I need to prove the following,
Let $f_n$ be a sequence of continuous functions defined in R. Suppose that $|f_{n+1}(x)−
f_n(x)| ≤ \frac {n^2}{1+2^n}$
for all x ∈ R and for all $n ≥ 1$.
Prove that $f_n$ converges uniformly to some continuous
function $f(x)$ defined in R.
Right, I am trying to prove that it is Uniformly Cauchy and this is what I have done until now:
Let $f_m$ be a sequence of continuous functions on R such that for all $m, m>n$, and let $\varepsilon >0$
then for
$$|f_m(x)-f_n(x)|$$
$$=|f_m(x)-f_{m-1}(x)+…+f_{n¬1}(x)-f_n(x)|$$
$$\leq |f_m(x)-f_{m-1}(x)|+…+|f_{n+1}(x)-f_n(x)| $$
$$\leq |f_m(x)-f_{m-1}(x)|+…+ \frac {n^2}{1+2^n}$$
I don't know how to continue from there. What value should $\varepsilon$ take such that there exist some N belonging to natural numbers $|f_m(x)-f_n(x)|< \varepsilon$?
Best Answer
First see if we can get an upper bound that is easier to manipulate.
Let $a_n = {n^2 \over 1+2^n}$, then ${a_{n+1} \over a_n} = (1+{1 \over n})^2 {2^{-n} + 1 \over 2^{-n} + 2}$ and $\lim_n {a_{n+1} \over a_n} = {1 \over 2}$.
So there is some $N$ such that for $n \ge N$ we have ${a_{n+1} \over a_n} \le {3 \over 4}$ and so $a_n \le ({ 3 \over 4})^{n-N} a_N$.
Then for $n \ge N$ you have $|f_{n+1}(x) - f_n(x)| \le ({ 3 \over 4})^{n} a_N ({ 3 \over 4})^{-N} $.
Then $|f_{n+p}(x)-f_n(x)| \le \sum_{k=0}^{p-1} ({ 3 \over 4})^{n} a_N ({ 3 \over 4})^{-N} \le (4 a_N ({ 3 \over 4})^{-N}) ({ 3 \over 4})^{n}$.
It follows from this that $f_n(x)$ is Cauchy and hence converges to some $f(x)$. Letting $p \to \infty$, we get $|f(x)-f_n(x)| \le (4 a_N ({ 3 \over 4})^{-N}) ({ 3 \over 4})^{n}$.
Continuity then follows from uniform convergence.