Given:
$f_1(x)=x$ if $x\le1/2$
$f_1(x)=1-x$ if $1/2\le x\le1$
$f_1(x+1)=f_1(x)$
$\forall n\ge2,f_n(x)=(1/2)*f_{n-1}(2x)$
Let $S_m(x)=\sum_{n=1}^m f_n(x)$
$S_m$ is a continuous function on $[0,\infty]$
Task:
Show that the sequence $(S_m)_{m \epsilonℕ}$ converges uniformly to a continuous function S.
My question:
I'm sorry, I don't really have one. I'm just totally clueless. I know I'm supposed to show the work that I've done up to this point, but I do not know how to work with uniform convergence. We were introduced to it it a few months ago, but it didn't make sense to me then, and it doesn't make sense to me now.
Here's my given definition for uniform convergence: Let $(f_n)_{n\epsilonℕ}$ be a sequence of functions in the real numbers. The sequence is said to converge uniformly to a function $f$ provided that, given $\epsilon\gt0$, there exists $N_\epsilon \in ℕ$ such that $sup_{x\in X}|f_n(x)-f(x)|\lt\epsilon$ for $n\ge N_\epsilon$
Note that the relation $f_n(x)=2^{-1}f_{n-1}(2x)$ implies that for every $n$, $f_n(x)=2^{1-n}f_1(2^{n-1}x)$. Now, note that for every real $x$, we have $\vert f_1(x)\vert \leq 1$. Hence, we obtain the relation $0\leq f_n(x)\leq 2^{1-n}$ for every $x$ (the first inequality follows by the definition of $f_n$). We can apply then the Weierstrass M-test to show that the sequence of functions $(S_n)_{n\in\mathbb{N}}$ converges uniformly in $\mathbb{R}$. Since all the $S_n$ are continuous the limit function (say $S$) is continuous.
Edit: A proof that doesn't use explicitly the M-test is as follows: We know that the limit function exist for every $x$. Now, note that $$ \vert S_n(x)-S(x)\vert=\sum_{m=n}^\infty f_m(x)\leq \sum_{m=n}^\infty 2^{1-m}= 2^{2-n} $$ Since this is valid for every $x$, then we must have that $\sup\limits_{x\in\mathbb{R}}\vert S_n(x)-S(x)\vert\leq 2^{2-n}$ Then, the supremun can be made as small as desired taking $n$ large enough. This implies that the convergence is uniform