Is the image of a closed subset in a Banach space under a lineal and continuous projection always closed? If the answer is no, is there any example?

Question

I was wondering if the following conjecture is true: let $X$ be a Banach space, let $P: X\to X$ be a continuous linear projection, and let $C$ be a closed subspace (unrelated to $P$) of $X$. Then the image $P[C]$ is closed. By saying that $C$ is not related to $P$, I mean that we do not suppose that $C$ is contained in the image of $P$, that is, we do not know any relationship between $C$ and $P$. I have thought that the continuity of $P$ implies the continuity of $I-P$. But it turns out that $Ker (I-P)$ doesn't depend on $C$. So I don't know how to proceed. Thanks for your help friends.

Answer 1

Let $X = \ell^2$ and $P$ is defined via $$ P(x_1, x_2, x_3, x_4, x_5, x_6, \ldots) = (0, x_2, 0, x_4, 0, x_6,\ldots).$$ Further, the subspace $C$ consists of all sequences in $\ell^2$ of the form $$ (1 z_1, z_1, 2 z_2, z_2, 3 z_3, z_3, 4 z_4, z_4, \ldots),$$ i.e. $$ C = \{ y \in \ell^2 \mid \forall n \in \mathbb N : y_{2n-1} = n y_{2n}\}. $$ It is easy to check that

• $C$ is closed
• $P(C)$ is dense in $P(X)$
• The following sequence is in $P(X) \setminus P(C)$: $$ (0, 1, 0, 1/2, 0, 1/3, \ldots). $$

Answer 2

Consider the Banach space of summable real sequences $X=\ell^1$. We have that $Y=\{x\in\ell^1:x_1=0\}$ is a closed subspace of $\ell^1$ and $P:X\ni x\mapsto (0,x_2,...)\in Y$ is a bounded projection onto $Y$. Consider the subspace $C=\langle\delta_1,\delta_2,...\rangle\subseteq X$, where $\delta_j\in\ell^1$ is the element $(\delta_j)_n=0$ for $n\neq j$ and $(\delta_j)_j=1$ and $\langle A \rangle$ of a subset $A$ denotes its linear span. Now we have that $$ P(C)=\langle\delta_2,\delta_3,...\rangle\subseteq Y. $$ If $P(C)$ was closed it would admit a countable (algebraic) basis, but this cannot happen in an infinite dimensional Banach space.