For the second one, you're almost there. If the inner product is given by , and , then . But this is the same as (since ). So T is selfadjoint.
The first one is a good deal more complicated than I first thought, and I wonder if there is something wrong with the question. My original idea was to use integration by parts to express in terms of . But when you do the integration by parts, with the limits 0 and 1, you actually get . The awkward terms f(1)g(1)-f(0)g(0) prevent you from getting a neat expression for the adjoint of the differentiation operator. That seems to have the effect of making this question unreasonably difficult.
I asked my professor, and he said we should use this theorem:
T is normal if and only if there exist an orthonormal basis for V consisting of eigenvectors of T.
So, to find eigenvectors of T, I have So I need to find some polynomials f, are there any?
the standard ordered basis of V is
So the matrix
So the eigenvalue of T is 0,1.
Well, then any polynomial to the first power would satisfy T, right?
To use the theorem here, you must first show that the complex space does not have a base consisting of eigenvectors of T, and then deduce that since T is not normal as an operator on that space it must also fail to be normal when we restrict to real scalars.