How can I show that it's a Banach space? 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. 
 A: 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 $\|\cdot\|_1$.
A: You need to show that if $f_n$ is a Cauchy sequence with respect to $\|f\|_1=\sup_{t \in I}|f(t)|+M_f$ then it converges to some $f$ in $Lip(\alpha)$. All proofs that something is a Banach space have the same form: 
First find $f$. Then show $f$ is in the space. Finally show convergence in norm $f_n \to f$.
I think you have a typo in your definition and $Lip(\alpha)$ should be 
$$\mathrm{Lip}(\alpha)=\left\{f:I \to \mathbb{C} \;\bigg|\; M_f=\sup_{s\neq t} \frac{|f(s)-f(t)|}{|s-t|^{\alpha}} \color{red}{< \infty } \right\}$$
Assume $f_n$ is Cauchy, let $\varepsilon > 0$ and let $N$ be s.t. $\|f_n - f_m\|_1 < \varepsilon$. Then for $t_0 \in I$: $|f_n(t_0) - f_m (t_0)| \le \|f_n - f_m\|_1 < \varepsilon$. Since $\mathbb C$ is complete, $f(t_0) = \lim_{n \to \infty} f_n (t_0) \in \mathbb C$. Let $f$ be the pointwise limit at every point. We now have found $f$. It remains for you to show that $f$ is in $Lip (\alpha)$ and that convergence happens in norm.
A: Clearly $\mathrm{Lip}(\alpha)$ is a vector space and $\left\|\:\right\|_{1}$ and  $\left\|\:\right\|_{2}$ are norms, attention should be centered on showing that this space is complete with these norms.
1). Let $\left\{f_{n}\right\}_{n\in\mathbb{N}}\subset \mathrm{Lip}(\alpha)$ be a Cauchy sequence with the norm $\left\|\:\right\|_{1}$, then for all $\varepsilon>0$ there is $N$ such that if $n,m\geq N$ then 
$$\sup_{t\in I }\left|f_{n}(t)-f_{m}(t)\right|+M_{f_{n}-f_{m}}=\left\|f_{n}-f_{m}\right\|< \frac{\varepsilon}{2}.$$
In particular, $\left|f_{n}(t)-f_{m}(t)\right|<\frac{\varepsilon}{2}$ for all $t\in I$, then $(f_{n}(t))$ is Cauchy sequence in  $\mathbb{C}$, then there exists $f(t)$ such that $\lim_{n\rightarrow \infty }f_{n}(t)=f(t)$.
We must show that $\left\|f_{n}-f\right\|_{1}\rightarrow 0$ and $f\in \mathrm{Lip}(\alpha)$. In fact, we knaw that 
$$\left|f_{n}(t)-f_{m}(t)\right|<\frac{\varepsilon}{2} \qquad \forall t\in I$$
then taking limit $m\rightarrow \infty$ we have
$$\left|f_{n}(t)-f(t)\right|<\frac{\varepsilon}{2} \qquad \forall t\in I$$
Therefore
$$\sup_{t \in I}\left|f_{n}(t)-f(t)\right|<\frac{\varepsilon}{2}. \tag{*}$$
For other hand, we also have $M_{f_{n}-f_{m}}<\frac{\varepsilon}{2}$, then 
$$\frac{\left|f_{n}(s)-f_{m}(s)-f_{n}(t)+f_{m}(t)\right|}{\left|s-t\right|}<\frac{\varepsilon}{2} \qquad \forall t\neq s$$
then taking limit $m\rightarrow \infty$ we have
$$\frac{\left|f_{n}(s)-f(s)-f_{n}(t)+f(t)\right|}{\left|s-t\right|}<\frac{\varepsilon}{2} \qquad \forall t\neq s$$
Therefore   $M_{f_{n}-f}<\frac{\varepsilon}{2}$. Hence, by (*) we have 
$$\left\|f_{n}-f\right\|_{1}=\sup_{t \in I}\left|f_{n}(t)-f(t)\right|+M_{f_{n}-f}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$
Note that $f\in \mathrm{Lip}(\alpha)$, in fact, for all $t\neq s\in I$
$$\begin{array}{rcl}\frac{\left|f(s)-f(t)\right|}{\left|s-t\right|^{\alpha}}&=&\frac{\left|f(s)-f_{n}(s)-f(t)+f_{n}(t)+f_{n}(s)-f_{n}(t)\right|}{\left|s-t\right|^{\alpha}}\\
&\leq& \frac{\left|f(s)-f_{n}(s)-f(t)+f_{n}(t)\right|}{\left|s-t\right|^{\alpha}}+\frac{\left|f_{n}(s)-f_{n}(t)\right|}{\left|s-t\right|^{\alpha}}.
 \end{array} $$
Therefore
$$M_{f}\leq M_{f_{n}-f}+M_{f_{n}}<\frac{\varepsilon}{2}+M_{f_{n}}<\infty.$$
2).  We will show that $\left\|\:\right\|_{1}$ and $\left\|\:\right\|_{2}$ are equivalent norms on $\mathrm{Lip}(\alpha)$ space, in fact, on the one hand it is clear that
$$ \left\|f\right\|_{2}=\left|f(a)\right|+M_{f} \leq \sup_{t\in I}\left|f(t)\right|+M_{f}= \left\|f\right\|_{1}. \tag{**}$$
For other hand, note that
$$\begin{array}{rcl} 
\sup_{t\in I}\left|f(t)\right|&\leq& \sup_{t\in I}\left|f(t)-f(a)\right| + |f(a)| \\
&\leq & \sup_{t\neq s}\left|f(s)-f(t)\right| +|f(a)| \\
&=& \sup_{t\neq s}\left\{|s-t|^{\alpha}\frac{\left|f(s)-f(t)\right|}{|s-t|^{\alpha}}\right\} +|f(a)|\\
&\leq& |b-a|^{\alpha}\sup_{t\neq s}\frac{\left|f(s)-f(t)\right|}{|s-t|^{\alpha}} +|f(a)| \\
&=&  |b-a|^{\alpha}M_{f} +|f(a)|. 
\end{array}$$
Then, we have
$$\begin{array}{rcl} 
\left\|f\right\|_{1}&=&\sup_{t\in I}\left|f(t)\right|+M_{f}\\
&\leq& |b-a|^{\alpha}M_{f} +|f(a)|+M_{f} \\
&=& \left(|b-a|^{\alpha}+1\right)M_{f}+|f(a)|\\
&\leq& \left(|b-a|^{\alpha}+1\right)M_{f}+\left(|b-a|^{\alpha}+1\right)|f(a)|\\
&=& \left(|b-a|^{\alpha}+1\right)\left\|f\right\|_{2}
\end{array}$$
Therefore
$$\left\|f\right\|_{2}\leq \left\|f\right\|_{1} \leq \left(|b-a|^{\alpha}+1\right)\left\|f\right\|_{2}.$$
Then, the norms are equivalents.
