How can I show this function sequence is Cauchy in $C([-1,1])$ with $||f|| = \max|f(x)|$?

Question

The function sequence is

\begin{equation} f_n(x)=x^{1+\frac{1}{2n-1}} \end{equation}

I know that, in order to prove it, I need to show that given $\varepsilon > 0, \exists N \in \mathbb{N}$ such that $||f_m -f_n|| < \varepsilon, \ \forall m, n \geq N$.

I've shown that

\begin{equation} ||f_m -f_n|| = \max_{x \in[-1,1]} |f_n(x)-f_m(x)| = \max_{x \in [0,1]} |x^{1+\frac{1}{2n-1}}-x^{1+\frac{1}{2m-1}}| \leq \max_{x \in [0,1]} |x^{\frac{1}{2n-1}}-x^{\frac{1}{2m-1}}| \end{equation}

and, from what it looks like in Desmos, the $\max$ of that function decreases as $N$ increases, but I don't know how to prove it. I've tried to work out the derivative of $g_{n,m}(x) = |x^{\frac{1}{2n-1}}-x^{\frac{1}{2m-1}}|$ and find the zero of that and see how it decreases with $N$, but there must be an easier way...

Answer 1

It is easier to prove that $(f_n)_{n\in\Bbb N}$ converges to some function in $f\in C([-1,1])$. And it does. Just take $f(x)=|x|$ and use the fact that$$f_n(x)=x^{\frac{2n}{2n-1}}=|x|^{\frac{2n}{2n-1}}=|x|^{1+\frac1{2n-1}}.$$So, since the sequence $(f_n)_{n\in\Bbb N}$ converges, it is a Cauchy sequence.

Answer 2

Hints: It is a mistake to drop the factor of $x$. Note that $|f_n(x)-f_m(x)| <2\epsilon $ if $|x| <\epsilon$. Observe also that $f_n(-x)=f_n(x)$ so we are left with the case $x \geq \epsilon$. Now $|x^{1/(2n-1)}-x^{1/(2m-1)}|\leq |x^{1/(2n-1)}-1|+|1-x^{1/(2m-1)}|$ and the inequality $e^{-t} \geq 1 -t$ show that $|x^{1/(2n-1)}-1| \leq \frac 1 {2n-1} \epsilon^{1/{2n-1}} \to 0 $ (and similarly for the second term).