= 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 with
$$
\overline B_1\subset V\cap G_1,\qquad\overline B_{n+1}\subset B_n\cap G_{n+1},\qquad r_n<2^{-n}.
$$
Each 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 $E_m=\{x:\sup_{T\in\mathcal T}\|Tx\|\leq m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\|h\|<r$, the <triangle inequality> applied to $T(x_0+h)-Tx_0$ gives $\|Th\|\leq2m$. Taking $h=(r/2)u$ for unit $u$ proves $\sup_T\|T\|\leq4m/r$.
For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_kx\to x$, the norms $\|P_kx\|$ form a bounded sequence with limit $\|x\|$. Also, with $P_0=0$,
$$
|\varphi_n(x)|=\|\varphi_n(x)x_n\|=\|(P_n-P_{n-1})x\|\leq2\sup_k\|P_kx\|.
$$
This proves both preliminary bounds. The supremum therefore defines a finite <norm> $\|\cdot\|'$, with $\|x\|\leq\|x\|'$.
To prove \b[(i) implies (ii)], the identity map from $(X,\|\cdot\|')$ to $(X,\|\cdot\|)$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the <bounded inverse theorem> makes its inverse bounded: $\|x\|'\leq C\|x\|$. This is precisely (ii).
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