# Banach space under the Lip norm

Let $(X,d)$ be a compact metric space. A function $f:x\to \Bbb R$ is said to be Lipschitz continuous if $$\|f\|_d = \sup\left\{\frac {|f(x)-f(y)|}{d(x,y)}:x,y\in X,x\neq y\right\}< \infty.$$ Denote by $\operatorname{Lip}(X,d)$ the collection of all lipschitz continuous functions on $X$.

a. Prove that $\operatorname{Lip}(X,d)$ is a Banach space under the norm

$$\|f\|=\|f\|_\infty +\|f\|_d$$

where $\|f\|_\infty$ the sup norm.

show that $\|fg\|\le \|f\|\cdot\|g\|$ is also true.

I know the case of sup norm only but when here comes the case of two variable x and y i got confused. Some what i did but i don't know if i am correct. I am preparing for Qualifying so if you do for me i would be thankful. Thanks in Advance

Let $(f_n)\subseteq\operatorname{Lip}(X,d)$ be a Cauchy sequence. In particular, $(f_n)$ is a Cauchy sequence of continuous functions with respect to the sup norm, so it converges uniformly to a (continuous) function $f$ (this is what you already knew). Let's show that $f_n\rightarrow f$ in $\operatorname{Lip}(X,d)$.

Given $\varepsilon>0$, choose $N\in\mathbb{N}$ such that $\Vert f_n-f_m\Vert<\varepsilon$ whenever $n,m\geq N$. Given $n,m\geq N$ and $x\neq y$ in $X$, we have $$\frac{|f_n(x)-f_m(x)-(f_n(y)-f_m(y))|}{d(x,y)}\leq \Vert f_n-f_m\Vert_d\leq \Vert f_n-f_m\Vert<\varepsilon.$$

Letting $m\rightarrow \infty$ and taking the sup on $x\neq y$, we obtain that $\Vert f_n-f\Vert_d<\varepsilon<\infty$ for any $n\geq N$, so (given one such $n$), we have $f_n-f\in\operatorname{Lip}(X,d)$, thus $f=f_n-(f_n-f)\in\operatorname{Lip}(X,d)$. Furthermore, we have just proved that $\Vert f_n-f\Vert_d\rightarrow 0$ as $n\rightarrow\infty$, and we already knew that $\Vert f_n-f\Vert_\infty\rightarrow 0$ as $n\rightarrow\infty$. This means exactly that $f_n\rightarrow f$ in $\operatorname{Lip}(X,d)$.

Therefore, $\operatorname{Lip}(X,d)$ is a Banach space.

Now, let $f,g:X\rightarrow\mathbb{R}$. Let $x\neq y$ in $X$. Then $$|(fg)(x)-(fg)(y)|\leq|f(x)||g(x)-g(y)|+|f(x)-f(y)||g(y)|$$ $$\leq\Vert f\Vert_{\infty}\Vert g\Vert_d d(x,y)+\Vert f\Vert_d d(x,y)\Vert g\Vert_\infty.$$

So given any $z\in X$, $$|(fg)(z)|+\frac{|(fg)(x)-(fg)(y)|}{d(x,y)}\leq\Vert f\Vert_\infty\Vert g\Vert_\infty+\Vert f\Vert_\infty\Vert g\Vert_d+\Vert f\Vert_d\Vert g\Vert_\infty$$ $$\leq (\Vert f\Vert_\infty+\Vert f\Vert_d)(\Vert g\Vert_\infty+\Vert g\Vert_d).$$ Taking the sup on $x\neq y$ and $z\in X$, we obtain $\Vert fg\Vert\leq\Vert f\Vert\cdot\Vert g\Vert$.

Remark: Maybe this exercise can be easier if you first show that $\Vert f\Vert_d=\inf\left\{K\geq 0:\forall x,y\in X,\ |f(x)-f(y)|\leq Kd(x,y)\right\}$, and then use that expression.