Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In $C[0,1]$ prove that the subset of Lipschitz functions is dense. I can't prove it.

share|cite|improve this question
up vote 5 down vote accepted

No need to pull out the heavy guns.. Suppose $f(x) \in C[0,1]$. For a given $n$ let $f_n(x)$ be the piece-wise linear function whose graph connects $(0,f(0))$ to $(1/n, f(1/n))$ to $(2/n, f(2/n))$ to ... to $(1,f(1))$. Then $f_n(x)$ is continuous, it's Lipschitz since it's piecewise linear, and $f_n \to f$ in $C[0,1]$ as $n \to \infty$.

share|cite|improve this answer
how do you prove that converges to f? – Gastón Burrull Sep 1 '12 at 5:17
Use the uniform continuity of $f(x)$... Given $\epsilon > 0$, if $\delta > 0$ is as in the definition of uniform continuity then if ${1 \over n} < \delta$ then $|f_n(x) - f(x)| < \epsilon$ for all $x$. – Zarrax Sep 2 '12 at 0:41

Polynomials are dense in $C[0,1]$ by Weierstrass. Those are Lipschitz (in $[0,1]$).

share|cite|improve this answer
... because their derivative is continuous on $[0,1]$, hence bounded. – Bookend May 16 at 21:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.