$Lip(\alpha)$ is a Banach space with respect to a certain norm In this link, How can I show that it's a Banach space?, the question is to show that Lip($\alpha$) is a Banach space under two different norms. The solutions provided show that this is the case for the first norm. 
I am wondering if there is a way to directly show it for the second norm, without proving the equivalence of the two norms. I am having a good deal of trouble with this for some reason. 
 A: Let $\{ f_n \}_n$ be a Cauchy sequence in Lip$(\alpha)$ with respect to the norm
\begin{align*}
\| f \|_2 &:= |f(a)| + M_f \\
&:= |f(a)| + \sup_{t \neq s} \frac{|f(s) - f(t)|}{|s -t|^\alpha}.
\end{align*}
We wish to show that there exists a function $f \in $Lip$(\alpha)$ such that 
$\lim_n \| f_n - f \|_2 = 0$. We first note that the sequence of real numbers 
$\{ f_n(a) \}_n$ is a convergent sequence.  Indeed, this follows immediately from the estimate 
\begin{align*}
|f_n(a) - f_m(a)| \leq |f_n(a) - f_m(a)| + M_{f_n - f_m} = \| f_n - f_m \|_2,
\end{align*}
the assumption that $\{ f_n \}_n$ is a Cauchy sequence in Lip$(\alpha)$, and 
the fact that $\mathbb R$ is complete.  We now claim that $\{ f_n(x) \}_n$ is a convergent sequence in $\mathbb R$ for every $x \in [a,b]$.  Indeed, this follows for $x = a$ by what we have already done, while for $x \neq a$, we have by the triangle inequality
\begin{align*}
|f_n(x) - f_m(x)| &\leq |(f_n(x) - f_m(x)) - (f_n(a) - f_m(a))| + |f_n(a) - f_m(a)| \\
&\leq |x -a|^\alpha M_{f_n - f_m} + |f_n(a) - f_m(a)| \\
&\leq (|x - a|^\alpha + 1) \| f_n - f_m \|_2.
\end{align*}
This, the Cauchy assumption on $\{ f_n \}_n$ in Lip$(\alpha)$, and the completeness of $\mathbb R$ yield the desired claim.  Since the sequence $\{ f_n(x) \}_n$ converges for every $x \in [a,b]$, we may define a function $f:[a,b] \rightarrow \mathbb R$ via 
$$
f(x) := \lim_{n \rightarrow \infty} f_n(x). 
$$
We now claim that $f \in$Lip$(\alpha)$.  To see this, observe that since $\{ f_n \}_n$ is Cauchy in Lip$(\alpha)$, we have that this sequence is bounded in Lip$(\alpha)$, i.e.
$\exists R > 0$ such that 
$$
\| f_n \|_2 \leq R, \quad \forall n \geq 1.
$$ 
By our definition of $f$, we thus have for $s \neq t$ 
\begin{align*}
\frac{|f(s) - f(t)|}{|s-t|^\alpha} &= \lim_{n \rightarrow \infty}
\frac{|f_n(s) - f_n(t)|}{|s-t|^\alpha} \\ 
&\leq \sup_{n \geq 1} M_{f_n} \\
&\leq \sup_{n \geq 1} \| f_n \|_2 \leq R. 
\end{align*}
Thus, $M_f < \infty$, so that $f \in$ Lip$(\alpha)$.  Finally, we show that 
$\lim_n \| f_n - f \|_2 = 0$.  Let $\epsilon > 0$.  Since $\{ f_n \}_n$ is Cauchy in Lip$(\alpha)$, there exists $N \geq 1$ such that 
$$
|f_n(a) - f_m(a)| + M_{f_n - f_m} < \epsilon, \quad n,m \geq N.
$$
Let $s \neq t$.  Then the previous implies that 
$$
|f_n(a) - f_m(a)| + \frac{|(f_n(s)- f_m(s)) - (f_n(t)-f_m(t))|}{|s-t|^\alpha} < \epsilon, \quad n,m \geq N.
$$
Letting $m \rightarrow \infty$ in the previous inequality implies that 
$$
|f_n(a) - f(a)| + \frac{|(f_n(s) - f(s)) - (f_n(t) - f(t))|}{|s-t|^\alpha} \leq \epsilon, \quad n \geq N.
$$
Taking the supremum over all $s \neq t$ in the previous expression yields
$\| f_n - f \|_2 \leq \epsilon$ for all $n \geq N$.  This completes the proof.
