im trying to solve this, Consider $H$ a Hilbert space and $(e_n)_{n\in\mathbb{N}}$ an orthonormal basis of $H$. Let's consider the sets:
$A = {a_n = e_{2n} : n \in \mathbb{N}}$. $B = {b_n = e_{2n} + \frac{1}{n} e_{2n+1} : n \in \mathbb{N}}$.
The closed vector subspaces of $H$ are defined by $V = \text{span}(A)$ and $W = \text{span}(B)$, where the closures are taken with respect to the norm in $H$.
i. Show that $V \cap W = {0}$.
ii. Define $E = V \oplus W$. Show that $V$ and $W$ are closed vector subspaces in $E$.
iii. Show that $E$ is not closed in $H$.
Hint: Consider the series $\sum_{n\in\mathbb{N}}(a_n - b_n)$, with $a_n \in A$ and $b_n \in B$.
iv. Show that the projection operator $PV : E \rightarrow E$ is not bounded.
Hint: Consider the sequence $x_n = a_n - b_n$, with $a_n \in A$ and $b_n \in B$."
i. To demonstrate $V \cap W = \{0\}$, without loss of generality, let's consider two arbitrary vectors $x = e_{2n}$ and $x = e_{2n+1}$ from the bases. Subtracting one from the other, we get $x - x = \frac{1}{n}e_{2n+1}$. The term $\frac{1}{n}$ is always nonzero. If $e_{2n+1}$ were equal to $0$, this would contradict the fact that $(e_n)_{n\in\mathbb{N}}$ forms a basis. Therefore, $V \cap W = \{0\}$.
ii. Let $E = V \oplus W$ be the direct sum of $V$ and $W$. To show that $V$ and $W$ are closed vector subspaces in $E$, we can use the fact that the direct sum of two closed subspaces is also a closed subspace. Since $V$ and $W$ are closed subspaces of $H$, their direct sum $E$ is also a closed subspace of $H$.
iii. To show that $E$ is not closed in $H$, we can consider the series $\sum_{n=1}^\infty (a_n - b_n)$. Since $a_n \in A$ and $b_n \in B$, it can be seen that $a_n - b_n$ converges to zero in $H$. However, this series does not converge in $E$, as the difference between the norms of consecutive terms does not converge to zero. Therefore, $E$ is not closed in $H$.
iv. Finally, i dont see how to conclude :c