In a infinite dimensional normed vector space $W=V \oplus U$ with $U$ finite dimensional, is the projection of $W$ on $V$ bounded?

Question

Let $W$ be a infinite dimensional normed vector space over $\mathbb{C}$

Let $V$ , $U \subset W$ be two subspaces of $W$ such that $U$ is finite dimensional and $W = V \oplus U$

Let $P : W \to V$ the projection on $V$ i.e. $\forall w \in W$ decomposing $w=v+u$ with $v \in V$ and $u \in U$ then $P(w)=v$

I would like to know if $P$ is bounded

Thanks.

Answer 1

Suppose that $V$ is not closed and $P$ is bounded implies that $P-Id$ is continuous, $Ker(P-Id)$ is $V$ contradiction. To find a counterexample, take $V$ to be the kernel of an unbounded linear function defined on $W$.

Answer 2

$P$ doesn't have to be bounded even if you assume that $U, V$ are closed subspaces of $W$.

For an explicit example, consider $c_{00}$ (the space of all real sequences with finite support) with the supremum norm and define

$$V = \{(a_n)_n \in c_{00} : a_{2n} = 0, \forall n \in \mathbb{N}\}$$ $$U = \{(a_n)_n \in c_{00} : a_{2n-1} + na_{2n} = 0, \forall n \in \mathbb{N}\}$$

$U$ and $V$ are closed in $c_{00}$ and we have $c_{00} = V \oplus U$. Indeed, $V \cap U = \{0\}$ and for any $(a_n)_n \in c_{00}$ we have

$$(a_n)_n = \underbrace{(a_1 + a_2, 0, a_3 + 2a_4, 0, a_5 + 3a_6, 0, \ldots)}_{\in V} + \underbrace{(-a_2, a_2 , -2a_4, a_4, -3a_6, a_6, \ldots)}_{\in U}$$

However, the projection $P$ onto $V$ is given by

$$P(a_n)_n = (a_1 + a_2, 0, a_3 + 2a_4, 0, a_5 + 3a_6, 0, \ldots)$$

so it is clearly not bounded as $\|Pe_{2n}\|_\infty = n$ for all $n \in \mathbb{N}$.