Lipschitz condition problem

Question

Consider the following function:

$$f(t,y)=6t\sqrt[3]{y^2}$$

I want to verify if $f$ is locally Lipschitz in $(t,y)=(0,0)$.

So, I need to see if $f$ verifies the Lipschitz condition:

A function is locally Lipschitz in $(t_0,y_0)$, if: $$\|f(t,x)-f(t,y)\|\leq L\|x-y\|$$ where $L \in \mathbb{R}$ is a constant, and $(t,x)\in V(t_0,y_0)$, $(t,y)\in V(t_0,y_0)$

Well, we have that:

$$|f(t,x)-f(t,y)|=6|t|\left|\frac{x^{\frac{2}{3}}-y^{\frac{2}{3}}}{x-y}\right||x-y|$$

So, we just need to see if the term $|t|\left|\frac{x^{\frac{2}{3}}-y^{\frac{2}{3}}}{x-y}\right|$ is or not limited when $(t,x)\to (0,0)$ and $(t,y)\to (0,0)$.

Fixing $y=0$ we have:

$$\left|\frac{x^{\frac{2}{3}}-y^{\frac{2}{3}}}{x-y}\right|=\left|\frac{1}{x^{\frac{1}{3}}}\right|\xrightarrow{x\to 0} \infty$$

and we have $|t|\to 0$

So, we have an indetermination for $|t|\left|\frac{x^{\frac{2}{3}}-y^{\frac{2}{3}}}{x-y}\right|$ when $(t,x)\to (0,0)$, $(t,y)\to (0,0)$.

What could I do to solve this?

Answer 1

@ninjaaa: I offer you here a HINT to edit your answer if desired. All norms are equivalent on $\mathbb R^2$. Take $||(t,y)||=|t|+|y|$ on $\mathbb R^2$ and the usual distance on $\mathbb R$; note $B(0;1)$ the unit ball so $|t|+|y|\leq1$. The function $f(t,y)$ goes from $\mathbb R^2$ to $\mathbb R$ and curves given by fixed $t$ are symmetric respect to the $f(t,y)$-axis each of them being symmetric with the corresponding to $-t$ curve respect to the $y$-axis. In order to prove the question it is clearly enough to prove it for $f(t,y)=t y^\frac 23$. Restrict attention inside the ball $B(0;1)$.
The two figures below clarify somewhat the behavior of the function $f(t,y)=t y^\frac 23$

Answer 2

If a differentiable function $f$ is Lipshitz in some neighbourhood of $0$, we have $|f(h)| < c \cdot |h| $ for all $0 < h < \epsilon $ for some $\epsilon > 0$. It means that $f$ has bounded derivative in $0$.

Hence $frac{|f(h)|}{ |h| } < c \implies f'(0) \leq c$ .

Let $g(y) := \sqrt[3] { y^2 }$. But we have: $ g'(y) = (\sqrt[3] { y^2 })' = ( y^{2/3} )' = \frac{2}{3} y^{-1/3} = \frac{2}{3} \frac{1} {\sqrt[3]{y}}$. So $g'(y) \rightarrow \infty $ where $y \rightarrow 0$. Hence function $g$ isn't locally Lipshitz in $0$, and $f(t,y) = 6t \cdot g(y) $ also.