# An application of Fredholm Alternative

I have just started reading the Fredholm Alternative for finite dimensional spaces and I came towards an excersise that it reads as follows: If $$H$$ is a Hilbert space and $$T:H\to H$$ is bounded, linear map, with $$\langle Tx,x\rangle >0$$ for $$x\neq0$$ then prove that $$T$$ is surjective. My way of thinking was to use the adjoint map $$T^*$$ like that:

$$\langle Tx,x\rangle = \langle x,T^* x\rangle >0$$ , for $$x\neq0$$. The last thing means that $$KerT^* =\{0\}$$ and therefor , $$({KerT^*})^\bot =H$$, which means that $$T$$ is surjective.

But I do have some considerations in case the dimension of $$H$$ is not finite:

1) Is the adjoint $$T^*$$ always defined ?

2) Why we have $$KerT^*\oplus({KerT^*})^\bot =H$$ ?

Propably the above hold as $$H$$ is Hilbert and $$T$$ is bounded, but I cannot understand how I can deduce those...

Any clarification or hint is really appreciated.

2) Because (see the FDVS book cited above for every italicized term) if $$U$$ is any subspace of an inner product space $$V$$, then $$V$$ is the direct sum of $$U$$ and the orthogonal complement of $$U$$. (For now, see the last bulleted item under "Inner Product Spaces, Properties" in the article Orthogonal complement.)
• Thanks a lot for your nice answer! As far as I can see that the linear operator $T$ is bounded, is irrelevant here to prove that it is surjeective ....Am I wrong? – dmtri Mar 26 '19 at 17:37
• After watching around some books in the net, at least the pages they let you see, about Hilbert spaces, I see that the boundedness is required for the existence of the $T^*$ , so you are right and thanks. – dmtri Mar 27 '19 at 9:27