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.
Best Answer
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.
See also https://math.stackexchange.com/questions/347599/questions-about-the-picard-lindelöf-theorem and the links there for all-at-once proof ideas.