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?
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2$\begingroup$ The dual of every separabel Banach space has an eqivalent strictly convex norm, but I understand that this is not you question. $\endgroup$– Jochen WengenrothCommented Jun 29, 2020 at 8:12
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$\begingroup$ 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 $X$. $\endgroup$– Tanmoy PaulCommented Jun 29, 2020 at 15:55
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$\begingroup$ 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. $\endgroup$– Tanmoy PaulCommented Jun 29, 2020 at 15:58
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1 Answer
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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 $!F! := \| F \|_{X^*} + \|T^*F\|_2$.
answered Jun 29, 2020 at 13:31
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4$\begingroup$ Bill, your exclamation marks would have surely won a contest for the funniest norm symbol, if only there were one! :-) $\endgroup$ Commented Jun 29, 2020 at 22:50
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1$\begingroup$ @TomaszKania. I think I first saw $! \cdot !$ in my undergraduate days when I took a course in functional analysis from Wilansky's book. $\endgroup$ Commented Jul 6, 2020 at 20:43