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Here is a quote from Calculus: Early Transcendentals written by James Stewart

Clairaut’s Theorem:Suppose $f$ is defined on a disk $D$ that contains the point $(a, b)$. If the functions $f_{xy}$ and $f_{yx}$ are both continuous on $D$, then $f_{xy} = f_{yx}$

Where $$f_{xy} \equiv \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right)$$

I don't know how to compose a function that $f_{xy}$ or $f_{yx}$ are not continuous on $(a, b)$. Can you give me an example?

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For example take $$ f(x,y)= xy\frac{x^2-y^2}{x^2+y^2} \quad ((x,y)\not=(0,0)), \quad \quad f(0,0)=0. $$ Here $f_x$ and $f_y$ both exist and are even continuous on $\mathbb{R}^2$, and $$ f_{xy}(0,0)=-1,\quad f_{yx}(0,0)=1. $$ The continuity of $f_x$ and $f_y$ in $(0,0)$ is a bit tricky, but it is easy to show that $f_x(0,y)=-y$ for all $y$ and $f_y(x,0)=x$ for all $x$.

Edit: You can see directly that $f_{xy}$ and $f_{yx}$ are not continuous in $(0,0)$. You have calculated $$ f_{xy}(x,y)=f_{yx}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3} \quad ((x,y) \not= (0,0)). $$ Inserting $(1/n,1/n)$ leads to $f_{xy}(1/n,1/n)= 0 \to 0$ $(n \to \infty)$. Inserting $(1/n,0)$ leads to $f_{xy}(1/n,0)= 1 \to 1$ $(n \to \infty)$. Thus $\lim_{(x,y)\to (0,0)} f_{xy}(x,y)$ does not exist.

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    $\begingroup$ I try to calculate $f_{xy}$ and $f_{yx}$. They are the same. $$ f_{xy} = \frac{x^2 - y^2}{x^2 + y^2} + 8xyf \frac{x^2 + y^2}{(x^2 + y^2)^3} = \frac{x^6 + 9x^4y^2 -9x^2y^4 - y^6}{(x^2 + y^2)^3} = f_{yx} $$ $\endgroup$
    – gyro
    Commented Dec 22, 2022 at 8:04
  • $\begingroup$ Of course they are the same on $\mathbb{R}^2 \setminus \{(0,0)\}$ according to Clairaut's Theorem. But $f_{xy}(0,0)$ and $f_{yx}(0,0)$ are different and have to be calculated by the definition of partial derivatives. This shows that $f_{xy}$ and $f_{yx}$ cannot both be continuous in an neighborhood of $(0,0)$. $\endgroup$
    – Gerd
    Commented Dec 22, 2022 at 8:23
  • $\begingroup$ I see. Thank you! $\endgroup$
    – gyro
    Commented Dec 22, 2022 at 11:52
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Just pick any function which isn't continuous and integrate it twice.

For example, let $D=(-1,1)\times (-1,1)$. Consider $$g(x,y)=\operatorname{sgn}(x)\operatorname{sgn}(y)$$ Which is not continuous on $D$. Now let $$f(x,y)=\int_0^y\int_0^x g(x',y')\mathrm dx'\mathrm dy' \\ =\int_0^y\int_0^x \operatorname{sgn}(x')\operatorname{sgn}(y')\mathrm dx'\mathrm dy' \\ =\int_0^y|x|\operatorname{sgn}(y')\mathrm dy' \\ =|xy|$$

Note that $f$ is continuous on $D$.

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  • $\begingroup$ I should add: $\operatorname{sgn}(x)=|x|/x$ and $\operatorname{sgn}(0)=0$ is the sign function. $\endgroup$
    – K.defaoite
    Commented Dec 21, 2022 at 14:44
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    $\begingroup$ How are $f_{xy}$ and $f_{yx}$ defined on a neighborhood of the origin? $\endgroup$ Commented Dec 21, 2022 at 15:45
  • $\begingroup$ @TedShifrin That wasn't stated as a requirement. $\endgroup$
    – K.defaoite
    Commented Dec 21, 2022 at 23:02
  • $\begingroup$ The hypothesis was “$f_{xy}$ and $f_{yx}$ are both continuous on the disk $D$,” so I believe you are in error. $\endgroup$ Commented Dec 22, 2022 at 0:26
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    $\begingroup$ Yes, OK, I responded to the wrong issue. You're claiming the equality of $f_{xy}(0)$ and $f_{yx}(0)$ fails to hold because neither is defined? Perhaps you win on a logical technicality, but I don't think you're enhancing the OP's understanding of the theorem. $\endgroup$ Commented Dec 22, 2022 at 3:44

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