Image of T

Jan 2010
32
1
If \(\displaystyle H\) is a Hilbert Space and \(\displaystyle T:H\rightarrow H\) a bounded operator. Given that \(\displaystyle T\) is bounded below \(\displaystyle \|Tx\|\geq\delta\|x\|\) for some \(\displaystyle \delta>0\) and that \(\displaystyle ker(T^*)=0\) show that the image of \(\displaystyle im(T)\) of \(\displaystyle T\) is closed.

This is a revision question to prove that if an operator is bounded below and the kernel of the adjoint is zero then it is invertible. The question requires that you prove \(\displaystyle T\) is bijective. I have proved injective and done most of the surjection proof, I just now need to show that the image of T is equal to the closure of the image of T i.e. \(\displaystyle im(T)\) is closed.

I am assuiming that you need to show that the compliment is open somehow, it seems clear that the compliment is the empty set so open and closed. Any help would be great.
 

Opalg

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If \(\displaystyle H\) is a Hilbert Space and \(\displaystyle T:H\rightarrow H\) a bounded operator. Given that \(\displaystyle T\) is bounded below \(\displaystyle \|Tx\|\geq\delta\|x\|\) for some \(\displaystyle \delta>0\) and that \(\displaystyle ker(T^*)=0\) show that the image of \(\displaystyle im(T)\) of \(\displaystyle T\) is closed.
(You don't need the information about \(\displaystyle \ker(T^*)\) for this part of the question.) If \(\displaystyle Tx_n\to y\) then \(\displaystyle (Tx_n)\) is a Cauchy sequence. The condition \(\displaystyle \|Tx\|\geq\delta\|x\|\) (applied to \(\displaystyle x=x_m-x_n\)) shows that \(\displaystyle (x_n)\) is also Cauchy and therefore converges to a limit z say. Then Tz = y, which shows that y is in im(T). Therefore im(T) is closed.
 
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