Why is ( U, $ \Vert . \Vert $ ) a Banach Space? The space of uniformly convergent Fourier series Consider $ U = \{ f \in \mathcal{C}_{2\pi}, \: S_n(f) \: $converges uniformly on $ \mathbb{R} \} $  with  $  \Vert f \Vert = $sup$_n \Vert S_n (f) \Vert_{\infty } $ 
Show that $ (U, \Vert  .\Vert ) $ is a Banach Space. 
I am not sure how to go about the first step : how does f has an existing limit, and if f is equal to its Fourier series or not. Thanks for any help on this.
$ \forall \epsilon > 0 \:  \exists N > 0, \: \forall p,q > N \: $sup$_n \Vert S_n (f_p - f_q) \Vert_{\infty }  < \epsilon $ 
How to show $ \exists f, \: f_n \rightarrow f $ ?
Then after we need to prove $ f \in U$
 A: First, we are going to see that the condition
$$ 
\| f \|_U := \sup_N \| S_N(f) \|_\infty < \infty \tag{U}
$$
is equivalent to the fact that $S_N(f)$ converge to $f$ uniformly. First, if $S_N(f) \to f$ uniformly, then, by continuity of the suppremum norm $\|S_N(f)\|_\infty \to \|f\|_\infty$ and therefore the sequence $(\| S_N(f) \|_\infty)_N$ is bounded. For the other direction we need to use that, for every $\epsilon > 0$ and function $f$ satisfying (U), there is a $g  \in C[0,2\pi]$ such that


*

*$\quad \displaystyle{\sup_N \| S_N(g -f)\|_\infty  < \epsilon}$

*$\quad S_N(g) \to g$ in the $\| \|_\infty$-norm. 


Think of $g$ as a Schwartz class function approximating $f$. Notice also that, for every continuous function $h$, we have that
$$
  \| h \|_\infty \leq \sup_N \|S_N(h) \|_\infty.
$$
Indeed, that holds for every trig. polynomial and you can extend by the Stone-Wierstrass theorem. The identity above implies that $\|f-g\|_\infty < \epsilon$.
Now, for every $f$ satisfying (U), you can pick $g_\epsilon$ satisfying the two properties above, therefore:
\begin{eqnarray}
  & & \|S_N(f) - f \|_\infty \\
  & \leq & \| S_N(f - g_\epsilon) \|_\infty + \| S_N(g_\epsilon) - g_\epsilon \|_\infty + \| g_\epsilon - f \|_\infty\\
  & \leq & \epsilon + \| S_N(g_\epsilon) - g_\epsilon \|_\infty + \epsilon \to 2 \, \epsilon.
\end{eqnarray}
Therefore the limit of the quantity above is smaller that $ 2  \epsilon$, but since $\epsilon > 0$ is arbitrary we have that the limit is $0$.
Obtaining that $U$ is a Banach space is a corollary, since for every sequence $f_j$ satisfying that
$$
  \sum_{j=1} \| f_j \|_U < \infty
$$
it also holds that the sum above converge uniformly. Therefore there is a limit $f = \sum_{j} f_j$. And checking that $f \in U$ is a trivial calculation.
