# 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.

• What have you done so far? Have you checked that the given norms are really norms? Probably you should exploit that such functions are continuous and due to the norm the limit of a sequence of such functions is continuous. The $M_f$ estimate is also maintained by such a limit. Do you really call denote that spaces with Lip? A simpler case could be starting with $\alpha=1$. – Quickbeam2k1 Mar 15 '13 at 6:33

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 $\|\cdot\|_1$.

• In fact, it may help to observe that since you're dealing with functions on $[0,1]$, you need only consider $\epsilon \leq 1$. – Branimir Ćaćić Mar 15 '13 at 21:16

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.

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.