In the middle of a proof I came across the following statement that I was not able to prove.
Let $\epsilon>0$, and consider the function $f(x,t) = |e^{-cx^2t}-1|$ for $x\in\mathbb{R},t>0$, where $c>0$ is a real constant. Then for all small t we have $\sup_{x\in[a,b]} |e^{-cx^2t}-1|<\epsilon$.
Maybe I'm confused because I don't have much background on analysis on $\mathbb R^n$. But I started thinking on it in an "uniform limit" sense, i.e., I should show that for every $\epsilon>0$ there exists $\delta>0$ (not dependant on $x$) such that $|t|<\delta$ implies $|e^{-cx^2t}-1|<\epsilon$ for all $x\in[a,b]$. I know that for fixed $x$, $f(x,t)\to0$ as $t\to0$ but I'm struggling to show that this is uniform in the way I mentioned. And after proving this I would just take the sup $x\in[a,b]$.
Additionally, does the statement hold if I take the sup of $x$ in a subset of $\mathbb R$ that isn't compact (e.g. $(a,b)$ or all of $\mathbb R$)? Does it hold as a consequence of $f$ being uniformly continuous on a compact?
Thanks in advance.
For fixed $c>0$ and $t>0$, the function $x \mapsto e^{-cx^2 t}$ has a maximum value of $1$ at $x=0$, and decreases symmetrically as $|x| \to \infty$. Thus, $$\sup_{x \in [a, b]} |e^{-cx^2 t} - 1| = \max\left\{1-e^{-ca^2 t}, 1-e^{-cb^2 t}\right\}.$$
If $0 < t < \delta$, then the above is further bounded by $$\max\left\{1-e^{-ca^2 \delta}, 1-e^{-cb^2 \delta}\right\}.$$ Thus, in principle you can choose $\delta$ small enough to make the above smaller than $\epsilon$.