The folloing question was asked in my analysis quiz and was hard nut to crack for me even after exam.
Suppose $\{f_n\}$ is a sequence of continuous real valued functions on $[0,1]$ satisfying :
(i) for each $x$ in reals $\{f_n(x)\}$ is decreasing sequence.
(ii) the sequence ${f_n}$ converges uniformly to 0.
Then let $g_n(x)=\sum_{k=1}^n (-1)^k f_k(x)$ for all $x\in \mathbb{R}$. Now, which of the following is correct:
(i) $\{g_n\}$ is Cauchy wrt sup norm.
(ii) $\{g_n\}$ is uniformly convergent.
(iii) $\{g_n\}$ need not converge pointwise.
(iv) $\{g_n\}$ is uniformly bounded.
I tried proving i and ii option but couldn't do using required definitions and gave a lot of time.
So, please tell how should I approach the question. I am stuck.
First, for each $x \in [0,1]$, the series defining $g_n$ is an alternating series, so the special criteria for alternating series tells that the series converges. So $(g_n)$ converges pointwise, so the option $(iii)$ is not true. Let $g$ be the limit pointwise of the sequence $(g_n)$.
For every $x \in [0,1]$, you have $$|g(x)-g_n(x)| = \left|\sum_{k=n+1}^{+\infty} (-1)^k f_k(x) \right| \leq f_{n+1}(x)$$
again by the criteria for alternating series (criteria on the remainders). So $$||g(x)-g_n(x)||_{\infty} \leq ||f_{n+1}||_{\infty}$$
so because $(f_n)$ converges uniformly to $0$, you deduce that $||g(x)-g_n(x)||_{\infty}$ converges to $0$, i.e. $(g_n)$ converges uniformly to $g$. In particular, you deduce that $(g_n)$ is Cauchy for $||.||_{\infty}$, and is bounded for $||.||_{\infty}$. So options $(i)$, $(ii)$ and $(iv)$ are correct.