What is the adjoint of this derivative operator?

Let $V$ be the vectorspace $\mathbb{C}[X]_{\leq 2}$ of polynomials of degree at most $2$. Consider the inner product $$\langle a_0 + a_1 x + a_2 x^2 , b_0 + b_1 x + b_2 x^2 \rangle = a_0 \overline{b}_0 + a_1 \overline{b}_1 + a_2 \overline{b}_2.$$

I'm being asked to find the adjoint of the derivative operator $$D: \mathbb{C}[X]_{\leq 2} \to \mathbb{C}[X]_{\leq 2}: f \mapsto f'.$$ By definition, the adjoint operator must satisfy $$\langle D(v), w \rangle = \langle v, D^{*}(w) \rangle$$ for all $v, w \in \mathbb{C}[X]_{\leq 2}$. I let $v = a_0 + a_1 x + a_2 x^2$ and $w = b_0 + b_1 x + b_2 x^2$. Then $$\langle D(v), w \rangle = \langle a_1 + 2a_2 x, b_0 + b_1 x + b_2 x^2 \rangle = a_1 \overline{b}_0 + 2a_2 \overline{b}_1.$$ Now, in order to find $D^{*}$, this must equal $$\langle v, D^{*}(w) \rangle.$$ But how can I compute this scalar product, when I don't know what $D^{*}$ does to $w$ yet? I have $D^{*}(w) = D^{*}(b_0 + b_1 x + b_2 x^2) = b_0 + b_1 D^{*}(x) + b_2 D^{*}(x^2)$ assuming linearity. But how can I work this out further?

Help/suggestions are appreciated!

2 Answers

You almost have it!

Define $c_0+c_1 x+ c_2 x^2=D^*(b_0+b_1x+b_2x^2)$. We have $$a_1\bar{b}_0+2a_2\bar{b}_1 = \left<D(v),w\right> = \left<v,D^*(w)\right> = a_0 \bar{c}_0 + a_1 \bar{c}_1 + a_2 \bar{c}_2$$ Thus, $c_0=0$, $c_1=b_0$, $c_2=2b_1$, which means that $$D^*(b_0+b_1x+b_2x^2)=b_0x+2b_1x^2$$

Note: $D^*$ can be written in terms of $D$. For any $p(x)=a_0+a_1x+a_2x^2$: $$D^*(p(x))=xD(a_0x+a_1 x^2)=xD(xp(x)\,\operatorname{mod} x^3)$$

You can calculate $D^\ast$ in a basis.

Describe $D$ in the Basis $(1,x,x^2)$ with a Matrix $A$. Then you get $$A=\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}.$$ And so you can discribe $D^\ast$ in the same basis by $$A^\ast=A^T=\begin{pmatrix}0&0&0\\1&0&0\\0&2&0\end{pmatrix}.$$

So you get $$D^\ast(b_0+b_1x+b_2x^2)=b_0x+2b_1x^2.$$