= Solution
\b[(vi) implies (iv)] without assuming beforehand that any coordinate functional is continuous. Define the everywhere-defined <linear map>
$$
\Phi:X\to\ell^\infty,\qquad\Phi(x)=(\varphi_n(x))_{n\geq1}.
$$
The 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 $\varphi_n$.
Together with the preceding implications this proves \b[all six conditions are equivalent]. No continuity of the coordinate functionals was assumed during the preliminary bounds.
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