# Space of Continuous 2$\pi$-periodic functions is Banach space

I want to proof that the space $(C^0_{2\pi}(\mathbb{R},\mathbb{R}^n), || \cdot||)$, where $|| \cdot||:= \underset{x \in \mathbb R}\max|f(x)|$ is a Banach space. (I showed that $(C^0_{2\pi}(\mathbb{R},\mathbb{R}^n)$ is a normed vector space.) First of all I know that $C^0_{2\pi}(\mathbb{R},\mathbb{R}^n) \subset C^0(\mathbb{R},\mathbb{R}^n)$ and $C^0(\mathbb{R},\mathbb{R}^n)$ is a Banach space. Thus, it is enough to show that $C^0_{2\pi}(\mathbb{R}, \mathbb R^n)$ is a closed subspace. Now I have difficulties to show that for a sequence $(u_n)_{n \in \mathbb N}\subset C^0_{2\pi}(\mathbb{R}, \mathbb R^n)$ the limit $u := \underset{n \to \infty}\lim u_n \in C^0_{2\pi}(\mathbb{R}, \mathbb R^n)$. I guess it is kind of trivial but I am not sure how to proceed. Can I use pointwise convergence of $||u_n||$ i.e. $\underset{n \to \infty}\lim ||u_n|| = \underset{n \to \infty}\lim \underset{x \in \mathbb R}\max|u_n| = \underset{x \in \mathbb R}\max|u|$ and that is it?

Any help is appreciated.

• you must show that if $f_n(x)\to f(x)$ then $f_n(x+2\pi)\to f(x)$, and because $f_n(x)=f_n(x+2\pi)$ for all $n\in\Bbb N$... – Masacroso May 24 '17 at 8:24

Your approach is correct. If $\lim_{n\in\mathbb N}f_n=f$ with respect to your norm then, for each $x\in\mathbb R$, $\lim_{n\in\mathbb N}f_n(x)=f(x)$. Therefore, if every $f_n$ is periodic with period $2\pi$, then $f$ also has that property.