Determine whether the linear operator $T(f) = f'$ (taking the derivative) has an adjoint or not.
Consider the inner product $\left<f,g\right> = \int_0^1 f(t)g(t)\ dt $ defined on the vector space $\mathbb{R}[x]$ of all real polynomials and let $T(f) = f'$. Now we need to decide whether $T$ has a adjoint or not
It can be checked that T is neither normal or self-adjoint, but I struggle in determining whether there is an adjoint or not. I also consider creating some examples but it is impossible to write out an orthonormal basis and plug in the formula since we're dealing with an infinite vector space...
*Thanks for pointing out my mistake. I have reworked the proof.
I came up with a rather simple method: consider $f = 1$ (or any constant) and $g=x^2$ we have
$$\langle T(f), x^2 \rangle = \int_{0}^{1}f'(t)x^2 dt = 0 = \langle f, T^*(x^2) \rangle = \int_{0}^{1}1 * T^*(x^2) dt $$
Now, observe that there exists F(x) where $ \frac{dF(t)}{dt} = T^*(x^2) $. $F(1)-F(0) = 0$.
Consider $\frac{dF(t^2)}{dt} = \frac{dF(t^2)}{dt^2} = T^*(x^4) * 2t$.
Now, $\langle T(x), x^4 \rangle = \int_{0}^{1}x^4dt = \frac{1}{5} $
But $\langle x, T^*(x^4) \rangle = \int_{0}^{1}t * T^*(x^4) dt = F(1)-F(0) =0$
This is a contradiction.