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I was having trouble finding the directional derivatives at the origin of the function $$f(x,y)=\begin{cases} [(2x^{2}-y)(y-x^{2})]^{1/4} & \text{for $x^{2} \leq y \leq 2x^{2}$} \\ 0 & \text{otherwise} \end{cases}$$

I understand the directional derivative in the direction $\xi=(v,w)$ at the origin to be \begin{align*} D_{\xi}(0) &= \lim_{h \to 0} \frac{f((0,0)+h\xi)-f(0,0)}{h} \\ &=\lim_{h \to 0} \frac{f(hv,hw)}{h} \end{align*} I'm a bit confused past this point. The condition for $x^{2} \leq y \leq 2x^{2}$ is what I find confusing, since it implies that $f(h(v,w))=0$ for $hv^{2} \leq w \leq 2hv^{2}$.

user135520
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    Draw the region $x^2\leq y\leq 2x^2$ near zero, and you'll see what happens along any line - no line through $(0,0)$ can stay inside that region as it gets close to $0$. So any line through $(0,0)$ has $f$ equal to zero in some neighborhood of $0$, and hence obviously derivative zero. – Thomas Andrews Dec 22 '15 at 15:42
  • Thanks, I think I understand why the directional derivatives are $0$. We can look at $(hv,hw)$ as a line through the origin but as we get closer to the origin as $h \to 0$, we move outside of that region.f – user135520 Dec 22 '15 at 16:24
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    Yes. I saw this trick recently showing that $(y-x^2)(y-2x^2)$ has a local minimum at $(0,0)$ along every line through $0$, but there are negative values arbitrarily close to $0$. It's much the same reasoning - when $x^2<y<2x^2$ the function is negative, positive elsewhere, but every line through $(0,0)$ is outside that region "near" $(0,0)$. http://www.mathcounterexamples.net/no-minimum-at-the-origin-but-a-minimum-among-all-lines/ – Thomas Andrews Dec 22 '15 at 16:39
  • Thanks for sharing, I don't think I would have thought of it independently. – user135520 Dec 22 '15 at 16:43

1 Answers1

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Pick any line through the origin, it will have the form $l(t) = (tv,tq)$, where $(v,q) \neq 0$.

Let $S = \{(x,y) | x^2 \le y \le 2 x^2 \}$ and $T= \{t | l(t) \in S \}$. Note that if $(0,y) \in S$, then $y = 0$.

Hence if $x=0$, then $T= \{0\}$ and if $x \neq 0$, then $T = \{0\} \cup [{q \over v^2}, {q \over 2 v^2}]$.

So, if $x = 0$, then $f(l(t)) = 0$, and if $x \neq 0$, then for $t< {|q| \over v^2}$ we have $f(l(t)) = 0$.

Since $f(l(0)) = 0$, we see that $df((0,0), (v,q)) = 0$.

copper.hat
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    In particular, the value of this function inside the region $S$ is a red herring. Indeed, you could pick whatever values you wanted, as long as $f(0,0)=0$, and you'd get directional derivatives at $(0,0)$ in every direction, even with discontinuities at the other boundary points of $S$. – Thomas Andrews Dec 22 '15 at 16:06
  • Would either of you two know of a way to show that this function is not differentiable at the origin. It looks to be continuous there. – user135520 Dec 22 '15 at 16:36
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    @user135520: If $f$ was differentiable at the origin, it would have to have derivative $(0,0)$ since it must match the directional derivative (in the appropriate sense). By noting that $f(t,{3 \over 2} t^2) = {1 \over \sqrt{2} } t$ you can show that it cannot have derivative zero at the origin. – copper.hat Dec 22 '15 at 18:38