Adjoint of derivative operator on polynomial space I was working on a problem when I made the following reasoning.
I know that every linear operator $T:V \longrightarrow V$ on a Hilbert space $(V,\langle.,.\rangle)$ such that $\dim(V)<\infty$ has one (unique) adjoint operator $T^*:V \longrightarrow V$ (that is, $\langle T u,v\rangle = \langle u, T^* v \rangle$ $\forall u,v \in V$).
So if $V:=P_n$ is the space of all polynomials with degree less than or equal to $n \in \mathbb{N}$ (which gives $\dim(V)=n+1<\infty$) and $\langle f,g \rangle := \int_0^1f(t)g(t) \,  dt$, what is the adjoint of the derivative operator $T=\dfrac{d}{dt}$?
I've tried to solve that, but still to no avail. I wonder if that is a silly question, but I haven't had any success searching for the answer either, so I apologize in advance if that's the case.
 A: Let's try this when $n=2$:  An orthonormal basis emerges from the Gram--Schmidt process:
$$
f_0=1,\qquad f_1=2\sqrt{3}\left(x-\frac12\right), \qquad f_2=6\sqrt{5}\left(x^2 - x + \frac16\right)
$$
Now observe that $f_2'=6\sqrt{5}(2)\left(x-\frac12\right) = 2\sqrt{15}f_1$ and $f_1'= 2\sqrt{3} f_0$ and $f_0'=0$, so the matrix is
$$
\begin{bmatrix}
0 & 2\sqrt{3} & 0 \\ 0 & 0 & 2\sqrt{15} \\ 0 & 0 & 0
\end{bmatrix}.
$$
The matrix of the adjoint ought to be the transpose of this:
$$
\begin{bmatrix}
0 & 0 & 0 \\
2\sqrt{3} & 0 & 0 \\
0 & 2\sqrt{15} & 0
\end{bmatrix}.
$$
So $f_0 \mapsto 2\sqrt{3}\,f_1$ and $f_1\mapsto 2\sqrt{15}\, f_2$ and $f_2\mapsto 0$.
A: In this business, to find the (formal) adjoint of a differential operator, integrate by parts. If $T = \frac{d}{dt}$, then
\begin{align*} 
\langle Tf, g\rangle &= \int_0^1 f'(t)g(t)dt \\
&= f(t)g(t)\bigg|_0^1 - \int_0^1 f(t)g'(t)dt \\
&= f(t)g(t)\bigg|_0^1 - \langle f,Tg\rangle \\
&= \bigg( f(1)g(1) - f(0)g(0)\bigg) - \langle f,Tg\rangle 
\end{align*}
This is easier if you restrict to the space of polynomials which have $f(0) = f(1)$ (often, both zero); then, $T^* = -T$. Otherwise, as Daniel Fischer points out, you need an operator $B$ which has $\langle f, Bg \rangle = f(1)g(1) - f(0)g(0).$
A: Let us think matrix representations with the basis $S=\{1,t,...,t^n\}$. Say $T$ represents derivative operator and $M$ represents inner product.
$\langle Df,g \rangle = \langle f , D^*g\rangle$ where $D$ is derivative operator and $f$ and $g$ are any polynomials in $P^n(\mathbb{R})$.
Let $u$ and $v$ be representations of $f$ and $g$. (if $f(t)=5t^3+2$, then $u=(2,0,0,5)^T$). Then $\langle Df,g \rangle = \langle f , D^*g\rangle$ is equivalent to (for all $u,v \in \mathbb{R}^{n+1}$)
$$(Tu)^TMv=u^TMT^*v \iff u^TT^TMv = u^TMT^*v \iff T^*=M^{-1}T^TM$$
Note that $T^*$ represents $D^*$. After finding the $(n+1)\times(n+1)$ matrix $T^*$, $D^*$ will be found.
