Symmetric operator (1 Viewer)

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1. The problem statement, all variables and given/known data
Let A be a linear operator on a Hilbert space X. Suppose that D(A) = X,
and that (Ax, y) = (x, Ay) for all x, y in H. Show that A is bounded.

3. The attempt at a solution

I've tried to prove it by using the fact that if A is continuous at
a point x implies that A is bounded.

Suppose that x_n converges to x.
|| Ax_n - Ax ||^2 = || A(x_n - x) ||^2 = ( A(x_n - x), A(x_n - x) ) =
(x_n - x, A^2 (x_n - x) ) <= ||x_n - x || || A^2(x_n - x) ||

But i don't think I can conclude from this that because ||x_n - x|| -> 0, this expression goes to zero, since || A^2(x_n - x) || may blow up.. Or doesn't it?

Please help me :)
 

Dick

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D(A)=domain of A? I don't see how you can prove this. Self adjointness doesn't imply boundedness. You must be assuming A is also continuous? Then there is not much to prove. Better check the question.
 
Last edited:

dextercioby

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1. The problem statement, all variables and given/known data
Let A be a linear operator on a Hilbert space X. Suppose that D(A) = X,
and that (Ax, y) = (x, Ay) for all x, y in H. Show that A is bounded.
In a Hilbert space bounded <=> continuous. If A is symmetric and everywhere defined, it's equal to its adjoint, hence self-adjoint hence closed. By the closed graph theorem, it is also continuous, end proof.

The theorem you needed to prove is called "The Hellinger-Toeplitz theorem" and proves that unbounded symmetric operators cannot be defined on all [itex] \mathcal{H} [/itex].
 

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