Space of Lipschitz Functions Complete? Consider the subspace of continuous, real-valued functions on $[0,1]$ that are Lipschitz. Is this subspace complete under the sup norm ($\Vert \cdot \Vert_{\infty} = \sup \{ |f(x)| : x\in S \}$)?
I would say yes, since all Lipschitz functions ($d_{Y}(f(x_{1}),f(x_{2}))\leq K d_{X}(x_{1},x_{2})$, $K \geq 0$, where here $X = [0,1]$, $Y = \mathbb{R}$) are uniformly continuous, functions of certain types tend to converge uniformly to functions of the same types (e.g. differentiable functions to differentiable functions), but it seems unlikely. 
Could somebody please help?

Theorem (Weierstrass). Any continuous $f:[0,1]\to R$ is the uniform limit of a sequence of real polynomials .....Now any continuously differentiable $g:[0,1]\to R$ is Lipschitz-continuous with Lipschitz constant $K=\max \{|g'(x)| :x\in [0,1]\}.$ Polynomials on $[0,1]$ are therefore Lipschitz. So with the  $\sup$ norm, the set of Lipschitz-continuous $g:[0,1]$ is dense in $C[0,1]$, the space of all continuous $f:[0,1]\to R.$
