MHB Troubling contradiction in Functional Analysis

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The discussion centers on the properties of continuous operators in Hilbert spaces, specifically focusing on idempotent operators where \( P^2 = P \). It is asserted that if \( P \) is idempotent, it should also be self-adjoint to be classified as an orthogonal projection. However, a counterexample is provided with a two-dimensional operator \( P(x,y) = (x+y,0) \), which is idempotent but not self-adjoint, as its range and null space are not orthogonal. The adjoint operator is identified as \( P^*(x,y) = (x,x) \), further illustrating the contradiction in the initial assumption. This highlights the nuanced relationship between idempotency and self-adjointness in functional analysis.
ModusPonens
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.
 
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ModusPonens said:
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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