Is Banach Space of Lipschitz Function Separable?

Question

To simplify the question, let $X=[0,1]$ with Euclidean distance. Denote $$L_f=\sup_{x,y\in X}\frac{|f(x)-f(y)|}{|x-y|},$$ and $$||f||_L=\max\{L_f,||f||_\infty\}.$$ Consider space $$Lip_0(X)=\{f: f(0)=0,||f||_L<\infty\}.$$ On $Lip_0(X)$, $||.||_L$ is equivalent to $L_f$, because $X$ is bounded. There are some famous results about this space, for example $Lip_0(X)$ is complete w.r.t. $||.||_L$ and separable w.r.t. $||.||_\infty$. I was wondering if $Lip_0(X)$ is separable under the norm $||.||_L$. And if not, what kind of conditions or modifications are needed to make it separable. Thanks a lot!

Answer 1

Oops, I lied. I said

"Certainly it's separable. For example, convolution with the Cesaro kernel shows that trigonometric polynomials are dense; hence trigonometric polynomials with rational coefficients are dense."

and that seemed right, but it was a little glib. The space $Lip_1$ is essentially an $L^\infty$ sort of thing, and convolving an $L^\infty$ funtion with an approximate identity does not give convergence in the $L^\infty$ norm. And of course $L^\infty$ is not separable...

No, actually $Lip_0$ is not separable. For $0<h<1$ define $f_h$ by $$f_h(0)=0,\quad f_h'=\chi_{(0,h)}.$$

Then $||f_h-f_{h'}||_{L}\ge1$ for $h\ne h'$. This is clear for various reasons; some readers may just consider the supremum of the derivative, other readers may prefer to just draw a picture.