Can someone help me proving this:

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$,

\begin{align} f(x,y) = \begin{cases} 0 & \text{ if } y \leq 0 \text{ or } y \geq x^2 \\ \\ 1 & \text{ if } 0 < y < x^2 \end{cases}. \end{align}

Show that $f$ is not continious in $(0,0)$ but that all directional derivates do exist in $(0,0)$.

For the first part I used $\xi_n := (x_n, y_n) = (\frac{1}{\sqrt{2n}}, \frac{1}{3n})$, so $\xi_n \rightarrow (0,0), (n \rightarrow \infty)$ but $\lim\limits_{n \to \infty} f(\xi_n) \neq f((0,0)) = 0$, but I struggle to show the second part with the directional derivates. We defined latter as:

\begin{align} f'(t) = \dfrac{\text{d}}{\text{dt}} (g(u(t))) =: \nabla g(u(t)) \cdot u'(t) \end{align} with $f(t) = g(u_1(t), ..., u_n(t)), g: \mathbb{R}^n \rightarrow \mathbb{R}, u: \mathbb{R} \rightarrow \mathbb{R}^n$.

I thought - as the image of the function only is $0$ or $1$ - that the gradient has to be $0$, as $\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial y} = 0$. So therefore the dircetional derivates should all exist.

It would be nice to know where my thinking mistake is. (The exercise is part of an old exam, which I'm preparing for. There are no solutions available, but I know that this problem had quite a lot of points, which usually means it's not that simple).

Thanks in advance.


The argument "only takes two values, so gradient is zero" only works at places where the function is (at least directionally) differentiable. Here you still need to show that differentiability.

So it is better to directly use the definition of directional derivative, which should be something like the following (in direction of a unit vector $v$):

$\frac{\partial f}{\partial v}(0,0):=\lim_{h\to 0} \frac1h(f(vh)-f(0,0))$

For $h$ sufficiently small (depending on $v$), note that $vh=(hv_1,hv_2)$ lies in the set where $f$ is zero, so the derivatives all exist and are zero.

  • $\begingroup$ That's also something I thought about. So I have $\lim_{h \to 0} 1/h \cdot f(vh)$. If $f(vh) = 0$, than I see your point but what I don't get is why $f(vh)$ should always be $0$. Can you explain that point a bit more? $\endgroup$ – offline May 31 '18 at 14:43
  • 1
    $\begingroup$ Draw a picture and colour the bit where $f$ is zero in white, the bit where $f$ is $1$ in some other colour. If you keep $v$ fixed, the line $[0,s]\ni t\mapsto tv$ completely lies in the white area for $s$ small enough. $\endgroup$ – Kusma May 31 '18 at 15:52

Draw a figure! One has $f(x,y)=0$ except when $0<y<x^2$. Then $f(x,y)=1\ne f({\bf 0})$ on all points $\ne{\bf 0}$ of the parabola $y={1\over2}x^2$. This implies that $f$ is not continuous at ${\bf 0}$. On the other hand, for any line $\ell$ through ${\bf 0}$ we have $f(x,y)\equiv0$ on all points $(x,y)\in\ell$ sufficiently near ${\bf 0}$. This shows that all directional derivatives of $f$ are $0$ at ${\bf 0}$.


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