$g$ not continuous in $(0,0)$, differentiable in every direction AND $|D_vg(x)| \leq |v|$ I have found plenty of simliar questions to mine, but in this case there is one more condition that needs to be satisfied, this is the problem: 
"Find a function $g:\mathbb{R}^2 \rightarrow\mathbb{R}$, so that all directional derivatives $D_v g(x)$ exist ($v\in\mathbb{R}^2$) but $g$ isn't continuous in $(0,0)$ AND $|D_vg(x)| \leq |v|$."
It's easy to find a function that satisfies the first two conditions, but I just don't know how to use the third one. I also have a problem in understanding it. For example, I can have 2 vectors which point in the same direction but have different length, for example: 
$\left(
\begin{array}{c}
1\\
0\\
\end{array}
\right)$ and $\left(
\begin{array}{c}
0.01\\
0\\
\end{array}
\right)$, both point into the same direction, but their length isn't the same. With this in mind, $|v|$ can become arbitrary small, meaning that  $|D_vg(x)|$ has to be $0$, so $g$ has to be a constant function. But that doesn't really help, since there is no constant function that is uncontinuous in $(0,0)$ (or is there?), so I guess that I didn't understand the $|D_vg(x)| \leq |v|$ condition properly. 
Could you give me some advice?
 A: The condition $|D_vg| \leq |v|$ is really a constraint on $g$ as the angle of $v$ varies.
Your observation that $|D_vg|$ has to go to zero as $|v|\to 0$ is correct but not really relevant here. In fact $|D_vg|$ is always proportional to $|v|$. As you scale $v$, the proportionality constant will stay the same. (In some sense the "real" directional derivative of $g$ is the ratio $|D_vg|/|v|$.)
Requiring that $|D_vg|\leq |v|$ says that proportionality constant must always be less than $1$. As the angle of $v$ varies, the slope of $g$ along the line $vt$ can't increase without bound -- for example, $g(r,\theta) = r^{1/(1+2\sin^2\theta)}$ does not meet this condition, since along the line $\theta = \pi/2$ the function behaves like a cube root.
A: Suppose $x,y\in \mathbb R^2.$ Define $f:\mathbb R\to \mathbb R^2$ to be the function $f(t)= g(x+t(y-x)).$ Then $f'(t) = D_{y-x}g(x+t(y-x))$ for all $t.$ By the MVT, $f(1)-f(0) = f'(c)$ for some $c\in (0,1).$ It follows that
$$|g(y)-g(x)|=|f(1)-f(0)| = |f'(c)| = |D_{y-x}g(x+c(y-x))|\le |y-x|.$$
Thus $g$ is Lipschitz on $\mathbb R^2,$ so is certainly continuous everywhere.
