Let $X=\left\{ f:[0,1]\to\mathbb R \;\big|\; \lVert f\rVert_X<\infty,\ f(0)=0\right\}$. Show $(X,\lVert\cdot\rVert_X)$ is complete.

Question

The following is a problem on an old Analysis preliminary exam at my institution; I'm prepping for the prelim. The problem is:

Let $$X=\left\{ f:[0,1]\to\mathbb R \;\big|\; \lVert f\rVert_X<\infty,\ f(0)=0\right\}, \quad \lVert\cdot\rVert_X =\sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|^{1/2}}.$$ Show $\left(X,\lVert\cdot\rVert_X\right)$ is complete.

For earlier parts of this problem, I have already shown that $X\subseteq \mathcal C[0,1]$, and that $\lVert\cdot\rVert_X$ is indeed a norm on $X$. We have that $\left(\mathcal C[0,1],\lVert\cdot\rVert_\infty\right)$ is complete, which should be helpful, though I can't see how exactly to use it here.

Answer 1

Hint regarding the usefulness of the completeness of $C[0,1]$:

Say $f_n$ is Cauchy in $X$. You can show that in general $||f||_\infty\le ||f||_X$, hence $f_n$ is also Cauchy in $C[0,1]$. So you have $f$ with $||f_n-f||_\infty\to0$. That does not show by itself that $||f_n-f||_X\to0$, but it does give you a handle on things - at least now you have the limit, you just have to show the sequence actually converges to $f$ in $X$.