MHB Lipschitz Condition: Does $f(t,y)$ Satisfy? Find Constant

[evinda]

Hello! (Wave)

Does the following $f(t,y)$ satisfy the Lipschitz condition as for $y$, uniformly as for $t$? If so, find the Lipschitz constant.

$$f(t,y)=\frac{|y|}{t}, t \in [-1,1]$$

I have tried the following:

$$\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}=\frac{|y_1|-|y_2|}{t|y_1-y_2|} \leq - \frac{|y_1|-|y_2|}{|y_1-y_2|}= \frac{|y_2|-|y_1|}{|y_1-y_2|} \leq \frac{|y_1-y_2|}{|y_1-y_2|}=1$$

Thus $f$ satisfies the Lipschitz condition and the Lipschitz constant is equal to $1$.

Is it right? (Thinking)

 

[Evgeny.Makarov]

$$\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}=\frac{|y_1|-|y_2|}{t|y_1-y_2|} \leq - \frac{|y_1|-|y_2|}{|y_1-y_2|}$$

These equality and inequality are incorrect.

 

[evinda]

These equality and inequality are incorrect.

Is it right now?

$\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}= \frac{|\frac{|y_1|}{t}-\frac{ |y_2|}{t} |}{|y_1-y_2|}= \frac{||y_1|-|y_2||}{|t| |y_1-y_2|} \leq \frac{|y_1-y_2|}{|t| |y_1-y_2|}=\frac{1}{|t|} \overset{t \to 0}{\to } +\infty$

So we deduce that $f(t,y)$ doesn't satisfy the Lipschitz condition.

 

[Evgeny.Makarov]

The formulas are correct, but the conclusion is incorrect without the word "uniformly".

 

[evinda]

The formulas are correct, but the conclusion is incorrect without the word "uniformly".

So do we want to prove that the $\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}$ converges uniformly to $+\infty$, i.e. for any $t$ we pick?
But doesn't it only happen for $t \to 0$? Or am I wrong?

Or did you mean with uniformly that the limit is $+\infty$ for any $y_1, y_2 \in \mathbb{R}$ and for $t \to 0$?

 

[Evgeny.Makarov]

This problem does not require talking about convergence, especially uniform convergence. It is about uniform Lipschitz property. For each particular $t\ne0$ the function $f(t,y)$ is Lipschitz.

 

[evinda]

This problem does not require talking about convergence, especially uniform convergence. It is about uniform Lipschitz property. For each particular $t\ne0$ the function $f(t,y)$ is Lipschitz.

Ah I see... Thanks a lot! (Smile)