# Easy way to prove Weierstrass M - test

I know that $(C(X \rightarrow \mathbb R),d_\infty)$ is a complete space with $$d_\infty ( f,g ) := \sup_{x \in X} |f(x)-g(x)|$$

Let $(f_n)_{n=0}^\infty$ be a sequence in $C(X \rightarrow \mathbb R)$ such that $M = \sum_{n=0}^\infty \sup_{x \in X} |f(x)|$ converges. Defining $\|f\| := \sup_{x \in X} |f(x)|$ we have also that $(C(X \rightarrow \mathbb R),\|\cdot\|)$ is a Banachspace with the usual definitions for addition and scalar multiplication for functions. I want to show that $(f_n)_{n=0}^\infty$ uniformly converges to a function $f$. My attempt:

Assume $N > M$. $$\left \| \sum_{n=1}^N f_n - \sum_{n=1}^M f_n \right \| = \left \| \sum_{n=M+1}^N f_n \right \| \leq \sum_{n=M+1}^N \|f_n\|$$ $$= \left | \sum_{n=0}^N \|f_n\| - \sum_{n=0}^M \|f_n\| \right | \leq \left | M - \sum_{n=0}^N \|f_n\| \right | + \left | M - \sum_{n=0}^M \|f_n\| \right|$$

By convergence of $\sum \|f_n\|$ we can conclude that the last two terms get small for $N,M$ large enough. Thus $$\left( \sum_{n=0}^N f_n \right )_{N=0}^\infty$$ is a Cauchysequence in $(C(X\rightarrow \mathbb R),\|\cdot\|)$ which proves convergens of the $\sum_n f_n$ to some function. But convergence of functions in $\sup$-norm is equivalent with uniform convergence, i.e. $\|f-g\| = d_\infty(f,g)$.

Is this an easy (and correct) way to prove it ?

We also see that $\sum_{n=0}^N f_n$ is an element of $C(X \rightarrow \mathbb R)$ for all $N$

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Are you supposed to show that $\sum_n f_n$ converges, or that $f_n$ converges? If the former, it seems as though you've proven it. If the latter, what do you think happens to $\|f_n\|$ as $n \rightarrow \infty$ if $\sum_n \|f_n\| < \infty$? –  A Blumenthal Jan 21 '13 at 22:05
Indeed my intention was to prove uniform convergence of the sum. And by the zero test we have that $\|f_n\| \rightarrow 0$ as $n \rightarrow \infty$. And I guess that this also happens uniformly –  Andre Jan 21 '13 at 22:08