# Does the norm of an operator on a $w^{*}$ dense subspace determine its norm?

Let $X$ be a (separable) Banach space, $T:X^{*}\to X^{*}$ a bounded operator, and $Y\subset X^{*}$ a norm closed, $w^{*}$ dense subspace of $X^{*}$. Is it true that $\|T\|=\|T_{|Y}\|$?

No, this is not true. For example, let $X=\ell_1$, $X^*=\ell_\infty$, and $Y=c_0\subset \ell_\infty$. Then $Y$ is a weak* dense subspace, but not norm-dense. Fix a nonzero vector $v\in \ell_\infty$ and define $T:\ell_\infty\to\ell_\infty$ as $T(x) = \phi(x)v$ where $\phi$ is a Banach limit. Then $T$ is identically zero on $Y$, but not on $X^*$.
• Thank you. do you have a quick counterexample when $X^{*}$ is separable? Jun 6 '16 at 17:24