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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.