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In almost all statements of Clairaut's theorem, the equality of the mixed partials is given only after the assumption that both mixed partials exist and are continuous. In this blog post here, the author proves a stronger statement

"At some point $\vec{x}=(x_0,y_0)$, if the partials $f_x$ and $f_y$ are both continuous and if $f_{xy}$ exists and is continuous, then $f_{yx}$ exists and we have $f_{yx} = f_{xy}$."

The proof goes as follows:

Using the mean value theorem, the author obtains the statement $$\lim_{h\rightarrow 0}\ \lim_{k\rightarrow 0}\ f_{xy}(x_0 + \overline{h}, y_0 +\overline{k}) = f_{yx}(x_0, y_0)$$

for some point $\overline{h} \in (0, h)$ and $\overline{k} \in (0, k)$. Since $\overline{h}$ depends on $k$, the limit cannot be taken trivially. From the continuity of $f_{xy}$ we get

$$\mid f_{xy}(x,y) - f_{xy}(x_0, y_0)\mid < \frac{\epsilon}{2}$$

for $(x,y)$ within $\delta$ of $(x_0, y_0)$. If we take $\mid h \mid$ and $\mid k\mid$ small enough, say less than $\frac{\delta}{2}$, it follows that the point $(x_0 + \overline{h}, y_0 +\overline{k})$ satisfies the limit. So we fix $h$ small enough to obtain

$$\mid f_{xy}(x_0 + \overline{h},y_0 + \overline{k}) - f_{xy}(x_0, y_0)\mid < \frac{\epsilon}{2}$$

for $\mid k\mid < \frac{\delta}{2}$.

What he does next is totally lost to me.

He says we can take the limit from $k\rightarrow 0$, and when we do so, the inequality may become an equality (why?). But since we chose $\frac{\epsilon}{2}$ initially, we obtain $$\mid \lim_{k\rightarrow 0}\left[f_{xy}(x_0 + \overline{h},y_0 + \overline{k})\right] - f_{xy}(x_0, y_0)\mid \le \frac{\epsilon}{2} < \epsilon$$

from which the limit is established.

My main questions concern what exactly he did in the last few steps. How could the inequality possibly become an equality? And it seems to me that what he did is no different than just trivially taking $k$ to zero and then $h$ to zero.

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1 Answer 1

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The fact under consideration is true. It amounts to Fubini's Theorem, in this shape: for some continuous $h(x,y)$ define $$ G(x,y) = \int_a^x \int_c^y \; h(u,v) \; dv \; du $$ Fubini says that we can interchange the order of integration. That and the fundamental theorem of calculus say that both $$ \frac{\partial}{\partial y} \left( \frac{\partial G}{\partial x} \right) $$ and $$ \frac{\partial}{\partial x} \left( \frac{\partial G}{\partial y} \right) $$ exist, are continuous and equal to $h.$

In particular, if $ \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) $ exists and is continuous, iterated integration tells us that $$ G(x,y) = f(x,y) - f(x,c) - f(a,y) + f(a,c), $$ with $$ \frac{\partial}{\partial y} \left( \frac{\partial G}{\partial x} \right) = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) $$ However, the other order exists for $G,$ so we get existence for $f$ and $$ \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial x} \left( \frac{\partial G}{\partial y} \right) $$

This argument is from AKSOY_MARTELLI. I was able to view the pdf on my computer screen but not print it out. We are talking about the first part of Theorem 3, (i) implies (ii), bottom of page 128 to the first paragraph on page 129.

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Do you think the proof provided in the blog incorrect? Because I would love to have a proof of this statement which is independent of Fubini's. –  EuYu Jan 12 '12 at 23:31
    
@EuYu I just looked at the blog for a few seconds. My main conclusion was that I did not recognize the author's name, and would rather see almost any published article that went through a referee process. How would you describe your own background? –  Will Jagy Jan 13 '12 at 0:33
    
I'm just an undergraduate student. I was simply curious about the proof. –  EuYu Jan 13 '12 at 2:46
    
@EuYu Curiosity is good. Also your skepticism about a proof that does not seem to hang together is healthy. Since it turns out the overall claim is true, we cannot find a flaw in the blog proof by simply finding a counterexample. –  Will Jagy Jan 13 '12 at 2:57
    
Thanks for all your help. –  EuYu Jan 13 '12 at 3:06
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