 Solution

The Baire category theorem says that a nonempty complete metric space is not a countable union of closed sets with empty interior; equivalently, every countable intersection of open dense sets is dense. To prove the latter, start with any nonempty open set $V$ and open dense sets $G_n$. Choose closed balls of positive radii withEach choice is possible by density and openness. Their centers form a Cauchy sequence; its limit exists by completeness and belongs to every closed ball, hence to $V\cap \bigcap G_n$. This proves density and the theorem.

The Uniform boundedness principle says that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Z$ has uniformly bounded operator norms. Put $E_m=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\Vert h\Vert <r$, the triangle inequality applied to $T(x_0+h)-T{x}_0$ gives $\Vert Th\Vert \le 2m$. Taking $h=(r/2)u$ for unit $u$ proves ${\sup }_T\Vert T\Vert \le 4m/r$.

For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_{k}x\to x$, the norms $\Vert P_{k}x\Vert$ form a bounded sequence with limit $\Vert x\Vert$. Also, with $P_0=0$,This proves both preliminary bounds. The supremum therefore defines a finite norm $\Vert \cdot {\Vert }^{\prime }$, with $\Vert x\Vert \le \Vert x{\Vert }^{\prime }$.

To prove (i) implies (ii), the identity map from $(X,\Vert \cdot {\Vert }^{\prime })$ to $(X,\Vert \cdot \Vert )$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the bounded inverse theorem makes its inverse bounded: $\Vert x{\Vert }^{\prime }\le C\Vert x\Vert$. This is precisely (ii).