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Let $g(t)=t^2$ if $t \ge 0$ and $g(t)=-t^2$ when $t < 0$.

Let $f(x,y)=g(x)+g(y)$.

I have so far been able to show that $f(x,y)$ is differentiable and has continuous partial derivatives.

Now let $A=\{(x,0)\} \cup \{(0,y)\}$. Show that $f(x,y)$ is not twice differentiable.

$\textbf{My Attempt:}$

I know that $f'(x,y)= \begin{bmatrix} g'(x) & g'(y) \end{bmatrix}$. How do I work with the matrix representation of the derivative of $f(x,y)$ to show that $f(x,y)$ is not twice differentiable?

I know this has to do with the fact that $g(t)$ is not twice differentiable in a neighborhood of zero. I would really appreciate some advice regarding the technical details.

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  • $\begingroup$ When do you say a function of two variables is twice differentiable? I.e: what is the definition you are using? $\endgroup$
    – gebruiker
    May 14 '14 at 13:26
  • $\begingroup$ @Aal, If f is twice differentiable, then $f''=\begin{pmatrix} \partial_{11} f & \partial_{21} f \dots \\ \vdots \\ \partial_{n1} f & \dots & \partial_{nn} f \end{pmatrix}$ $\endgroup$
    – user7090
    May 14 '14 at 13:49
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The fact that $g''(0)$ is not properly defined means that there is an element in your matrix that is not defined as well, making the whole matrix undefined.

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  • $\begingroup$ NB. I'm not shure what your intensions are for the region $A$ you defined (the axis without the origen), but it excludes the origen (our troublemaker). So if you want to show differentiability on $A$ then you don't run in to any errors and everything is defined. $\endgroup$
    – gebruiker
    May 14 '14 at 19:12

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