# Prove $\left(\operatorname{Lip}\left([0,1]\right),\lvert\lvert\cdot\rvert\rvert\right)$ is a Banach Space

We denote by $\operatorname{Lip}\left([0,1]\right)$ the collection of all Lipschitz functions on $[0,1]$. We know that a function $f:[0,1] \to \mathbb R$ is called Lipschitz if there exists $K>0$ such that $|f(x)-f(y)| \leq K|x-y|$ for all $x,y \in [0,1]$. For $f \in \operatorname{Lip}\left([0,1]\right)$, set $$||f||=|f(0)|+\sup_{0 \leq x,y \leq 1, x \neq y}\frac{|f(x)-f(y)|}{|x-y|}$$ I have already proven that $||f||$ is a norm on $C\left([0,1]\right)$. Now, I must show that $\left(\operatorname{Lip}\left([0,1]\right),\lvert\lvert\cdot\rvert\rvert\right)$ is a Banach space. By definition, this is true if $(X,d)$ is complete, where $X=\operatorname{Lip}\left([0,1]\right)$ and $$d=d(f,g)=||f-g||=|f(0)-g(0)|+\sup_{0 \leq x,y \leq 1, x \neq y}\frac{|f(x)-g(x)-(f(y)-g(y))|}{|x-y|}$$ We also know that a metric space $(X,d)$ is complete if every Cauchy sequence in $X$ is convergent, where a Cauchy sequence $(x_n)$ is defined as $$(\forall \epsilon>0)(\exists N>0)(\forall m,n \geq N)(d(x_m,x_n)<\epsilon)$$ Also, according to a theorem, $\left(C\left([a,b]\right),||\cdot||\right)$, where $||f||=\max_{a \leq t \leq b}|f(t)|$ is a Banach space. Is this result applicable to my proof? I have had trouble lately with the concept of completeness in metric spaces, and how to show it in a general sense. Any help with this proof, and pertinent advice would be appreciated. Thank you.

Here is my attempt:

Let $f_n \in X$ be a Cauchy sequence, and denote $||f||_d=\sup_{0 \leq x,y \leq 1, x \neq y}\frac{|f(x)-f(y)|}{|x-y|}$. In particular, $f_n$ is a Cauchy sequence of continuous functions, so it converges to a continuous funtion $f$. We must show that $f_n \to f$ in $X$. Given $\epsilon>0$, choose $N \in \mathbb{N}$ such that $||f_n-f_m||<\epsilon$ whenever $m,n \geq N$. Given $m,n \geq N$ and $x \neq y$ in $[0,1]$, we have $$\frac{|f_n(x)-f_m(x)-(f_n(y)-f_m(y))|}{|x-y|} \leq ||f_n-f_m||_d \leq ||f_n-f_m||< \epsilon$$ Letting $m \to \infty$, and taking the supremum on $x \neq y$, we obtain that $||f_n-f||_d<\epsilon$ for any $n \geq N$. Given such $n$, we have $f_n-f \in X \implies f=f_n-(f_n-f) \in X$. We have just proven that $||f_n-f||_d$ converges to $0$ as $n \to \infty$, and it follows that $|f_n(0)-f(0)| \to 0$ as $n \to \infty$. Thus, $f_n \to f$ in $X$, and so, $(X,d)$ is complete.

Is my reasoning correct here?

• A small issue: if you have $\operatorname{foo}(n,m) < \varepsilon$ for all large enough $m$, and take the limit as $m\to\infty$, in the limit you only know that you have the weak inequality $\leqslant \varepsilon$. The bigger issue: what is $f$? Before you show that $(f_n)$ converges to $f$, you first must establish the existence of some $f$, and show that it belongs to the space. Then you have a candidate for the limit, and if you have that, your argument shows that it is indeed the limit. – Daniel Fischer Feb 22 '15 at 23:31
• Is it not sufficient that $f$ is a function on $\operatorname{Lip}([0,1])$? How else would I define this? – Douglas Fir Feb 22 '15 at 23:36
• You need $f\in \operatorname{Lip}([0,1])$. But what is $f$? – Daniel Fischer Feb 22 '15 at 23:38
• I have changed my proof based on your correction. Is this a better argument? – Douglas Fir Feb 23 '15 at 0:35
• The question is whether "In particular, $f_n$ is a Cauchy sequence of continuous functions, so it converges to a continuous function $f$" is a known fact you can use without explanation, or whether it needs an argument here. If you can use it, the only issue is that you only have the weak inequality $\lVert f_n - f\rVert_d \leqslant \varepsilon$, not the strict inequality, guaranteed. – Daniel Fischer Feb 23 '15 at 9:59