Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2009 / A / 3 / vi / Solution

(vi) implies (iv) without assuming beforehand that any coordinate functional is continuous. Define the everywhere-defined linear mapThe preliminary coordinate bound ensures it really takes values in ${\ell }^{\infty }$. Condition (vi) states exactly that its graph is sequentially closed: $y_r\to x$ and $\Phi (y_r)\to \beta$ in the supremum norm imply $\Phi (x)=\beta$. The product is a metric space, so sequentially closed means closed. The closed graph theorem therefore makes $\Phi$ bounded. Every coordinate projection on ${\ell }^{\infty }$ is bounded, hence so is each ${\phi }_n$.

Together with the preceding implications this proves all six conditions are equivalent. No continuity of the coordinate functionals was assumed during the preliminary bounds.