Does $f(t,x)=tx^2-tx$ satisfy Lipschitz condition?

Question

I have function $f: [0,T] \times \mathbb{R} \rightarrow \mathbb{R}$, $T \in \mathbb{R_+}$ $$f(t,x)=tx^2-tx$$ I want to check if it satisfies Lipschitz condition and I got $$|f(t_1, x_1)-f(t_2, x_2)|=|t_1x_1^2-t_1x_1-t_2x_2^2+t_2x_2| \leq T |x_1^2-x_2^2 -(x_1-x_2)|= \\T |x_1+x_2-1||x_1-x_2| \leq T |x_1+x_2-1|(|t_1-t_2|+|x_1-x_2|)$$ and I guess since $x_1, x_2$ are real then $|x_1+x_2-1|$ is unbounded so it doesn't satisfy Lipschitz condition. Is this enough of a proof though? I thought of checking if this function is uniformly continuos but coudn't quite get anywhere.

Answer 1

No, this is not enough. What you should be trying to prove is that for every $K > 0$ there exist $(t_1,x_1)$ and $(t_2, x_2) \in [0,T]\times \mathbb{R}$ such that $$ |f(t_1,x_1) - f(t_2,x_2)| > K|(t_1, x_1) - (t_2, x_2)|. $$

For that, pick $(t_1, x_1) = (T, x + 1)$ and $(t_2, x_2) = (T, x)$. Then

$$ |f(T, x+1) - f(T, x)| = T|(x+1)^2 - (x+1) - (x^2-x)| = T|2x| > K|(T, x+1) - (T, x)| $$ for $x$ large enough.