# If every rearrangement of the series converges uniformly then the series converges absolutely uniformly

Let $$I \subset \mathbb{R}$$ and for $$\forall n \in \mathbb{N}: f_n \in C(I, \mathbb{R})$$. Prove that if for any $$\sigma:\mathbb{N} \rightarrow \mathbb{N}$$ bijection, the series $$\sum_n f_{\sigma(n)}$$ converges uniformly on $$I$$, then $$\sum_n |f_n|$$ also converges uniformly on $$I$$. I'm thinking of a proof by contradiction: Let's suppose that the latter series is not uniformly convergent, that is: $$\exists \varepsilon >0:\forall N\in\mathbb{N}:\exists n>N: \exists x \in I:$$ $$\sum_{k=N}^{n}|f_k(x)|\ge \varepsilon$$ Then $$\prod_{N \in \mathbb{N}} \{n|n>N \land \exists x \in I: \sum_{k=N}^{n}|f_k(x)|\ge \varepsilon \}\ne \emptyset$$ but then I stuck, because an element of the latter set is not necessarely a bijection, because it might fail to be injective. My thought was to construct somehow a rearrangement of the series that would not converge uniformly hence a contradiction. I also fail to see where the continuity of the functions $$f_n$$ comes in. I would appreciate any suggestions.

Assume otherwise. Then there exist $$\epsilon > 0$$ such that

$$\forall n \in \mathbb{N}, \quad \exists x \in I \quad \text{s.t.} \quad \sum_{i=n}^{\infty} |f_i(x)| > \epsilon. \tag{*}$$

Now we would like to construct $$\sigma$$ which violates the assumption. To this end, we recursively define the triple $$(A_j, n_j, x_j)_{j=1}^{\infty}$$ as follows:

Construction. Assume that $$(A_j, n_j, x_j)_{j=1}^{k-1}$$ is well-defined so that $$A_j$$'s and $$\{n_j\}$$'s are mutually disjoint. Pick $$n$$ so that it is larger than any elements in $$\bigcup_{j=1}^{k-1}A_j \cup \{n_j \}$$. In the following picture, elements chosen up to the $$(k-1)$$-th stage are represented by black dots.

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By $$\text{(*)}$$, there exists $$x_k \in I$$ such that $$\sum_{i=n}^{\infty} |f_i(x_k)| > \epsilon$$. So, either the sum of positive parts or the sum of negative parts must exceed $$\epsilon/2$$, and in particular, there exists a finite subset $$A_k \subset \mathbb{N} \cap [n, \infty)$$ so that

$$\left| \sum_{i \in A_k} f_i(x_k) \right| > \epsilon / 2. \tag{2}$$

Then pick $$n_k$$ as the smallest element in $$\mathbb{N}\setminus\left(\bigcup_{j=1}^{k-1}A_j \cup \{n_j \} \cup A_k\right)$$. In the following figure, elements of $$A_k$$ are represented by red dots and $$n_k$$ is represented by the blue dot.

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By the construction, it is clear that $$\mathbb{N} = \bigcup_{j=1}^{\infty} A_j \cup \{n_j\}$$. From this, we may define $$\sigma : \mathbb{N} \to \mathbb{N}$$ as the function that enumerates elements in the sets of

$$(A_1, \{n_1\}, A_2, \{n_2\}, \cdots)$$

in order of appearance. In other words, if we regard $$A_k$$'s as ordered lists, then $$\sigma$$ is an infinite ordered list obtained by concatenating $$A_1$$, $$\{n_1\}$$, $$A_2$$, $$\{n_2\}$$, $$\cdots$$. Now, if we write $$N_k = \#\big( \bigcup_{j=1}^{k} A_j \cup \{n_j\} \big)$$, then

$$\sup_{x \in I} \left| \sum_{i = N_{k-1} + 1}^{N_k} f_{\sigma(i)}(x) \right| \geq \left| \sum_{i \in A_k} f_i (x_k) \right| - |f_{n_k}(x_k)| \geq (\epsilon/2) - |f_{n_k}(x_k)|.$$

But it is easy to check that $$f_n \to 0$$ uniformly, and so, it follows that this lower bound is at least as large as $$\epsilon/3$$ for all sufficiently large $$k$$. This proves that partial sums of $$(f_{\sigma(i)})$$ cannot converge uniformly, contradicting the assumption. $$\square$$