Let $g_n(x)=\sum_{k=1}^n (-1)^k f_k(x) \forall x\in \mathbb R.$ Then which one of the following are correct answers?

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.

• The signs in $|f_m(x)+f_{m+1}(x)+...+f_n(x)|$ should be alternating. (So, the corrected sum is at most $|f_m(x)|)$. – David Mitra Apr 22 at 7:36
• 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. – Dbchatto67 Apr 22 at 7:40

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 $$M=1+\sup_{x\in[0,1]}\left\lvert\sum_{k=1}^\infty(-1)^kf_k(x)\right\rvert$$.
• Here, $a_n(x)=(-1)^n$. right? – Unknown x Apr 22 at 9:01
• Yes, that is our $a_n$ in this context. – José Carlos Santos Apr 22 at 9:04