$|g(t_{1}) e^{-(t_{1}-x)^{2}}- g(t_{2})e^{-(t_{2}-x)^{2}}|\leq |f(t_{1}) e^{-(t_{1}-x)^{2}}- f(t_{2})e^{-(t_{2}-x)^{2}}| $?

Question

Suppose $f, g: \mathbb R \to \mathbb C$ such that $|g(t_{1}) -g(t_{2})| \leq |f(t_{1})- f(t_{2})| $ for every $t_{1}, t_{2} \in \mathbb R.$ Take any $x\in \mathbb R$ and fix it. Edit: We also assume that $f,$ and $g$ are not constant functions; but continuous and vanishing at infinity.

My Question is: (1) Can we expect, $|g(t_{1}) e^{-(t_{1}-x)^{2}}- g(t_{2})e^{-(t_{2}-x)^{2}}| \leq |f(t_{1}) e^{-(t_{1}-x)^{2}}- f(t_{2})e^{-(t_{2}-x)^{2}}| $ ?

(2) If the answer of (1) is negative, then can we expect, to get some $0\neq h\in \mathcal{S} (\mathbb R)$(the Schwartz space) such that $$|g(t_{1})h_{x}(t)- g(t_{2}) h_{x}(t_{2}) |\leq |f(t_{1})h_{x}(t)- f(t_{2}) h_{x}(t_{2}) |$$ , for every $t_{1}, t_{2} \in \mathbb R$? (where $h_{x}(t)= h(t-x), t\in \mathbb R$ (translation of h))

Answer 1

I can come up with a trivial counterexample. Let $g(t) = 2$ for all $t\in \mathbb{R}$ and let $f(t) = 1$ for all $t\in \mathbb{R}$. Then $$|g(t_1) - g(t_2)| = |f(t_1) - f(t_2)| = 0$$ for all $t_1,t_2\in \mathbb{R}$ so the condition holds trivially. But $$|g(t_1)e^{-(t_1-x)^2} - g(t_2)e^{-(t_2-x)^2}| = 2|e^{-(t_1-x)^2} - e^{-(t_2-x)^2}|$$ $$ > |e^{-(t_1-x)^2} - e^{-(t_2-x)^2}| = |f(t_1)e^{-(t_1-x)^2} - f(t_2)e^{-(t_2-x)^2}|$$ if $|e^{-(t_1-x)^2} - e^{-(t_2-x)^2}| > 0$ which is true for most pairs $t_1,t_2$.

So I think you may need to revise the assumptions in your problem a bit.