first,

what bruno suggests is "trivially" true (because he proposes a constant sequence) so i think what cp05 needs to prove is that there exists a "non constant" sequence of continuous f_n... bla bla bla

second,

good try cp05, still you need to be sure exactly what and why you are doing the things you are doing :

take f continuous on [0,1].

Define a sequence by \(\displaystyle f_n(x)=\sum_{k=1}^n \frac{f(x)}{2^k}\)

Since f is continuous and f_n is a finite sum of continuous functions then each f_n is continuous on [0,1].

By what i said earlier f is bounded on [0,1] by, lets say, M. So \(\displaystyle |f(x)| \leq M\) \(\displaystyle \forall x\in [0,1]\)

Now lets see the uniform convergence

Given \(\displaystyle \epsilon\) we need to find an adequate N which does not depend on x, only on \(\displaystyle \epsilon\).

Take ANY x\in [0,1]

\(\displaystyle |f(x)-f_n(x)|=|f(x)||1-\sum_{k=1}^n \frac{1}{2^k}|\)

We know that \(\displaystyle \sum_{k=1}^n \frac{1}{2^k}\to 1 \) when \(\displaystyle n\to \infty\)

so for \(\displaystyle \frac{\epsilon}{M} \) there is an N for which

\(\displaystyle |1-\sum_{k=1}^n \frac{1}{2^k}|<\frac{\epsilon}{M} \) if \(\displaystyle n\geq N\)

Using the first equalities we have that for \(\displaystyle n\geq N\)

\(\displaystyle |f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon\)

where we used that f is bounded by M. Take a good look, N does not depend on x, only on M (and on the bound but that is fixed anyway).

If you want to get the supremum way of seing it then think like this:

We just showed that for any \(\displaystyle \epsilon\) there is an\(\displaystyle N(\epsilon)\) such that

\(\displaystyle |f(x)-f_n(x)|<|f(x)|\frac{\epsilon}{M}<\epsilon\) if \(\displaystyle n\geq N\) for any \(\displaystyle x\in [0,1]

\)

Taking supremum over [0,1] grants you that

\(\displaystyle ||f-f_n||_{sup}<\epsilon \) if \(\displaystyle n\geq N\)

i hope i didnt make any mistake