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Feb 11, 2019 at 13:00 comment added Dirk Werner @Tanmoy: Let $K=\alpha(\mathbf{N})=\mathbf{N}\cup\{\infty\}$. Let $f(n)=1-\frac1n$ and $f(\infty)=1$. Then $f\in C(K)$ is a smooth point of the unit ball, normed by the limit functional. However, in the bidual ($= \ell_\infty( \mathbf{N}\cup\{\infty\} )$) this function is normed by every Banach limit, so $f$ is not smooth in the bidual.
Feb 10, 2019 at 16:12 comment added Tanmoy Paul Can we say anything about the space of type $C(K)$? If we follow the above argument, if $f\in C(K)$ is a smooth point then is it necessary that $f$ remains smooth in $C(K)^{**}$? As we know bi-dual of $C(K)$ is of the form $C(\Omega)$ for some compact $T_2$ $\Omega$.
Feb 8, 2019 at 4:38 vote accept Tanmoy Paul
Feb 5, 2019 at 17:01 history answered Dirk Werner CC BY-SA 4.0