Riesz isomorphism and dual map 
Let $V=\mathbf{R}[X]_{\leqslant 1}$ be equipped with inner product $\langle f,g\rangle=\int_{[-1,1]} f(x)x^2 g(x)\,dx$. Let $J:V\to V^*:u\mapsto\ell_u$ where $\ell_u(x)=\langle u,x\rangle$ be the Riesz isomorphism.
Let $\phi:V\to V:f\mapsto \frac{df}{dx}$ and let $\phi^*$ be the dual mapping. Show that $J^{-1}\phi^* J$ is the mapping $f(x)\mapsto \frac56 x\int_{[-1,1]} f(x)\,dx$.

Let $f\in V$, then $$(J^{-1}\phi^* J)(f)=J^{-1}[\phi^*(\ell_f)]=J^{-1}[\ell_f\circ \phi],$$ which is the Riesz representant of $\ell_f\circ \phi$. Let $\beta=\{1,X\}$ be a basis for $V$ and $\beta^*$ be the corresponding dual basis. Then the coordinate vector $x=\operatorname{co}_{\beta}(J^{-1}[\ell_f\circ \phi])$ satisfies $$ J_{\beta}^{\beta*}x=\operatorname{co}_{\beta*}(\ell_f\circ \phi)$$
where $J_{\beta}^{\beta*}=\begin{bmatrix}\langle 1,1\rangle & \langle X,1\rangle \\ \langle 1,X\rangle & \langle X,X\rangle \end{bmatrix}=\begin{bmatrix} \frac23 & \\ & \frac25\end{bmatrix}$and $\operatorname{co}_{\beta*}(\ell_f\circ \phi)=\begin{bmatrix}(\ell_f\circ \phi)(1) \\ (\ell_f \circ \phi)(X) \end{bmatrix}=\begin{bmatrix} 0 \\ \int_{-1}^{1}f(x)x^2dx\end{bmatrix}$
This gives after solving $x=\begin{bmatrix}0 \\ \frac{5}{2}\int_{-1}^1 f(x)x^2\,dx \end{bmatrix}$ and thus $J^{-1}[\ell_f\circ \phi]=\frac52 x\int_{[-1,1]} f(x)x^2dx$, which is not the right answer.
I can't seem to find my mistake. Where did I go wrong?
 A: Recall the adjoint map $\phi' : V \to V$ (usually also denoted by $\phi^*$) defined as the unique linear map such that $$\langle \phi(p), q\rangle = \langle p, \phi'(q)\rangle, \forall p,q \in V$$
Now for $p \in V$ we have
$$(J^{-1}\phi^*J)(p) = J^{-1}\phi^*(\langle \cdot, p\rangle) = J^{-1} (\langle \cdot, p\rangle \circ \phi) = J^{-1}(\langle \phi(\cdot), p\rangle) = J^{-1}(\langle\cdot, \phi'(p)\rangle) = \phi'(p)$$
so $J^{-1}\phi^*J = \phi'$.
For $ax + b, cx + d \in V$ we have
\begin{align}
\langle \phi(ax + b), cx + d\rangle &= \langle a, cx + d\rangle \\
&= \int_{[-1,1]} a(cx + d) x^2\,dx\\
&= ad \int_{[-1,1]} x^2\,dx\\
&= \frac23 ad\\
&= \frac53 \int_{[-1,1]} ad \cdot x^4\,dx\\
&= \int_{[-1,1]} (ax + b) \left( \frac53 d\cdot x\right) \cdot x^2\,dx\\
&= \left\langle ax + b, \frac53d\cdot x\right\rangle
\end{align}
while using that $\int_{[-1,1]} f = 0$ when $f$ is odd. We conclude that $$\phi'(ax + b) = \frac53 b\cdot x = \frac56 \int_{[-1,1]} (ax + b)\,dx$$
Alternatively, notice that $B = \left\{\sqrt{\frac32}, \sqrt{\frac52}x\right\}$ is an orthonormal basis for $V$ and the matrix of $\phi$ with respect to $B$ is $\pmatrix{0 & \sqrt{\frac53} \\ 0 & 0}$. Its conjugate transpose is $\pmatrix{0 & 0 \\ \sqrt{\frac53} & 0}$, which represents $\phi'$ in the same basis $B$. The formula $\phi'(ax + b) = \frac53 b\cdot x$ readily follows. 
