Let $n \in \mathbb{N}$. For the following, consider the vector space of polynomials of degree less than or equal to $n$ with complex coefficients, $\mathcal{P}_n(\mathbb{C})$.
a) If we equip the space with the inner product defined by $$<p,q> = \int^1 _0 p(t)\overline{q(t)} dt,$$ is the differentiation operation $\mathcal{D}$, self-adjoint? Explain.
b) Prove that $$<p,q> = \sum^n _{j=0} p \left( \frac{j}{n} \right)\overline {q \left(\frac{j}{n}\right)}$$ defines an inner product on $\mathcal{P}_n(\mathbb{C}).$
I am pretty lost on this problem so if anybody could help point me at the right answer and offer some guidance it would be greatly appreciated. I have three more like this and would like to use this one as the a base to work through the others on my own.
For part (a), if $\mathcal{D}$ is self-adjoint then we must have $$ \langle \mathcal{D}p,q\rangle=\langle p,\mathcal{D}q\rangle $$ for all $p,q\in\mathcal{P}_n$.
But if we take $p(x)=x$ and $q(x)=x^2$, then $$\langle \mathcal{D}p,q\rangle=\int_0^1x^2\;dx=\frac{1}{3}$$ while $$\langle p,\mathcal{D}q\rangle=\int_0^1x\cdot 2x\;dx=\frac{2}{3} $$ so $\mathcal{D}$ is not self-adjoint.
For part (b), it's fairly clear that $\langle p,q\rangle$ is symmetric, linear in the first variable, and conjugate-linear in the second. Moreover, $$ \langle p,p\rangle=\sum_{j=0}^n\Big|p\Big(\frac{j}{n}\Big)\Big|^2\geq 0$$ and if $\langle p,p\rangle =0$ then $p(\frac{j}{n})=0$ for $0\leq j\leq n$. Since $p$ is a polynomial of degree at most $n$, if it is zero at $n+1$ points then it must be the zero polynomial. So $\langle \cdot,\cdot\rangle$ is positive-definite.