I know that if $X$ is a Banach space, then, the direct sum of two closed subspace $X_1$ and $X_2$ is not necessarily closed. But what if $X$ is Hilbert?
I assume there is something to do with the orthonormal basis, since this is something you can ask for a Hilbert space but not a Banach space.
Moreover, is the projection to $X_1$ bounded? Namely, $P$ is defined on $X_1\oplus X_2$:
$$P(x)=x\quad \text{on $X_1$} \qquad P(x)=0\quad \text{on $X_2$}$$
is $P$ bounded?
Best Answer
In the sequence space $\ell^2$, let $X_1$ be the subspace of sequences $x = (x_0,x_1,x_2, \ldots)$ such that $x_{2n} = 0$ for all natural numbers $n$, and $X_2$ the subspace of sequences $x$ such that $x_{2n+1} = n x_{2n}$ for all $n$. It is easy to see that $X_1 \cap X_2 = \{0\}$, and that $X_1 + X_2$ is dense (e.g. it contains all sequences with finite support). However, $X_1 + X_2$ is not all of $\ell^2$, e.g. it can't contain the sequence $x_n = 1/(n+1)$: if this was $u + v$ with $u \in X_1$ and $v \in X_2$, we'd need $v_{2n} = x_{2n} = 1/(2n+1)$ and $v_{2n+1} = n/(2n+1)$, but then $\sum_n |v_n|^2 = \infty$.
When $X_1 \oplus X_2$ is not closed, the projection $P$ of $X_1 \oplus X_2$ onto $X_1$ must be unbounded: otherwise it would extend to a bounded linear operator $Q$ from $\overline{X_1 \oplus X_2}$ onto $X_1$; since $(I - P)x \in X_2$ for $x \in X_1 \oplus X_2$, we'd also have $(I-Q) x \in X_2$ for $x \in \overline{X_1 \oplus X_2}$. But then $x = Qx + (I-Q)x \in X_1 \oplus X_2$ for $x \in \overline{X_1 \oplus X_2}$, contradiction.