Show that $T\neq{T^*}$ Let $V=P_2(\mathbb{R}), T\in \mathcal{L}(P_2(\mathbb{R})),$ where $T(p)=(a_1x)$.  Make $V$ an inner product space by defining
$$\langle p,q\rangle=\int_0^1{p(x)q(x)\,dx}$$
So I calculate $$\langle T(p),q\rangle=\int_0^1{(a_1b_0x+a_1b_1x^2+a_1b_2x^3)\,dx}$$
But now I have to get $(a_1b_0x+a_1b_1x^2+a_1b_2x^3)=p(x)T^*q(x)=(a_0+a_1x+a_2x^2)T^*q(x)$.  This will give me $T^*$, and then I just have to show that $T\neq{T^*}$ 
But I can't get $T^*$ this way, or if I can I'm just not seeing it.
 A: Following up my first comment above, trying to make two integrands equal (for all $x$) is an overconstrained problem.  If you could, then yes, that would be an easy path to finding the adjoint, but this shortcut is  not available here.
You have that $\langle Tp, q\rangle = \tfrac12 a_1 b_0 + \tfrac13 a_1 b_1 + \tfrac14 a_1 b_2$.  Whatever $T^*\!q$ is, it must satisfy $\langle p,T^*\!q\rangle = \tfrac12 a_1 b_0 + \tfrac13 a_1 b_1 + \tfrac14 a_1 b_2$.  In particular:
$$\langle 1, T^*\!q \rangle = \langle x^2, T^*\!q \rangle = 0,\quad \langle x, T^*\!q\rangle = \tfrac12 b_0 + \tfrac13 b_1 + \tfrac14 b_2.$$
The first two constraints imply that for any $q\in V$, $T^*\!q$ is a scalar multiple of the polynomial $3 - 16x + 15x^2$ (this isn't immediately obvious, but can you calculate why?).  Computing the behaviour of $T^*\!$ then amounts to seeing how the constant in front of $(3-16x+15x^2)$ depends on $b_0$, $b_1$ and $b_2$.
To some extent, the calculations in this problem look more complex because the natural basis $\{1,x,x^2\}$ is not orthogonal with respect to the chosen inner product $\int_0^1$.
