This is my real analysis homework.


$$\mathcal{L} = \{f : [a,b] \to \mathbb{R} : f \text{ is a Lipschitz function }\}$$ 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?

  • 2
    $\begingroup$ I am unable to find any flaw. $\endgroup$ – José Carlos Santos May 19 at 11:13
  • 1
    $\begingroup$ Another counter example is given here: math.stackexchange.com/a/1584116/42969 $\endgroup$ – Martin R May 19 at 11:13
  • 1
    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy May 19 at 11:46
  • $\begingroup$ Thanks a lot guys! My doubt is cleared. $\endgroup$ – Ricky The Ising May 19 at 14:35

Your Answer

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

Browse other questions tagged or ask your own question.