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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.

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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.

$\hspace{5em}$ Step 1

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.

$\hspace{5em}$ Step 2


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. $\square$

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