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Suppose $f\colon\Bbb R^2\to\Bbb R$ is continuous. Prove that if, say, $D_1f(a,b)$ exists and there's a neighborhood $V$ of $(a,b)$ so that $D_2f$ exists on $V$ and is continuous., then that it implies condition (1)

(1) For every $\epsilon>0$ there is $\delta>0$ so that $|s|,|t|<\delta$ imply $$|f(a+s,b+t)-f(a+s,b)-f(a,b+t)+f(a,b)|\le \epsilon(|s|+|t|).$$

I tried to prove it as follows:

$$|f(a+s,b+t)-f(a+s,b)-f(a,b+t)+f(a,b)|$$ $$ < |t|\;|D_2f(a+s, y_1) - D_2(a,b)| + |f(a, b+t)-f(a,b) -tD_2f(a,b)| $$ Here the first is by MVT, (and $b < y_1 < b+t$) and now using the existence and continuity of $D_2f(a,b)$, I get

a $\delta_1 > 0$ so that $|D_2f(a+s, y_1) - D_2(a,b)| < \epsilon/2$ whenever $||(s, y_1 -b)|| < \delta_1$, i.e., $||(s, t)|| < \delta_1$. (since y1 lies between b and b+t)

Also, I get by existence of $D_2f(a,b)$, that there is $\delta_2 > 0$ so that $|f(a, b+t)-f(a,b) -tD_2f(a,b)| < \epsilon/2$ whenever $|t| < \delta_2$ Choose $\delta$ to be the smaller value, and we have $$|f(a+s,b+t)-f(a+s,b)-f(a,b+t)+f(a,b)| < |t| \epsilon < \epsilon(|t| + |s|)$$ whenever $|s|, |t| < \delta$

My worry is, I have not used the information about $D_1f$ at all. What have I done wrong, and if so please guide me towards a correct proof.

This question grew out of Professor Shifrin's answer here-What regularity conditions on partial derivatives are equivalent to differentiability?

me10240
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  • I can't follow this at all, to be honest. Where have you had to choose $\delta$ that gives the estimate with a particular $\epsilon$? As it stands, this seems to be meaningless. Did you work out the example of $f(x,y)=\sqrt{|x|}(y+1)$, as I suggested? It shows your claimed estimate can't possibly give the result as Dieudonné stated it. – Ted Shifrin Oct 28 '18 at 21:12
  • @TedShifrin: I added some more detail about how I got my $\delta$. I did not understand what I should conclude from the example. I see that I am looking for $|\sqrt{|s|}t| < \epsilon(|t| + |s|)$, but I can show that $|\sqrt{|s|}t| > 2/(1/s + 1/t^2)$ and hence I can not bound this above with the $\epsilon$. Is this what you were getting at ? – me10240 Oct 28 '18 at 22:12
  • @TedShifrin: I can abandon this "proof", could you instead show me how to prove the estimate the right way, so that it uses information about both derivatives ? – me10240 Oct 28 '18 at 22:16
  • To be honest, I don't yet see how to prove Dieudonné's remark directly. I guess we could just follow the proof that the remark $\implies$ differentiability $\implies$ the criterion. ... I had intended that you use my example function to find the flaw in your proof, as the assumption that $\partial f/\partial x(0)$ exists (which you never used) fails. – Ted Shifrin Oct 28 '18 at 23:45
  • @TedShifrin: Perhaps I was hasty in declaring your function a counterexample earlier. $\sqrt{|s|}|t| \leq \sqrt{|t|}(|s| + |t|)$ Does this now mean that for $\epsilon > 0, |s|, |t| < \delta = \epsilon^2$, I have $LHS < \epsilon(|s| + |t|)$ ? Let me know if I am right in this, in that case I will just give this up for now. – me10240 Oct 29 '18 at 00:14

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