Bounded operator on Hilbert space is orthogonal projection

Question

Let ${H}$ be a Hilbert space, $P:H \to H$ be a bounded operator such that $P^2 =P^* = P$.

Show that $P$ is an orthogonal projection for some closed subspace $S \subseteq H$, meaning for closed subspace $S$, $P(x) = \begin{cases}x, x\in S \\ 0, x \in S^{\perp}\end{cases}$

I'm trying to understand a given proof. The proof goes along something like:

Define $S = Im(P)$. First we show that $S$ is closed.

$P$ is bounded, so it is continuous, so if $x_n \to x$ then $P(x_n) \to P(x)$, but $P(x_n) = x_n$ so really $x_n \to P(x)$ which proves that $x = P(x)$ hence $x \in S$.

Stop right here. This assumes many things which were not given to us.

$P$ being bounded implies continuity only if $P$ is linear. We were not given that.

$P(x_n) = x_n$ only if we assume that $P$ is an orthogonal projection - Which is what we wanted to show in the first place?!

This proof in my eyes is very clearly false right from the get go. But is the statement I am trying to prove correct?

Answer 1

This part wants to show that $S$ is closed.
Indeed, 'operator' here refers to linear operator, and the $x_n$'s are taken from $S=\mathrm{im}\, P$, so $P^2 =P$ implies $P(x_n)=x_n$.

Answer 2

Incase someone is interested in a full proof:

Define $S = Im(P)$, first step is to prove $S$ is a closed subspace. Let $x_n \in S$, such that $x_n \to x$.

$P$ is bounded, therefor continuous, so $P(x_n) \to P(x)$. Notice however that since $x_n \in S$ then there is a $y_n \in H$ such that $P(y_n) = x_n$ which implies $P^2(y_n) = P(x_n)$. Since we were given $P^2 = P$, we have $P(y_n) = x_n = P(x_n)$, so overall, $x_n \to P(x)$, so $P(x) = x$ and so $x \in S$ and $S$ is closed.

Now let $w \in S^{\perp}$ and $x \in S$. Notice that $\langle x,Pw\rangle = \langle P^*x, w \rangle = \langle Px, w\rangle = \langle x,w\rangle = 0$

Hence $P(w) \in S^{\perp}$. But by definition $P(w) \in Im(P) = S$ so we must have $P(w) = 0$. for all $w \in S^{\perp}$.

Finally then, let $y = y_{S} + y_{S^\perp}$, then from linearity $P(y) = P(y_S)+P(y_{S^{\perp}}) = P(y_S) = y_S$

So $P$ is the orthogonal projection unto $S$.

Answer 3

The way I prove this goes as follows:

First, given that $P$ is bounded, i.e., continuous, we also have that $I - P$ is bounded, i.e., continuous; for if $C_P$ bounds $P$, that is,

$\Vert Px \Vert \le C_P \Vert x \Vert, \forall x \in H, \tag 1$

then

$\Vert (I - P)x \Vert = \Vert Ix - Px \Vert = \Vert x - Px \Vert$ $\le \Vert x \Vert + \Vert Px \Vert \le \Vert x \Vert + C_P\Vert x \Vert = (1 + C_P) \Vert x \Vert, \tag 2$

so $I - P$ is bounded by $1 + C_P$; hence $I - P$ is also continuous.

Next, note that

$\text{Im}\;P = \ker(I - P), \tag 3$

$\text{Im}(I - P) = \ker P, \tag 4$

since

$y \in \text{Im}\;P \Longrightarrow \exists x \in H, \; y = Px$ $\Longrightarrow (I - P)y = (I - P)Px = (P - P^2) = 0 \Longrightarrow y \in \ker(I - P), \tag 5$

and

$y \in \ker(I - P) \Longrightarrow$ $y - Py = (I - P)y = 0 \Longrightarrow y = Py \Longrightarrow y \in \text{Im}\;P; \tag 6$

(5) and (6) together establish (3); also,

$y \in \text{Im}(I - P) \Longrightarrow \exists x \in H, \; y = (I - P)x$ $\Longrightarrow Py = P(I - P)x = (P - P^2)x = 0 \Longrightarrow y \in \ker P, \tag 7$

$y \in \ker P \Longrightarrow (I - P)y = y - Py = y \Longrightarrow y \in \text{Im}\;(I - P); \tag 8$

thus (7) and (8) demonstrate (4) which, though not essential to this answer, provides insight into a certain symmetry 'twixt $P$ and $I - P$.

Now in light of (2) and (3) we see that

$S = \text{Im}\;P = \ker(I - P) \tag 9$

is closed since $I - P$ is continuous; also,

$x \in S \Longleftrightarrow \exists y, \; x = Py \Longleftrightarrow \exists y, \; Px = P^2y = Py = x; \tag{10}$

furthermore

$S^\bot = (\text{Im}\;P)^\bot = \ker P, \tag{11}$

for

$z \in (\text{Im}\;P)^\bot \Longleftrightarrow \forall y \in H, \; \langle z, Py \rangle = 0$ $\Longleftrightarrow \forall y \in H, \; \langle Pz, y \rangle = \langle P^\dagger z, y \rangle = 0 \Longleftrightarrow Pz = 0 \Longleftrightarrow z \in \ker P; \tag{12}$

thus we see that

$x \in S^\bot \Longleftrightarrow x \in \ker P \Longleftrightarrow Px = 0; \tag{13}$

(10) and (13) show that $P$ is an orghogonal projection in the sense given in the text of the question.

Answer 4

Assume $P^2=P$ and $P^*=P$. Then $P$ is the orthogonal projection onto $PH$.

1. $PH$ is closed.

Let $y_n=P x_n\in PH$ be a Cauchy sequence in $PH$. Then $y_n\to y\in H$, that is $Px_n\to y$. But $P(Px_n)=Px_n$. We have $$y=\lim Px_n=\lim P(Px_n)=Py,$$ that is $y=Py\in PH$ and $PH$ is closed.

2. $P|_{PH}=id$.

Since $P^2=P$.

3. $P|_{PH^\perp}=0.$

Let $y\in PH^\perp$. We have $(Px,y)=(x,Py)=0$ for all $x\in H$. Hence $Py=0$.