# Troubling contradiction in Functional Analysis

#### ModusPonens

##### Well-known member
Hello

I was doing an exercise that said: "If $P$ is a continuous operator in a Hilbert space $H$ and $P^2=P$ then the following five statements are equivalent". The first statement was that P is an orthogonal projection. Now this was suposed to be equivalent, under the condition of $P^2=P$, to $P^*=P$. However, I was able to prove that P is always an orthogonal projection, or so I think I did. I don't know of any mistake I've done in the proof. So what I ask is if there is a continuous operator in a Hilbert space that is idempotent, but not self adjoint.

#### Opalg

##### MHB Oldtimer
Staff member
Hello

I was doing an exercise that said: "If $P$ is a continuous operator in a Hilbert space $H$ and $P^2=P$ then the following five statements are equivalent". The first statement was that P is an orthogonal projection. Now this was suposed to be equivalent, under the condition of $P^2=P$, to $P^*=P$. However, I was able to prove that P is always an orthogonal projection, or so I think I did. I don't know of any mistake I've done in the proof. So what I ask is if there is a continuous operator in a Hilbert space that is idempotent, but not self adjoint.
Yes: in a two-dimensional space $P(x,y) = (x+y,0)$. The range is the $x$-axis, but the null space is the line $x+y=0$, which is not orthogonal to the range. The adjoint operator is given by $P^*(x,y) = (x,x)$.

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