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The Exercise

Let $f(x,y)=x$ if $|y|>x^2$

and $f(x,y)=0$ otherwise.

Show that all the directional derivatives of $f$ exist at the origin but there does not exist a linear map $D$ such that $D_vf(0)=D(v)$ for each $v\in R^n$. Show that $lim_{t\to 0}(f(tv)-f(0))/t)$ does not converge to $D_vf(0)$ uniformly over $v\in S^1$.

My Attempt

I have no idea yet about the second and third part of the problem, but I'm working on showing that all the directional derivatives of $f$ exist at the origin.

$D_vf(x)=lim_{t\to 0}(f(x+tv)-f(x))/t$

Take any $v\in R^n$

$D_vf(0,0)=lim_{t\to 0}(f((0,0)+tv)-f(0,0))/t=lim_{t\to 0}f(tv)/t$

How do I know that this limit exists?

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  • $\begingroup$ Seemingly the function vanishes on the parabolas $y=x^2$ and $y = -x^2$. Any small neighbourhood around the origin would have to contain a part of these parabolas. Does that help? $\endgroup$ – Vishesh Feb 5 '14 at 5:39
  • $\begingroup$ Actually it vanishes at many more points than that. Try drawing a picture. It should become visually clear as to why the convergence cannot be uniform. $\endgroup$ – Vishesh Feb 5 '14 at 5:56
  • $\begingroup$ But I'm still trying to do the first part of the problem, showing that all directional derivatives of $f$ exist at the origin. $\endgroup$ – Jeff Feb 5 '14 at 7:41
  • $\begingroup$ Well let $v=(v_1,v_2)$.Then if $|tv_2|>t^2v_1{^2}$, $f(tv) = tv_1$ and the limit is $v_1$ else it is 0. So the limit exists, right?? $\endgroup$ – Vishesh Feb 5 '14 at 15:01

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