# Is the metric space of Lipschitz Function with $d_{\infty}$ complete?

This is my real analysis homework.

Define

$$\mathcal{L} = \{f : [a,b] \to \mathbb{R} : f \text{ is a Lipschitz function }\}$$ and the metric $$d_{\infty}(f,g) = \sup\{|f(x)-g(x)| : x \in [a,b]\}$$

Prove that $$(\mathcal{L},d_{\infty})$$ is a complete metric space.

After giving the problem a try for a while, I begin to think that the problem is actually incorrect and begin to look for counterexample. Here is what I found.

Define $$f(x) = \sqrt{x}, x \in [0,1]$$. Since $$f$$ is continuous on $$[0,1]$$, then according to Weierstrass Theorem, there exist a sequence of polynomial $$(P_n(x))$$ such that $$P_n$$ converge uniformly to $$f$$. By Cauchy criterion for uniform convergence, we conclude that $$(P_n(x))$$ is a Cauchy sequence on $$(\mathcal{L},d_{\infty})$$. Also, polynomials are Lipschitz function, but since it does not converge to a Lipschitz function, then we conclude that $$(\mathcal{L},d_{\infty})$$ is not complete.

Is my counterexample correct?

• I am unable to find any flaw. – José Carlos Santos May 19 at 11:13
• Another counter example is given here: math.stackexchange.com/a/1584116/42969 – Martin R May 19 at 11:13
• Your example is valid. With the metric $d(f,g)=\{\sup \frac {|f(x)-f(y)} {|x-y|}:x \neq y\}$ the space is complete but not with the sup metric. – Kavi Rama Murthy May 19 at 11:46
• Thanks a lot guys! My doubt is cleared. – Ricky The Ising May 19 at 14:35