\b[(v) implies (iv).] Fix $n$ and consider any sequence $y_r\to y$ with $\varphi_n(y_r)\to\gamma$. Apply (v) to $y_r-y$ and $\beta=\gamma-\varphi_n(y)$. It gives $\gamma=\varphi_n(y)$. Thus the graph of $\varphi_n:X\to\mathbb K$ is closed. Both spaces are Banach, so the <closed graph theorem> makes $\varphi_n$ bounded.

Solution

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

 Solution (source code)