# Finding norm on $C[0,1]$ , which is not equivalent to the supremum norm, but which still makes $C[0,1]$ into a separable Banach space

We know that $$C[0,1]$$ with the sup norm $$||f||_{\infty}:=\sup_{x\in [0,1]} |f(x)|$$ is a separable Banach space.

My question is , does there exist a norm on $$C[0,1]$$ , which is not equivalent to the sup norm, but which still makes $$C[0,1]$$ into a separable Banach space ?

• If you forget the norm, then $C[0,1]$ is just some vector space of dimension $2^{\aleph_0}$, just like any other separable infinite-dimensional Banach space... – Eric Wofsey Nov 4 '18 at 23:13

As vector spaces, all separable infinite-dimensional Banach spaces are isomorphic (i.e. they have Hamel bases of cardinality $$\mathfrak c$$). Let $$\Phi$$ be a vector-space isomorphism from $$C[0,1]$$ to some other separable Banach space $$X$$. Then $$\|f\|_1 = \|\Phi(f)\|_X$$ defines a norm that makes $$C[0,1]$$ into a Banach space isomorphic to $$X$$.