Proof: If a function is in the Schwartz Space, then this function is uniformly continuous

Question

I don't have this really clear, I want to justify if $f\in\mathcal{S}(\mathbb{R})$ then $f$ is uniformly continuous. So far, I know how can I bound $|x|$ for $f$ is in the Schwartz space, but I can't proceed with the uniformly continuous proof because I don't know how to bound $|y-x|$ to find a $\delta$ which depends on an $\varepsilon>0$ such as $|f(y)-f(x)|<\varepsilon$. Thank you so much

Answer 1

I'll prove a result slightly stronger. Consider the space $C_0(\mathbb R)$ of the continuous functions from $\mathbb R$ to $\mathbb C$ such that $\lim_{|x|\to \infty} f(x) = 0$. The functions from this space are uniformly continuous.

Consider $\epsilon >0$ and $f\in C_0(\mathbb R)$. There exist, from the limit above, a radius $R>0$ such that $|f(x)|<\epsilon/2$ when $x\in \mathbb R$ and $|x|>R$. (Therefore, $f$ is small far from the point 0)

Now, let $B=\{x\in \mathbb R:|x|\leq R+1\}$. Since $B$ is compact and $f|_B:B \to \mathbb C$ is continuous we have that $f|_B$ is uniformly continuous (On compact spaces, to be continuous = to be uniformly continuous). So, there exist $\delta\in (0,1)$ such that $|f(x)-f(y)|<\epsilon/2$ for every $x,y\in B$ where $|x-y|<\delta$. (Therefore, $f$ is uniformly continuous near 0)

If $x,y\in \mathbb R$ and $|x-y|<\delta$, we have the following cases:

1. If $x,y \in B$ then $|f(x)-f(y)|<\epsilon/2<\epsilon,$
2. If $x\not \in B$ then $|x|>R+1$ and therefore $|y|>R$, because $|x-y|<\delta<1$. So we obtain $|f(x)|<\epsilon/2$ and $|f(y)|<\epsilon/2$. Therefore $$|f(x)-f(y)|\leq|f(x)|+|f(y)|<\epsilon.$$

From 1. and 2. we conclude that $|f(x)-f(y)|<\epsilon$ when $|x-y|<\delta$.

Now, observe that $S(\mathbb R) \subset C_0(\mathbb R)$. Indeed, if $f\in S(\mathbb R)$ then $(1+|x|)f(x)$ is bounded, that is, there is $B>0$ such that $(1+|x|)|f(x)|<B$ for all $x\in \mathbb R$. Observe that $$ |f(x)|<\frac{B}{1+|x|}$$ and as consequence $\lim_{|x|\to \infty}f(x) = 0$. Therefore $f\in C_0(\mathbb R)$.

Then, every $f\in S(\mathbb R)$ is uniformly continuous.