Finding the adjoint of a differentiation map The integral inner product is defined as 
$$\langle p(x),q(x) \rangle = \int_{-1}^1 p(t)q(t)dt$$ 
on both $\textsf{P}_2(\mathbb{R})$ and $\textsf{P}_1(\mathbb{R})$. 
Find the adjoint of the differentiation map 
$$\begin{align}
\textsf{T} : \textsf{P}_2(\mathbb{R}) & \to \textsf{P}_1(\mathbb{R}) \\
p(x) & \mapsto p'(x)
\end{align}$$
Any help on finding the adjoint of $\textsf T$ above is appreciated.
 A: Let $p\in P_2(\Bbb R)$ and $q\in P_1(\Bbb R)$ so that
\begin{align*}
p(t) &= a_{2} t^{2} + a_{1} t + a_{0} & q(t) &= b_{1} t + b_{0}
\end{align*}
Then
$$
\langle Tp, q\rangle
= \int_{-1}^1 p^\prime(t)\cdot q(t)\,dt
= 2 \, a_{1} b_{0} + \frac{4}{3} \, a_{2} b_{1}
$$
We may also write $(T^\ast q)(t)=c_{2} t^{2} + c_{1} t + c_{0}$ so
$$
\langle p, T^\ast q\rangle
= \int_{-1}^1 p(t)\cdot\{c_{1} t + c_{0}\}\,dt
= 2 \, a_{0} c_{0} + \frac{2}{3} \, a_{2} c_{0} + \frac{2}{3} \, a_{1} c_{1} + \frac{2}{3} \, a_{0} c_{2} + \frac{2}{5} \, a_{2} c_{2}
$$
Since $\langle Tp, q\rangle = \langle p, T^\ast q\rangle$, we can compare our two expressions and obtain
\begin{align*}
0
&= \langle Tp, q\rangle - \langle p, T^\ast q\rangle \\
&= 2 \, a_{1} b_{0} + \frac{4}{3} \, a_{2} b_{1} - 2 \, a_{0} c_{0} - \frac{2}{3} \, a_{2} c_{0} - \frac{2}{3} \, a_{1} c_{1} - \frac{2}{3} \, a_{0} c_{2} - \frac{2}{5} \, a_{2} c_{2} \\
&= a_0\cdot\left\{-2\,c_0-\frac{2}{3}\,c_2\right\}+a_1\cdot\left\{2\,b_0-\frac{2}{3}\,c_1\right\}+a_2\cdot\left\{\frac{4}{3}\,b_1-\frac{2}{3}\,c_0-\frac{2}{5}\,c_2\right\}
\end{align*}
This gives the system of equations $A\vec{x}=\vec{b}$ where
\begin{align*}
A &= \left[\begin{array}{rrr}
-2 & 0 & -\frac{2}{3} \\
0 & -\frac{2}{3} & 0 \\
-\frac{2}{3} & 0 & -\frac{2}{5}
\end{array}\right] & \vec{x} &= \left[\begin{array}{r}
c_{0} \\
c_{1} \\
c_{2}
\end{array}\right] & \vec{b} &= \left[\begin{array}{r}
0 \\
-2 \, b_{0} \\
-\frac{4}{3} \, b_{1}
\end{array}\right]
\end{align*}
Solving this system gives
\begin{align*}
c_0 &= -\frac{5}{2} \, b_{1} & c_1 &= 3 \, b_{0} & c_2 &= \frac{15}{2} \, b_{1}
\end{align*}
So, our formula for $T^\ast$ is
$$
T^\ast(b_{1} t + b_{0})
= -\frac{5}{2} \, b_{1}
+3 \, b_{0}\,t
+\frac{15}{2} \, b_{1}\,t^2
$$
