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.
Best Answer
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 $$ T^\ast(b_{1} t + b_{0}) = -\frac{5}{2} \, b_{1} +3 \, b_{0}\,t +\frac{15}{2} \, b_{1}\,t^2 $$