy}(x,y)$ and $f_{yx} (x,y)$ ), which is one of the required conditions for $f_{xy} (x,y)$ to be equal to $f_{yx} (x,y)$. However, I can't find a justification for the other necessary condition - that the functions must be continuous.
Finally, I want to make sure this proof is formal (although it is admittedly very trivial in nature), so let me know if there's any step which seems a little hard to digest mathematically.

Any help is appreciated.

Last edited: Oct 18, 2015
2. Oct 18, 2015

### geoffrey159

Schwart's theorem assumes $f\in{\cal{C}}^2(U)$, $U$ open set of $\mathbb{R}^2$.

3. Oct 18, 2015

### PWiz

Why is this assumption needed? Does the proof in the OP not work if we ignore this? I know counterexamples exist, but I want to know which step in the proof requires us to assume continuity.

4. Oct 18, 2015

### geoffrey159

It needs to be open because it makes legal such expressions as $f(x+h,y+h)$, which you use a lot for the demo. That means you are sure that there is an open ball centered in $(x,y)$ that is included in $U$, and it guarantees that their exist $h$ and $k$ small enough so that $(x+h,y+k)$, $(x+h,y)$, and $(x,y+h)$ are in $U$.

Last edited: Oct 18, 2015
5. Oct 18, 2015

### geoffrey159

In fact, you can lower the hypothesis to $f\in{\cal C}^1(U)$, $f_{xy}$ and $f_{yx}$ exist and are continuous.
I've just read the demo, it is more subtle than I thought.

Try to find two distrinct single variable functions $\phi$ and $\psi$ such that

$f(x+h,y+k)-f(x,y+k) - f(x+h,y) + f(x,y) = \phi(1) - \phi(0) = \psi(1) - \psi(0)$

Then with the mean value theorem, find a way to express these differences as a function of $f_{xy}$ and $f_{yx}$. Finish with the continuity hypothesis

6. Oct 19, 2015

### PWiz

This makes sense. What about interchanging the order of the limits though? Under which circumstances is that a legal operation?

7. Oct 19, 2015

### geoffrey159

This is THE problem in your demo. If it was possible to interchange the limits, the hypothesis of continuity for $f_{xy}$ and $f_{yx}$ would be useless.

8. Oct 27, 2015

9. Nov 8, 2015

### PWiz

Let $ψ(1)= f(x+h,y+k)-f(x,y+k)$ and $ψ(0) =f(x+h,y) - f(x,y)$
Then $$f_{yx} = \lim_{k \rightarrow 0} \left ( \lim_{h \rightarrow 0} \left ( \frac{ψ(1)-ψ(0)}{hk} \right ) \right )$$
Applying the mean value theorem, we get $$f_{yx} = \lim_{k \rightarrow 0} \left ( \lim_{h \rightarrow 0} \left ( \frac{ψ'(u)}{hk} \right ) \right )$$ where $y<u<y+k$
What now?

10. Nov 9, 2015

### Erland

It's hardly logical to call these expressions $\psi(1)$ and $\psi(0)$. Where did you get 1 and 0 from?
The logical would be to call them $\psi(y+k)$ and $\psi(y)$, viewing $\psi$ as a function of $y$, right?
Try this and it'll be quite easy...

11. Nov 10, 2015

### PWiz

You're right, I think I got it.
$$f_{yx} = \lim_{k \rightarrow 0} \left ( \lim_{h \rightarrow 0} \left ( \frac{ψ'(u)}{h} \right ) \right ) = \lim_{k \rightarrow 0} \left ( \lim_{h \rightarrow 0} \left ( \frac{\frac{∂f(x+h,u)}{∂y} - \frac{∂f(x,u)}{∂y}}{h} \right ) \right ) = \lim_{k \rightarrow 0} \lim_{h \rightarrow 0} f(g,u)_{xy}$$ where $x<g<x+h$. Taking the $h$ limit, $g \rightarrow x$, and taking then taking the $k$ limit, $u \rightarrow y$, giving $f(x,y) _{yx} = f(x,y)_{xy}$

12. Nov 10, 2015

### Erland

Basically correct, PWiz, but the treatment of limits needs to be improved. You're actually taking the $k$-limit first, when calculating $\psi'$, so it's not meaningful to have this limit at the outside.

Instead, I would suggest this: For $h$ and $k$ both $\neq 0$ and sufficiently small, begin with $\frac {f(x+h,y+k) - f(x,y+k) - f(x+h,y) + f(x,y)}{hk}$ and rewrite this in essentially the same way you already did, using $\psi$ and two applications of the Mean Value Theorem, but no limits at this stage, to obtain $f_{yx}(c,d)$, for some $c$ between $x$ and $x+h$ and some $d$ between $y$ and $y+k$. Taking the limit and using the continuity of $f_{yx}$ at $(x,y)$ we obtain that the first quotient tends to $f_{yx}(x,y)$ as $(h,k) \to (0,0)$.
By a similar calculation, the first quotient also tends to $f_{xy}(x,y)$, so the two mixed derivatives are equal at $(x,y)$.

13. Nov 16, 2015

### PWiz

Um, I'm just using the mean value theorem by stating that $\frac{ψ(y+k) - ψ(y)}{k} = ψ'(u)$ for some $u$ in the open interval $(y,y+k)$. I'm not taking any limit here.

14. Nov 17, 2015

### Erland

You're right. I was unclear. What I meant was that the derivative $\psi'(c)$ is itself a limit, of a function of $y$, and then there is no point in later taking the $k$-limit, which is associated to the variable $y$, since the expression to which you apply this limit does not depend upon $y$.
I know I wrote "no limits at this stage" in my reply, but since I tell you to use the mean value theorem, the same remark applies to this, so I didn't express it in an entirely consistent way. On the other hand, I do tell you not to take a $k$-limit later.

15. Nov 17, 2015

### PWiz

I'm afraid this no clearer to me than before. Can you explicitly show what you mean?

16. Nov 17, 2015

### Erland

Sorry PWiz, it seems that I have not been thinking clearly...

17. Nov 18, 2015

### geoffrey159

From the demo I read :

Set
$\phi(t) = f(x+th,y+k) - f(x+th,y)$
and
$\psi(t) = f(x+h, y+tk) - f(x,y+tk)$

Since these two functions are ${\cal C}^1([0,1])$, the mean value theorem says there exists $\alpha,\beta\in[0,1]$ such that
$\phi(1) - \phi(0)= \phi'(\alpha)$ and $\psi(1) - \psi(0) = \psi'(\beta)$.

Adapt this idea to show that there exists two functions $\tilde \phi$ and $\tilde \psi$ and $\alpha_1,\beta_1 \in [0,1]$ such that
$\phi(1) - \phi(0) = h ( \tilde \phi(1) - \tilde \phi(0) ) = hk f_{xy} (x+\alpha h, y+ \alpha_1 k)$
and
$\psi(1) - \psi(0) = k ( \tilde \psi(1) - \tilde \psi(0) ) = hk f_{yx} (x+ \beta_1 h, y+\beta k)$

You can now conclude by continuity of the mixed partial derivatives at $(x,y)$ and uniqueness of the limit.