# 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 $$f \in U$$

• (I made a mistake in my first comment, here is the ediited version): The "first step" is an ambiguous term. Please clarify what you mean. I understand that you want to prove that uniform boundedness of the partial sums is equivalent to uniform convergence of the Fourier series. You can use convergence in a "good" dense subspace, plus the uniform condition. – Adrián González-Pérez Jan 21 at 12:48
• Thanks for your answer. I tried the method of "finding" a suitable limit for a sequence $f_j$ but as you did it's better to prove that absolute convergence of a series in U implies simple convergence and this proves that U is indeed a Banach. – Psylex Jan 22 at 7:31

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.