# Renorming of $C[0,1]$ for a strictly convex dual

Let $$C[0,1]$$ be the space of all Real valued continuous functions on $$[0,1]$$ with the usual supremum norm. Does there exist an equivalent renorming on $$C[0,1]$$ such that the corresponding dual norm is strictly convex?

• The dual of every separabel Banach space has an eqivalent strictly convex norm, but I understand that this is not you question. – Jochen Wengenroth Jun 29 '20 at 8:12
• Thank you Jochen for pointing out. I was a bit confused about Hahn-Banach smoothness and smoothness. A Banach space $X$ is said to be Hahn-Banach smooth if every $x^*\in X^*$ has unique norm preserving extension to $X^{**}$. Similarly a weaker version of this property can also be defined for the norm attaining functionals. If $X^*$ is strictly convex then any subspace of $X$ is Hahn-Banach smooth BUT this property does not imply (weakly) Hahn-Banach smoothness of $X$. – Tanmoy Paul Jun 29 '20 at 15:55
• My earlier question can be reformulated as follows. Does there exist an equivalent renorming on $C[0,1]$ which makes it (weakly) Hahn-Banach smooth? Probably the answer to this problem is 'No' because the dual of any weakly Hahn-Banach smooth space has RNP. – Tanmoy Paul Jun 29 '20 at 15:58

One typically equivalently renorms a Banach space $$Y$$ to be strictly convex by finding an injective operator $$S$$ from $$Y$$ into some strictly convex space $$Z$$ and defining the new norm on $$Y$$ by $$\|y\| +\|Sy\|$$. When $$Y$$ and $$Z$$ are dual spaces and $$S$$ is weak$$^*$$ to weak$$^*$$ continuous, the new norm is a dual norm (the new unit ball of $$X^*$$ is weak$$^*$$ closed because $$T^*$$ is weak$$^*$$ continuous).
So let $$X$$ be any separable space and take an operator $$T:\ell^2 \to X$$ that has dense range. Renorm $$X^*$$ by $$!F! := \| F \|_{X^*} + \|T^*F\|_2$$.
• @TomaszKania. I think I first saw $! \cdot !$ in my undergraduate days when I took a course in functional analysis from Wilansky's book. – Bill Johnson Jul 6 '20 at 20:43