# Thread: Not C^2, but mixed partials exist and are equal.

1. ## Not C^2, but mixed partials exist and are equal.

I'm looking for a function f:R^2 --> R with the following properties:

1. f is C^2 everywhere except (0,0).
2. fxy and fyx exist, are equal, and are continuous everywhere.
3. fxx and fyy do not exist at (0,0).

Or a proof that no such function exist.

2. ## Re: Not C^2, but mixed partials exist and are equal.

Originally Posted by isterling
I'm looking for a function f:R^2 --> R with the following properties:

1. f is C^2 everywhere except (0,0).
2. fxy and fyx exist, are equal, and are continuous everywhere.
3. fxx and fyy do not exist at (0,0).

Or a proof that no such function exist.
Let $\boxed{f(x,y) = x\sqrt{x^2+y^4} + y\sqrt{x^4+y^2}}$. Then f is $C^2$, in fact $C^\infty$, everywhere except at the origin.

Also, $f_x(x,y) = \begin{cases}\sqrt{x^2+y^4} + \frac{x^2}{\sqrt{x^2+y^4}} + \frac{2x^3y}{\sqrt{x^4+y^2}} &\bigl((x,y)\ne(0,0)\bigr), \\ 0&\bigl((x,y)=(0,0)\bigr). \end{cases}$

Thus $f_{xy}(0,0) = \lim_{y\to0}\frac{f_x(0,y) - f_x(0,0)}y = \lim_{y\to0}\frac{y^2}y = 0$

and $f_{xx}(0,0) = \lim_{x\to0}\frac{f_x(x,0) - f_x(0,0)}x = \lim_{x\to0}\frac{2|x|}x$, which does not exist.

Since f(x,y) is symmetric in x and y, it follows that $f_{yx}(0,0) = f_{xy}(0,0) = 0$ and that $f_{yy}(0,0)$ does not exist.

Edit. I forgot that you also wanted $f_{xy}$ and $f_{yx}$ to be continuous at the origin. I'll leave you to figure out whether or not that is the case for this example. (I think it's okay, but I haven't checked carefully.)

3. ## Re: Not C^2, but mixed partials exist and are equal.

Thank you Opalg. I did check that they are continuous. In fact its easy to explicitly calculate fxy and see that it is continuous. I'm a newbie and don't see how I can change the setting to "solved".

4. ## Re: Not C^2, but mixed partials exist and are equal.

Ok, I marked it as solved!