By defining $$ C_0(\mathbb{R}^n):=\{u:u\in C(\mathbb{R}^n),\quad\mathtt{and}\quad\lim_{|x|\rightarrow\infty}u(x)=0\} $$ normed with $||u||:=\sup_{x\in\mathbb{R}^n}|u(x)|$. As far as I can remember, this is a Banach space.
My question: Is this ture or there are counterexamples for this?
Attempts: I think it is easy to give counterexaple for $C_c(\mathbb{R}^n)$ where the condition of "limit at infinitty" is changed into "having compact support". In fact by considering $$ f_n(x):=f_{n-1}(x)+\dfrac{1}{n} \chi_{x\in B_n\setminus B_{n-1}},\quad f_0(x):=0,\ B_0:=\varnothing. $$ which is a Cauchy sequence in $C_c$ but the limit doesn't have compact support. But the limit satisfies $\lim_{|x|\rightarrow\infty}f(x)=0$.
Thanks for reading and any idea will be helpful.
$C_0(\mathbb{R}^n)$ is isometric to the space $\mathbb{E}$ of continuous functions on $S^n$ (one-point compactification of $\mathbb{R}^n$) which vanish on $\infty$. The latter is the kernel under the evaluation map \begin{align*} \mathrm{ev}_{\infty}:C(S^n) &\to \mathbb{R}\\ f &\mapsto f(\infty). \end{align*} Since the evaluation map is continuous, $\mathbb{E}$ is closed, therefore Banach (since $C(S^n)$ is Banach) and it follows that $C_0(\mathbb{R}^n)$ also is.
As was said in the comments, this holds for any locally compact space Hausdorff $X$ in place of $\mathbb{R}^n$. The proof is the same.