# Bounded operator on Hilbert space is orthogonal projection

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?

• I think the word "operator", otherwise unqualified, generally refers to a linear map. Commented Jun 20, 2019 at 15:52
• Fair enough. Still doesn't account for the $P(x_n) = x_n$ part in my eyes. Commented Jun 20, 2019 at 15:55
• We want to show $S$ is closed, which is equivalent to showing it contains the limit of any of its Cauchy sequences. So I think it is safe to assume the $x_n \in S$, and thus $Px_n = x_n$. Commented Jun 20, 2019 at 15:58
• The author of that proof forgot to say $x_n\in S$ at the outset. Commented Jun 20, 2019 at 15:58
• OMG Ted Shifrin! I have your book! Big fan! Commented Jun 20, 2019 at 15:59

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 $$P(x_n)=x_n$$.

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 $$S$$.

• Are there continuous projections which are not orthogonal? It seems a simple constant multiplication will do. Commented Aug 23, 2023 at 17:52

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.

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.

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

Since $$P^2=P$$.

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

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