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Math Help - Not C^2, but mixed partials exist and are equal.

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

    Quote Originally Posted by isterling View Post
    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.)
    Last edited by Opalg; December 15th 2011 at 11:57 PM.
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  3. #3
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    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".
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  4. #4
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    Re: Not C^2, but mixed partials exist and are equal.

    Ok, I marked it as solved!
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