Check if the norm space is complete

Question

Question Let $X$ consist of all real values functions $f$ on $[0,1]$ such that

1. $f(0)=0$,
2. $\|f\|=\sup \left\{\frac{|f(x)-f(y)|}{|x-y|^{1/3}}:x\neq y\right\}$ is finite.

Prove that $\|\cdot\|$ is a complete norm on $X$.

---

It is easy to show that $\|\cdot\|$ is indeed a norm. Also, if $(f_n)$ is a Cauchy sequence in $X$, then $f_n\to f$ uniformly for some $f$. However, I am stumped at showing that $f\in X$ and $\|f_n-f\|\to 0$ as $n\to \infty$.

Answer 1

$(\|f_n\|)$ is a Cauchy sequence of real numbers, so it is bounded. Let $M$ be an upper bound. Then $|f_n(x)-f_n(y)| \leq M|x-y|^{1/3}$ for all $x,y$. Letting g $n \to \infty$ shows that $f \in X$. Use $\|f_n-f_m||<\epsilon$ for $n,m$ sufficiently large and let $m \to \infty$ to see that $\|f_n-f\| \to 0$.