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Yes to your first question. As for the second, regard the $x_j$-s as linear functionals on Tell Joel hello for me. |
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Yes to your first question. As for the second, regard the $x_j$-s as linear functionals on $X$*. If you have $x_j(x$$)=0$ for all $j$, then every multiple of $x$ is in the the first set you have in your second paragraph; i.e., $|\phi(tx$$)| <1$ for all $t$ and hence $\phi(x$$)=0$. Thus $\phi$ is a linear combination of the $x_j$-s and hence is continuous. Tell Joel hello for me. |
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