Here is the function : $f(t,x)=e^{tx}$ on $\mathbb{R\times R}$.
First the function is $\mathcal{C^1}$ so it's locally Lipschtiz in $x$ on $\mathbb{R\times R}$. Then I try to prove if it's globally Lipschitz in $x$ on $\mathbb{R\times R}$.
I want to use the argument of uniform continuity with sequences. So here I pick $x_n = n$ and $y_n= n +\frac{1}{n}$ and let $t \in \mathbb{R}$.
So I got : $\vert x_n -y_n\vert \le \frac{1}{n}$ and $\vert f(t,x_n)-f(t,y_n)\vert =\vert e^{tn}\vert \vert 1- e^{\frac{t}{n}}\vert$ which is equivalent for $n\to \infty$ to $\vert \frac{te^{tn}}{n} \vert \to +\infty$ if $t\ne 0$.
So $\vert f(t,x_n)-f(t,y_n)\vert > \varepsilon$
If $t=0$ then $\vert f(t,x_n)-f(t,y_n)\vert =0$ indeed $\vert e^{tn}\vert \vert 1- e^{\frac{t}{n}}\vert \to 0$
That means that there is no uniform continuity for $f$ on $\mathbb{R-\{0\}\times R}$. So it's not globally Lipschitz on $\mathbb{R-\{0\}\times R}$.
I don't know if it's the right method to answer because I was wondering if I used correctly the parameter $t$. What can I say for $t=0$ ?
Thanks in advance.
Your argument, in particular the sentence containing "equivalent", does not seem stringent to me.
It is true that proving the function is not uniformly continuous in $x$ for given $t$ settles the claim. But in this simple example theres is no gain: We would then have to argue with $\epsilon$s and $\delta$s, whereas the Lipschitz property is in terms of simple inequalities about $f$.
When $t=0$ then $\phi(x):=f(0,x)\equiv1$; whence $f$ is globally Lipschitz in $x$ for this value of $t$.
Let a $t>0$ be given (the case $t<0$ is similar). Assume $x<y$. Then by the mean value theorem there is a $\xi\in\ ]x,y[\ $ with $$e^y-e^x=t e^{t\xi}(y-x)\geq t e^{tx}(y-x)\ .$$ It follows that $${f(t,y)-f(t,x)\over y-x}\geq t e^{tx}\ ,$$ and here the right hand side is unbounded when $x\to\infty$. This implies that there is no constant $L_t$ such that $$|f(t,y)-f(t,x)|\leq L_t\>| y-x|$$ for all $x$, $y\in{\mathbb R}$.