I was reading an exercise, and supposedly this function: $$\chi \colon \Bbb R\times\Bbb R\to\Bbb R, \quad \chi (t,x)=3x^{2/3}$$ is not locally Lipschitz (in the second variable). In the notes this isn't proved, so I assumed that it was easy to see, but I've been trying to prove it and I can't get very far.
What I've done is suppose that it is locally Lipschitz, in particular we can take $x_0=0$ and $t_0=0$, then exists $\delta_0>0$ and $C>0$ such that, if $|t-t_0|=|t|\le\delta_0$ and $x_1,x_2\in \overline B(x_0,\delta_0)=\overline B(0,\delta_0)$, then $\|\chi(t,x_1)-\chi(t,x_2)\|\leqslant C\|x_1-x_2\|$.
Also, because we are taking $x_1$ and $x_2$ very close to the origin (and because the way the function behave) we have that $||\chi(t,x_1)-\chi(t,x_2)||\geqslant ||x_1-x_2||$, I don't know if this is actually helpful, but I got two characteristics of the constant $C$, which I'm not quite sure if they make any sense or if they are worth something.
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Because we have this: $\|\chi(t,x_1)-\chi(t,x_2)\|\geqslant \|x_1-x_2\|$, then this happens: $C\|\chi(t,x_1)-\chi(t,x_2)\|\leqslant C\|x_1-x_2\|$ only if $C<0$, wich isn't true since $C$ is a Lipschitz constant.
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We know that $\|\chi(t,x_1)-\chi(t,x_2)\|\leqslant C\|x_1-x_2\|$ and $\|\chi(t,x_1)-\chi(t,x_2)||\geqslant \|x_1-x_2\|$, then $1\leqslant\frac{\|\chi(t,x_1)-\chi(t,x_2)\|}{\|x_1-x_2\|}\leqslant C$ hence $1\leqslant C$.
What am I doing wrong? How can I prove that $\chi$ is in fact, not locally Lipschitz?
Best Answer
The function $x \mapsto \chi(t,x)$ is not Lipschitz at $x=0$. (Note: Being locally Lipschitz is a stronger condition.)
You can see that the derivative becomes unbounded near $x=0$:
Suppose $x \ge 0$. Then $\chi(t,x)-\chi(t,0) = 3 x^{\frac{2}{3}}= 3 \frac{1}{\sqrt[3]{x}}x$, and so $|\chi(t,x)-\chi(t,0)| = 3 \frac{1}{\sqrt[3]{x}}|x-0|$. Hence for any $L>0$, if we choose $0\le x \le (\frac{3}{L})^3$, then $|\chi(t,x)-\chi(t,0)| \ge L |x-0|$.
Consequently, $x \mapsto \chi(t,x)$ is not Lipschitz at $x=0$.