MY NOTATIONS. $\mathcal{P}_3(\mathbb{R})$ is the vector space of all polynomials with real coefficients up to degree 3. $D^*$ is the adjoint operator of $D$ defined as the unique (since $\mathcal{P}_3(\mathbb{R})$ has finite dimension) operator that satisfies $\forall p,q \in \mathcal{P}_3(\mathbb{R}) \ \langle Dp,q \rangle =\langle p,D^*q \rangle$.
THE PROBLEM. Consider in $\mathcal{P}_3(\mathbb{R})$ the inner product
$$\langle p,q \rangle =\int_0^1 p(x)q(x) \ dx$$
and let $D$ be the derivation operator over $\mathcal{P}_3(\mathbb{R})$. Find $D^*$.
MY IDEAS. I know I can solve this by appealing to a orthonormal basis of $\mathcal{P}_3(\mathbb{R})$. If $\mathcal{B}=\{ v_1,...,v_4 \}$ is one of this kind, then $\forall p \in \mathcal{P}_3(\mathbb{R})$
$$ D^*=\sum_{i=1}^4 \langle Dv_i, p \rangle v_i$$
(where I used some known propositions and the fact that $\mathcal{P}_3(\mathbb{R})$ is a real vector space). Besides the fact that determining orthonormal basis is always rather boring, I find this approach exceedingly complicated. My professor gave me a hint that it is possible to work a solution using only the definition of adjoint operators – something I couldn’t manage to do. I tried to work with general polynomials $p$ and $q$ and the given inner product, but got nowhere. Does anyone has some idea to share?
By integration by parts, we get: $$\int_0^1 (D^* f)(x) g(x) \, dx = \int_0^1 f(x) g'(x) \, dx = f(1) g(1) - f(0) g(0) - \int_0^1 f'(x) g(x) \, dx.$$
So, the last term gives a contribution of $-f'$ to $D^* f$. For the other terms, you need to find $a$ and $b$ such that $\langle a, g \rangle = g(1)$ and $\langle b, g \rangle = g(0)$ for all $g \in \mathcal{P}_3(\mathbb{R})$, and then $D^*f = f(1) a - f(0) b - f'$. (Note that the condition on $b$ is equivalent to $\langle b, x^i \rangle = \delta_{i, 0}$ for $i=0,1,2,3$, and then by symmetry $a(x) = -b(1-x)$.)