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I'm really asking sorry for this question because I know it's a poor question, but I need a final anwer about.

Suppose I have a function $f(x, y)$, defined in some way when $(x, y) \neq (0, 0)$ and defined as $0$ (or a constant $k$) when $(x, y) = (0, 0)$.

  • When I have to check for the partial derivatives (first derivatives) to be continuous, (at the origin) what I have to do is to show that

$$\lim_{(x, y) \to (0, 0)} f'_x (x, y)$$ must be equal to

$$\lim_{h\to 0} \frac{f(h, 0) - f(0, 0)}{h}$$

right?

If the two values are the same, then $f'_x$ is continuous (at the origin), otherwise they are not.

The same speech holds for $f'_y$.

This clearly generalizes to mixed derivatives, and second derivatives, right?

Heidegger
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  • Yes, that's it. You apply the definition of continuity to the function $f'_x(x,y)$. – Matthew Leingang Mar 08 '23 at 22:00
  • @MatthewLeingang Thank you! Instead for what concerns the mixed derivatives, in addition to that I also have to get $f''{xy} = f''{yx}$, right? (Meant as their value at the origin, after having showed the values of $f''_{xy}$ are the same when using the definition of continuity, and the limit) – Heidegger Mar 08 '23 at 22:08
  • @MatthewLeingang I mean, for the first derivatives it's not necessary that $f'x(0, 0) = f'_y(0, 0)$ but for the mixed derivatives it's necessary that $f''{xy}(0, 0) = f''_{yx}(0, 0)$, am I right? – Heidegger Mar 08 '23 at 22:12
  • If all you are doing is verifying continuity at a point, you don't additionally need to check that equality. – Matthew Leingang Mar 08 '23 at 22:18
  • @MatthewLeingang So if I get that, say, at the origin $f''{xy}(0, 0) \neq f''{yx}(0, 0)$, the mixed derivatives are anyway continuous? – Heidegger Mar 08 '23 at 22:19
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    I'm saying if you want to show $f''{xy}$ is continuous at $(0,0)$, all you have to show is $\lim{(x,y) \to (0,0)} f''{xy}(x,y) = f''{xy}(0,0)$. If you want to show $f''{yx}$ is continuous at $(0,0)$, all you have to show is $\lim{(x,y) \to (0,0)} f''{yx}(x,y) = f''{yx}(0,0)$. You do not need to additionally check that $f''{xy}(0,0) = f''{yx}(0,0)$; this is neither necessary nor sufficient for $f''{xy}$ and $f''{yx}$ to be continuous at $(0,0)$. – Matthew Leingang Mar 08 '23 at 22:53
  • @MatthewLeingang Are you sure about? Sorry for stressing you, but in this question https://math.stackexchange.com/questions/219759/show-that-both-mixed-partial-derivatives-exist-at-the-origin-but-are-not-equal a guy answers by concluding that since $f''{xy}(0, 0) \neq f''{yx}(0, 0)$, the mixed partial derivatives are not continuous... – Heidegger Mar 08 '23 at 22:58
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    If $f''{xy}$ and $f''{xy}$ are continuous in a neighborhood of $(0,0)$, then $f''{xy}(0,0)=f''{yx}(0,0)$. This is sometimes called Clairaut's theorem. Contrapositively, if $f''{xy}(0,0) \neq f''{yx}(0,0)$ then at least one of these second derivatives is not continuous in a neighborhood of $(0,0)$. That leaves some room for the possibility that $f''{xy}$ and $f''{yx}$ are continuous at $(0,0)$ but not in a neighborhood of $(0,0)$, and therefore could possibly not satisfy $f''{xy}(0,0) = f''{yx}(0,0)$. – Matthew Leingang Mar 08 '23 at 23:16

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