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.
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.
