orthogonality exercise in Hilbert space

Question

Let X a Hilbert space, S is a closed vector subspace of X. Suppose that $(x_n)$ is a Cauchy sequence in X. Prove the following statements:

• The sequence $(P_S(x_n))$ is a Cauchy sequence in S.

• $x_n - P_S(x_n) \in S^{\perp}$

Answer 1

First point

The projection $P_S$ onto $S$ is a contracting map. Therefore $$\Vert P_S(x) - P_S(y) \Vert \le \Vert x-y\Vert$$ for all $x,y \in X$. This implies that $\{P_S(x_n)\}$ is a Cauchy sequence if $\{x_n\}$ is Cauchy.

Second point

By definition of a projection, we have $\langle P_S(x) , y \rangle = \langle x , P_S(y) \rangle$ for all $x,y \in X$.

As for $y \in S$, $P_S(y) =y$, we get that for all $y \in S$:

$$\langle P_S(x_n) - x_n, y \rangle = 0$$

which proves that $x_n - P_S(x_n) \in S^{\perp}$.

Answer 2

My answer is the following (I'm not sure if my answer is correct):

• I have no answer.

• If $x_n - P_S(x_n) \in S^{\perp}$, then it is true that:

  $(x_n-P_S(x_n))P_S(x_n) \in S^{\perp}$

  $(P_S(x_n)+P_S^{\perp}(x_n)-P_S(x_n))P_S(x_n) \in S^{\perp}$

  $P_S^{\perp}(x_n)P_S(x_n)=0$

  Then it is proved that $x_n - P_S(x_n) \in S^{\perp}$