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$

  • $\begingroup$ (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. $\endgroup$ – Adrián González-Pérez Jan 21 '19 at 12:48
  • $\begingroup$ 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. $\endgroup$ – Psylex Jan 22 '19 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.


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