# the space of Lipschitz functions is complete metric space

I want to show the space of Lipschitz functions $$f : [a, b] → {\rm I\!R}$$ with the following metric is complete.

$$d(f, g) = \underset{xin[a,b]}{\sup} |f(x) − g(x)| + \underset{{x,y\in[a,b],x\neq y}}{\sup}\frac{|[f − g](x) − [f − g](y)|}{|x − y|}$$

I tried to proceed but stucked in the middle; $$\\$$

$$\textbf{Attempt:}$$ Assuming a Cauchy sequence. Then $$d(f_n, f_m)<\epsilon$$. i.e.

$$d(f_n, f_m) = \underset{x\in[a,b]}{\sup} |f_n(x) − f_m(x)| + \underset{{x,y\in[a,b],x\neq y}}{\sup}\frac{|[f_n − f_m](x) − [f_n − f_m](y)|}{|x − y|}<\epsilon$$

the goal is to show that $$|f_n(x)-f(x)|<\epsilon$$.

so; $$|f_n(x)-f(x)|=|f_n(x)-f_n(y)+f_n(y)-f(x)|\leq |f_n(x)-f_n(y)|+|f_n(y)-f(x)|$$

by Lipschitz continuity we have that $$|f_n(x)-f_n(y)|\leq \frac{\epsilon}{2}|x-y|$$ so;

$$|f_n(x)-f(x)|\leq \frac{\epsilon}{2}|x-y|+|f_n(y)-f(x)|$$

Idk how to show that $$|f_n(y)-f(x)|< \frac{\epsilon}{2}$$

• On which sets are those functions defined? – Jakob Werner Sep 2 '19 at 21:12
• @JakobWerner On $[a,b]$, of course. – amsmath Sep 2 '19 at 21:17
• stat, my friend, what is $f$? You have not defined it... – amsmath Sep 2 '19 at 21:17
• I edited the question, sorry I forgot $f : [a, b] → {\rm I\!R}$ – stat Sep 2 '19 at 21:23
• A Cauchy sequence is in particular a Cauchy sequence in the sup norm, hence has a uniform limit function. Try to show that this function is Lipschitz continuous and is a limit in the space of Lipschitz functions. – Jakob Werner Sep 2 '19 at 21:29

There are a couple of errors in your arguments as pointed out in the comments. Here is a proof: given $$\epsilon >0$$ there exists $$N$$ such that $$d(f_n,f_m) <\epsilon$$ for $$n , m>N$$. This gives $$|f_n(x)-f_m(x)| <\epsilon$$ for all $$x$$ for $$n , m>N$$ $$\cdots (1)$$.
Hence $$(f_n(x))$$ is a Cauchy sequence for each $$x$$. Define $$f(x)$$ as $$\lim_{n\to \infty} f_n(x)$$.
By letting $$m \to \infty$$ in (1) we get $$|f_n(x)-f(x)| \leq \epsilon$$ for all $$x$$ for $$n >N$$. Similarly, by letting $$m \to \infty$$ in the inequality $$|(f_n(x)-f_n(y))-(f_m(x)-f_m(y))| \leq \epsilon |x-y|$$ we get $$|(f_n(x)-f_n(y))-(f(x)-f(y))| \leq \epsilon |x-y|$$ for $$n >N$$ for all $$x,y$$. Fixing $$n=N+1$$ and using the fact that $$f_{N+1}$$ is Lipschitz you see that $$f$$ is also Lipschitz. [Use triangle inequality]. Putting these two facts together we get $$d(f_n,f) \leq \epsilon$$ for $$n >N$$. This completes the proof.