# Globally Lispchitz function has a unique solution defined for all $t\in \mathbb{R}$

I am trying to justify a result that I saw that says

Let $$f:\mathbb{R}\times \mathbb{R}^n$$ be a globally Lipschitz function. Then the Cauchy problem $$x'=f(t,x), x(t_0)=x_0$$ has a unique solution defined for all $$t\in \mathbb{R}$$.

Is this going to be true ?

First I saw that if we have a linearly bounded function $$f(t,x)$$ that is there exists a constant $$C$$ such that $$|f(t,x)|\leq C(1+|x|), \forall (t,x)\in \mathbb{R}\times\mathbb{R}^n$$ then the cauchy problem has a unique solution defined for all $$t\in \mathbb{R}$$.

I saw this by using Gronwall's lemma.Since $$x(t)=x_0+\int_{t_0}^tf(s,x(s))ds$$ we will have that $$|x|\leq |x_0|+\int_{t_0}^t|f(s,x(s))|ds\leq |x_0|+\int_{t_0}^tC(1+|x|)ds\leq |x_0|+C(t-t_0)+C\int_{t_0}^t|x|$$. We have that for all $$t'\in [0,t]$$ there exists $$T$$ such that $$T\geq t'$$ and so $$C(t-t_0)\leq C(T-T_0)$$. Letting $$|x_0|+C(T-t_0)=M$$, we have $$|x(t)|\leq M+\int_{t_0}^t|x|ds$$, and now using Gronwall's lemma we have that $$|x(t)|\leq Me^{C(t-t_0)}$$, and so the function is defined $$\forall t\in \mathbb{R}$$. Now suppose the maximal interval of existence $$(\alpha,\beta)$$ isn't $$\mathbb{R}$$, without loss of generality we can assume that $$\beta <\infty$$. We would have $$\lim_{x\rightarrow \beta}x(t)$$ will be in the border of $$\mathbb{R}^n$$, this is a contradiction since we know that $$x(t)\in \mathbb{R}^n, \forall t\in \mathbb{R}$$.

Now with this I have tried to show that a globally Lipschitz function is Linearly bounded but all I got is that if we assume that $$f(t,x)=f(t+1,x), \forall (t,x)\in \mathbb{R}\times \mathbb{R}^n$$ , then I can prove that it is Linearly bounded. I can't seem to do this for the more general case, any help is aprecciated. Thanks in advance.

New edit : I think I was able to see in the general case but want to make sure this works, I am not interely sure it does because my constant is changing. Any $$t$$ is a compact interval of the form $$[0,T']$$. We have that $$|f(t,x)|-|f(t,x_0)|\leq |f(t,x)-f(t,x_0)|\leq L|x-x_0|\leq L|x|+L|x_0|$$. So we have that $$|f(t,x)|\leq |f(t,x_0)| +L|x|+L|x_0| \leq C(1+|x|)$$, but here we have that $$C$$ is changing if choose another $$t$$ the maximum of the function $$|f(t,x_0)|$$ can change.

As you did in the periodic case or similarly over the time interval $$[0,1]$$, you can construct a linear bound over any time interval $$[-N,N]$$. The coefficients may grow with $$N$$, but by the continuity you know that the bounds exist. Uniqueness tells us that the restriction of the $$N+1$$ solution to the $$N$$ interval has to coincide with the $$N$$ solution.
• Ah ok now I got it , the coefficients may change and so by that reason we consider the funciton only on each interval of the compact covering of $\mathbb{R}$. Thanks. Oct 14 '20 at 13:14