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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.

Michael Hardy
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Wheepy
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    "I wonder if that is a silly question" definitely ... not. – Daniel Fischer Jul 18 '13 at 03:03
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    That said, by partial integration, you get $\langle \frac{df}{dt},, g\rangle = [fg]^1_0 - \langle f,, \frac{dg}{dt}\rangle$, so it's $T^\ast = -T + B$, where $B$ is the operator that gets the boundary terms. I have no sleek idea how to determine that, other than computing it via an orthonormal base. – Daniel Fischer Jul 18 '13 at 03:12
  • $B = T - T*$ Because $[fg]_0^1 = <f',g> +<f,g'>$ – Henrique Tyrrell Jul 18 '13 at 03:15
  • @DanielFischer So that implies that $T^$ is "different" for each $n$? (By "different" I mean that the matrix for $T^$ in $P_n$ is not a submatrix of the matrix for $T^$ in $P_{n+1}$.) Edit: I'm saying this because I figure that the use of an orthonormal base -- which is different for each $n$ -- will affect $B$ in a manner that will make $T^$ "different" for every $n$. – Wheepy Jul 18 '13 at 03:20
  • No, not if you extend an ONB in $P_n$ to one in $P_{n+1}$, then the matrix of $T^\ast$ in $P_n$ is the $n\times n$ submatrix of $T^\ast$'s matrix in $P_{n+1}$. But you still need to figure out how $T^\ast$ resp. $B$ behaves for the new basis element. If you use an entirely different ONB for $P_{n+1}$ than for $P_n$, the matrices will probably not be as tightly connected. – Daniel Fischer Jul 18 '13 at 03:25
  • @DanielFischer Yeah, now I figure I messed up about the different bases part. =P – Wheepy Jul 18 '13 at 03:27
  • @DanielFischer For clarification of my doubt, could you please tell me how B is acting let say on polynomial? – Happy Dec 13 '18 at 18:09

3 Answers3

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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 $\langle f, Bg \rangle = f(1)g(1) - f(0)g(0).$

Neal
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    I liked your answer very much (and it's very well written), but I still think the form of $B$ is needed for it to be a complete answer... – Wheepy Jul 18 '13 at 03:35
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    $B= \delta(t-1)-\delta(t)$ – Herman Jaramillo Mar 20 '17 at 05:55
  • @HermanJaramillo What is delta? Could you please specify please? How delta act on a function! – Happy Dec 13 '18 at 17:59
  • @Happy : The delta operator acts on a function by the inner product operation. That is $\langle \delta(x-x_0), f(x) \rangle = f(x_0)$. It picks up the $x_0$ argument on $f$. $\delta$ goes from the space of real (or complex) functions to the space of real (or complex) numbers. – Herman Jaramillo Dec 27 '18 at 20:41
  • @HermanJaramillo This will not work for higher order derivatives, would it? For instance: $$\langle v, D^{\alpha_ie_i}u\rangle_{\Omega} = (-1)^{\alpha_i}\langle D^{\alpha_ie_i}v, u\rangle_{\Omega} +\sum_{\beta_i=0}^{\alpha_i-1} (-1)^{\beta_i}\langle n_iD^{\beta_ie_i}v, D^{(\alpha_i-\beta_i)e_i}u \rangle_{\partial\Omega},$$ where $e_i$ is the vector of zeros with $1$ at the $i$-th location, $\Omega\subset \mathbb{R}^d$, and $n$ is the outward normal on the boundary. The issue is that parts of the derivative operator remain on the second argument in the boundary terms. – lightxbulb Jul 15 '23 at 21:44
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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 $f_2\mapsto 0$.

Michael Hardy
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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.

Red Phoenix
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