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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 ?

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    $\begingroup$ 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... $\endgroup$ – Eric Wofsey Nov 4 '18 at 23:13
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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 $X$.

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