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Suppose that $\{f_n\}$ is a sequence of continuous real-valued functions on $[0,1]$ satisfying the following:

(A)$\forall x\in \mathbb R,\{f_n(x)\}$ is a decreasing sequence.

(B)the sequence $\{f_n\}$ converges to uniformly to $0$.

Let $g_n(x)=\sum_{k=1}^n (-1)^k f_k(x) \forall x\in \mathbb R. $ Then

(1)$\{g_n\}$ is cauchy with respect to the sup norm.

(2)$\{g_n\}$ is uniformly convergent.

(3) $\{g_n\}$ need not be pointwise convergent.

(4)$\exists M>0$ such that $|g_n(x)|\leq M, \forall n\in \mathbb N, \forall x\in \mathbb R$.

My attempt:- I know that (1),(2),and(4) are the correct answers. From (A), we get $x\in [0,1].$ $$...f_{n+1}(x)\leq f_{n}(x)\leq f_{n-1}(x)\leq...f_{2}(x)\leq f_1(x)$$ From(B), we get For a given positive real number $\epsilon$ there is a nutural number ($N=N(\epsilon)$) depends only on the chosen $\epsilon:\forall n\geq N(\epsilon)\implies |f_n(x)|<\epsilon(\forall x\in [0,1])...................(1)$

My aim is to prove that $\{g_n\}$ is Cauchy. We are choosing $\epsilon \in \mathbb R^+$. Enough to prove that there is a Natural number $N\in \mathbb N: \forall n>m\ge N\implies ||g_m-g_n||_{\infty}<\epsilon.$ $|g_n(x)-g_m(x)|\leq|f_m(x)+f_{m+1}(x)+...+f_n(x)| $ I am not able to proceed further.

When I took $f_n(x)=\frac{x^n}{n!}$, We can eliminate(3). Hence (3) is false. Please help me to conclude the answer.

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    $\begingroup$ The signs in $|f_m(x)+f_{m+1}(x)+...+f_n(x)| $ should be alternating. (So, the corrected sum is at most $|f_m(x)|)$. $\endgroup$ – David Mitra Apr 22 at 7:36
  • $\begingroup$ You have defined $f_n$'s on $[0,1].$ Then in $(A)$ suddenly you extend the domains of $f_n$'s to $\Bbb R.$ How is that possible? You have messed up $\Bbb R$ and $[0,1]$ in the entire question. $\endgroup$ – Dbchatto67 Apr 22 at 7:40
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Dirchlet's test tells us that $(g_n)_{n\in\mathbb N}$ converges uniformly. Therefore, (1) and (2) hold and (3) is false. On the other hand, since each $g_n$ is continuous and the convergence is uniform, $\sum_{k=1}^\infty(-1)^kf_k$ is a continuous function. Since its domain is $[0,1]$, its image is bounded. And, since $(g_n)_{n\in\mathbb N}$ converges uniformly to $\sum_{k=1}^\infty(-1)^kf_k$, (4) holds; just take $M=1+\sup_{x\in[0,1]}\left\lvert\sum_{k=1}^\infty(-1)^kf_k(x)\right\rvert$.

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  • $\begingroup$ Here, $a_n(x)=(-1)^n$. right? $\endgroup$ – Unknown x Apr 22 at 9:01
  • $\begingroup$ Yes, that is our $a_n$ in this context. $\endgroup$ – José Carlos Santos Apr 22 at 9:04

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