If $g(t,x) \rightarrow 0$ as $x \rightarrow \infty$, then decays uniformly in $t$?

Question

Assume we have continuous $g:[0,T]\times \mathbb{R}\rightarrow \mathbb{R}$ such that $$ \lim_{x \rightarrow \pm \infty} g(t,x)=0 $$ for every $t \in [0,T]$.

Then let $(t_k)_{k\in\mathbb{N}}\subseteq [0,T]$ converge to some $t \in [0,T]$ and $(x_k)_{k \in \mathbb{N}}\subseteq\mathbb{R}$ diverge to $\infty$. Is it then true that $$ \lim_{k \rightarrow \infty} g(t_k,x_k) =0? $$

Initially, I thought yes. I tried to use continuity to make $\lvert g(t_k,x_k)-g(t,x)\rvert$ small. But continuity is not uniform, such that there is noch guarantee the above expression goes to $0$ as $k\rightarrow \infty$.

So there might be a counterexample , although I do not see it…

Answer 1

Consider the function $$g(t,x) = (1+t)^{-e^{|x|}}-e^{-|xt|}.$$ It follows that $g:[0, T]\times \mathbb{R} \to \mathbb{R}$ is a continuous function, $\lim_{x\to \pm \infty} g(t,x) = 0$ for all $t$.

Now, we take the sequence $x_{k} = \ln(k)$, and $t_{k} = \frac{1}{k}$. We have that $x_{k} \to \infty$ and $t_{k} \to 0 \in [0,T]$. However, $$\lim_{k \to \infty} g(t_{k},x_{k}) = \lim_{k\to \infty} \left(1+\frac{1}{k}\right)^{-e^{\ln(k)}} -\frac{1}{e^{\ln(k)\frac{1}{k}}} = \lim_{k \to \infty}\frac{1}{\left( 1 + \frac{1}{k} \right)^{k}}-\frac{1}{k^{\frac{1}{k}}} = \frac{1}{e}-1 \not= 0.$$

Answer 2

Counter-example: Let $h:\mathbb{R}\rightarrow\mathbb{R}$ be any continuous function that satisfies $h(0)=1$ and $h(x)=0$ if $|x|\geq 1$. Then define $g:[0,1]\times \mathbb{R}\rightarrow\mathbb{R}$ by $$\boxed{g(t,x)=\left\{\begin{array}{cc} h(x-1/t) & \mbox{ if $t>0$} \\ 0 & \mbox{else}\end{array}\right.} $$ Then we have for all $t \in [0,1]$: $$ \lim_{x\rightarrow\infty} g(t,x) = 0 $$ $$\lim_{x\rightarrow-\infty} g(t,x)=0$$ Using $t_k=1/k$ and $x_k=k$ for $k \in \{1, 2, 3, ...\}$ we have: $$\lim_{k\rightarrow\infty} g(1/k,k) = h(0)=1$$ It remains to show $g$ is continuous. It suffices to check it is continuous at $(0,x)$ for all $x \in \mathbb{R}$. In case this is a homework question, I leave that as an exercise.