Does $P \circ P =P$ and $\langle Px, y \rangle = \langle x, Py \rangle$ imply $P$ is linear?

Question

Let $(H, \langle \cdot, \cdot \rangle)$ be a Hilbert space and $P:H \to H$. Suppose that

• $P \circ P =P$
• $\langle Px, y \rangle = \langle x, Py \rangle$ for all $(x,y) \in H^2.$

I would like to ask if $P$ is linear.

I tried some functions such as absolute value, identity, and constant. But none of them satisfy the second condition.

Please don't give me the proof, in case this statement is correct. I would like to give it a shot by myself.

Answer 1

The second condition alone is sufficient to get linearity.

Hint:

The right-hand side is linear in $x$, hence the left-hand side is also linear in $x$.

Answer 2

In fact, you don't need $P \circ P=P.$

Hint: Show $$\langle P(ax_1+x_2),y\rangle=\langle aP(x_1)+P(x_2),y\rangle$$ for every $y \in H$ using the second condition.

Answer 3

After many days of intermittent thoughts, I have come up with my own proof.

1. $P(\alpha x) = \alpha Px$ for all $(\alpha,x) \in \mathbb R \times H$

It suffices to show that $\langle P(\alpha x) - \alpha Px, y \rangle = 0$ for all $y \in H$. This is equivalent to $\langle P(\alpha x) , y \rangle = \langle \alpha Px, y \rangle = \alpha\langle Px, y \rangle$. This is actually true because $\langle P(\alpha x) , y \rangle = \langle \alpha x , Py \rangle = \alpha \langle x , Py \rangle = \alpha\langle Px, y \rangle$.

1. $P( x + y) =Px + Py$ for all $(x,y) \in H^2$

It suffices to show that $\langle P( x + y) - Px - Py,z \rangle = 0$ for all $z \in H$. This is equivalent to $\langle P( x + y), z \rangle = \langle Px + Py,z \rangle$. This is actually true because $\langle P( x + y), z \rangle = \langle x + y, P z \rangle = \langle x, P z \rangle + \langle y, P z \rangle = \langle Px, z \rangle + \langle Py, z \rangle= \langle Px + Py,z \rangle$.