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

  • $\begingroup$ On which sets are those functions defined? $\endgroup$ – Jakob Werner Sep 2 '19 at 21:12
  • $\begingroup$ @JakobWerner On $[a,b]$, of course. $\endgroup$ – amsmath Sep 2 '19 at 21:17
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    $\begingroup$ stat, my friend, what is $f$? You have not defined it... $\endgroup$ – amsmath Sep 2 '19 at 21:17
  • $\begingroup$ I edited the question, sorry I forgot $f : [a, b] → {\rm I\!R}$ $\endgroup$ – stat Sep 2 '19 at 21:23
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    $\begingroup$ 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. $\endgroup$ – 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.


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