Limit of a sequence in direct sum of normed vector spaces Let $W$ a normed vector space over $\mathbb{C}$
Let $V,U \subset W$ be two subsets of $W$ such that $W = V \oplus U$
Let $\{w_m\}_{m \in \mathbb{N}} \subset W$ be a sequence in $W$ such that $w_m \to w_0 \in W$
$\forall m \in \mathbb{N}: w_m =v_m + u_m$ with $v_m \in V$ and $u_m \in U$
I would like to know if is it true that $v_m \to v_0 \in V$ and $u_m \to u_0 \in U$ and $w_0 =v_0+u_0$
 A: The conclusion is certainly valid if $W$ is assumed to be a Banach space and $U, V$ its closed subspaces.
Define $P : W \to U$ as the projection on $U$, meaning if $x \in W$ has the unique decomposition $x= u + v$, with $u \in U$, $v \in V$ then $Px = u$.
Notice that $I - P : W \to V$ is the projection on $V$, i.e. $(I - P)x = v$.
$P$ is obviously linear. Is is also bounded by the Closed Graph Theorem:
Assume $w_m \xrightarrow{m\to\infty} w_0$ and $Pw_m \xrightarrow{m\to\infty} u' \in U$ (since $U$ is closed).
Then by linearity of the limit we get
$$\underbrace{(I-P)w_m}_{\in V} = w_m - Pw_m \xrightarrow{m\to\infty} w_0 - u' \in V$$
The sequence $((I-P)w_m)_{m=1}^\infty$ is in $V$ then so is its limit $w_0 - u' \in V$ because $V$ is closed.
Hence $w_0 = \underbrace{u'}_{\in U} + \underbrace{(w_0 - u')}_{\in V}$. By the uniqueness of the decomposition, we conclude $u' = u$, therefore the graph of $P$ is closed.
We conclude that $P$ is a bounded linear map, and so is $I - P$. They are therefore continuous so
$$w_m \xrightarrow{m\to\infty} w_0$$
implies
$$u_m = Pw_m \xrightarrow{m\to\infty} Pw_0 = u$$
$$v_m = (I-P)w_m \xrightarrow{m\to\infty} (I-P)w_0 = v$$
