# In $C[0,1]$ prove that the subset of Lipschitz functions is dense

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

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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 $n \to \infty$.
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 $[0,1]$).