Let $I=[a,b]$ (where $a<b$) be a compact interval on $\mathbb{R}$, $0<\alpha\leq1$.
and $$\mathrm{Lip}(\alpha)=\left\{f:I \to \mathbb{C} \;\bigg|\; M_f=\sup_{s\neq t} \frac{|f(s)-f(t)|}{|s-t|^{\alpha}} < \infty \right\}$$
1) Show that $\mathrm{Lip}(\alpha)$ space is a Banach space with the norm $\|f\|_1=\sup_{t \in I}|f(t)|+M_f$.
2) Show that $\mathrm{Lip}(\alpha)$ space is also a Banach space with the norm $\|f\|_2=|f(a)|+M_f$.
Hint: Show that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent norms on $\mathrm{Lip}(\alpha)$ space.
Thanks.
PD: I must show the first one. So, if someone can explain me this, I'll be great.
Best Answer
So, you want to show that $\operatorname{Lip}(\alpha)$ is a Banach space for the norm $\|\cdot\|_1$.
Let $C[0,1]$ denote the Banach space of continuous complex-valued functions on $[0,1]$, equipped with the sup-norm $\|\cdot\|_\infty$. Then, in particular, for any $f \in \operatorname{Lip}(\alpha)$, $\|f\|_\infty \leq \|f\|_1$.
Now, let $\{f_n\}_{n=1}^\infty$ be a Cauchy sequence in $\operatorname{Lip}(\alpha)$. By the observation above, $\{f_n\}$ is also a Cauchy sequence in $C[0,1]$, and thus converges to some $f \in C[0,1]$ in the sup-norm; it suffices to show that $f \in \operatorname{Lip}(\alpha)$ and that $f_n \to f$ in the norm $\|\cdot\|_1$.
In order to show that $f \in \operatorname{Lip}(\alpha)$, observe that $$ M_f := \sup_{\left|s-t\right| \neq 0} \frac{\left|f(s)-f(t)\right|}{\left|s-t\right|^\alpha} = \sup_{\epsilon > 0} \sup_{\left|s-t\right|=\epsilon} \frac{\left|f(s)-f(t)\right|}{\left|s-t\right|^\alpha} = \sup_{\epsilon > 0} \epsilon^{-\alpha} \sup_{\left|s-t\right|=\epsilon} \left|f(s)-f(t)\right|, $$ so that if you can find an upper bound, independent of $\epsilon$, for $$ M_f^\epsilon := \epsilon^{-\alpha} \sup_{\left|s-t\right|=\epsilon} \left|f(s)-f(t)\right|, $$ you're in business. Roughly, for any given $\epsilon > 0$, you'll want to find a suitable $f_n$ to make applying the triangle inequality for absolute values work out; it will also help to remember that $\{M_{f_n}\}$ is bounded from above, since $\{\|f_n\|_1\}$ is bounded from above (as $\{f_n\}$ is Cauchy in the norm $\|\cdot\|_1$) and $M_{f_n} \leq \|f_n\|_1$ for each $n$.
Once you do know that $f \in \operatorname{Lip}(\alpha)$, you can probably use similar tricks to get that $f_n \to f$ in the norm $\|\cdot\|_1$.