I have a sequence of continuous functions that, $f_{J}: [0,1] \rightarrow \mathbb{R}, J=1,2,3, \ldots$ converges to a function $f$. $f_{J}$ is continuous function and is piecewise linear on each interval in $\mathcal{P}^{J}$ (i.e. the powerset) and therefore completely defined by its values in $X^{J}$. First of all, it is defined that $f_{J+1}(t)=f_{J}(t)$ $\forall$ $t \in X^{J}$, where set $X^{J}=\{t_{0},t_{1},\ldots,t_{N}\}=\{0,2^{-J},2\times 2^{-J},3\times 2^{-J} \ldots,1\}$. For $ t=\left(2k-1 \right) 2^{-\left(J+1\right)} \in X^{J+1}\setminus X^{J}, k=1,\ldots,2^{J}.$ It is also defined that, $$f_{J+1}(t)=f_{J}(t)+\left( -1\right)^{k+J+1}2^{-J/(2-1)}$$
This definition of $f_{J+1}$ is extended to the whole of [0,1].
I need to prove that the sequence $f_{1},f_{2},\ldots$ converges uniformly to a continuous function, $f: [0,1] \rightarrow \mathbb{R}$, by showing that is it is a Cauchy sequence in the sup-norm.
$\textbf{My approach:}$
If $X$ is a complete metric space, then every Cauchy sequence $\{x_{n}\} \subset X$ converges to a limit $x$ since the function provided is complete with respect to the metric induced by the sup-norm. So, all I want to do is show that this is a Cauchy and by completeness it has to converge. I am not sure how can I go about proving it is a Cauchy, and that is what I need help with.
Best Answer
A sequence {$f_n$} is said to be Cauchy if for every $\epsilon >0$ there is an integer $N$ such that for $n,m \in \Bbb N$ then $|f_n - f_m| < \epsilon$ when $n>m>N$
So we know that $f_J \to f$
Then by the triangle inequality $|f_n - f_m| \le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < \frac{\epsilon }{2}$ where $\frac{\epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| \le |f_n - f| + |f_m - f| < \frac{\epsilon }{2} + \frac{\epsilon }{2} \to |f_n - f_m| < \epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < \frac{\epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $\frac{\epsilon }{2}$ which will show that it too converges.