# 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.

• "I wonder if that is a silly question" definitely ... not. – Daniel Fischer Jul 18 '13 at 3:03
• 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 3:12
• $B = T - T*$ Because $[fg]_0^1 = <f',g> +<f,g'>$ – Henrique Tyrrell Jul 18 '13 at 3:15
• @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 3:20
• 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 3:25

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

• 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 3:35
• $B= \delta(t-1)-\delta(t)$ – Herman Jaramillo Mar 20 '17 at 5:55
• @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

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 $f_2\mapsto 0$.

• For $n = 3$, your ordered orthonormal basis extends to $\left\{1, \sqrt3(2x - 1), \sqrt5\left(6x^2 - 6x + 1\right), \sqrt7\left(20x^3 - 30x^2 + 12x - 1\right)\right\}$, which one gets from the Gram-Schmidt process starting with $\{1, x, x^2, x^3\}$. Then the matrix of the adjoint of the differentiation operator is$$\begin{bmatrix} 0 & 0 & 0 & 0\\ 2\sqrt3 & 0 & 0 & 0\\ 0 & 2\sqrt{15} & 0 & 0\\ 2\sqrt7 & 0 & 2\sqrt{35} & 0\end{bmatrix}.$$ – Maurice P Apr 17 at 18:06

Reading Neal's answer and the comments to your question, it appears there is no simple way to express $$D^*$$. Furthermore, Michael Hardy's answer and my comment are not satisfying because we are all accustomed to working in the standard basis $$\left\{1, x, x^2, x^3, \dots\right\}$$. Following is another answer, while not neat, at least uses the standard basis. I will first show the method for $$n = 3$$ and for polynomials over $$\mathbb R$$ and then explain how to generalize.

Starting with $$\langle f, D^*g \rangle = \langle Df, g \rangle$$, our goal is to express $$D^*g(x) = c_3 x^3 + c_2 x^2 + c_1 x + c_0$$ in terms of the coefficients $$b_k$$ of $$g$$.\begin{alignat*}{2} \langle f, D^*g \rangle =\ & \left\langle a_0 + a_1 x + a_2 x^2 + a_3 x^3, c_0 + c_1 x + c_2 x^2 + c_3 x^3\right\rangle &\quad \langle Df, g \rangle =\ & \left\langle a_1 + 2 a_2 x + 3 a_3 x^2, b_0 + b_1 x + b_2 x^2 + b_3 x^3\right\rangle\\ =\ & a_0(c_0 / 1 + c_1 / 2 + c_2 / 3 + c_3 / 4)\ +\\ & a_1(c_0 / 2 + c_1 / 3 + c_2 / 4 + c_3 / 5)\ + & =\ & a_1(1 b_0 / 1 + 1 b_1 / 2 + 1 b_2 / 3 + 1 b_3 / 4)\ +\\ & a_2(c_0 / 3 + c_1 / 4 + c_2 / 5 + c_3 / 6)\ + && a_2(2 b_0 / 2 + 2 b_1 / 3 + 2 b_2 / 4 + 2 b_3 / 5)\ +\\ & a_3(c_0 / 4 + c_1 / 5 + c_2 / 6 + c_3 / 7) && a_3(3 b_0 / 3 + 3 b_1 / 4 + 3 b_2 / 5 + 3 b_3 / 6)\end{alignat*} where we used an algebraic formula $$\langle f, g \rangle := \int_0^1 f(t)\overline{g(t)}\, dt = \sum_{j,k}\frac{a_j\overline{b_k}}{j + k + 1}$$ for the inner product and more algebraic manipulation to get the above expressions for $$\langle f, D^*g \rangle$$ and $$\langle Df, g \rangle$$. Because the $$a_j$$ are arbitrary, we can equate their coefficients to get a system of four equations, which we put in matrix form and solve using the inverse of the coefficient matrix on the left (see Why does the inverse of the Hilbert matrix have integer entries?): \begin{align*} \begin{bmatrix} 1/1 & 1/2 & 1/3 & 1/4\\ 1/2 & 1/3 & 1/4 & 1/5\\ 1/3 & 1/4 & 1/5 & 1/6\\1/4 & 1/5 & 1/6 & 1/7\end{bmatrix} \begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\end{bmatrix} & = \begin{bmatrix} 0 & 0 & 0 & 0\\ 1/1 & 1/2 & 1/3 & 1/4\\ 2/2 & 2/3 & 2/4 & 2/5\\ 3/3 & 3/4 & 3/5 & 3/6\end{bmatrix} \begin{bmatrix} b_0\\ b_1\\ b_2\\ b_3\end{bmatrix}\\ \begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\end{bmatrix} & = \begin{bmatrix} 16 & -120 & 240 & -140\\ -120 & 1200 & -2700 & 1680\\ 240 & -2700 & 6480 & -4200\\ -140 & 1680 & -4200 & 2800\end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0\\ 1/1 & 1/2 & 1/3 & 1/4\\ 2/2 & 2/3 & 2/4 & 2/5\\ 3/3 & 3/4 & 3/5 & 3/6\end{bmatrix} \begin{bmatrix} b_0\\ b_1\\ b_2\\ b_3\end{bmatrix}\\ \begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\end{bmatrix} & = \begin{bmatrix} -20 & -5 & -4 & -4\\ 180 & 60 & 58 & 60\\ -420 & -180 & -180 & -183\\ 280 & 140 & 140 & 140\end{bmatrix} \begin{bmatrix} b_0\\ b_1\\ b_2\\ b_3\end{bmatrix}.\end{align*} Thus,\begin{alignat*}{6} D^*g(x) =\ && -20b_0 &\ -\ & 5b_1 &\ -\ & 4b_2 &\ -\ & 4b_3\hphantom{)} && +\\ &(& 180b_0 &\ +\ & 60b_1 &\ +\ & 58b_2 &\ +\ & 60b_3) & x & +\\ &(& -420b_0 &\ -\ & 180b_1 &\ -\ & 180b_2 &\ -\ & 183b_3) & x^2 &\ +\\ &(& 280b_0 &\ +\ & 140b_1 &\ +\ & 140b_2 &\ +\ & 140b_3) & x^3.\end{alignat*}

To generalize to any finite $$n$$ and polynomials over $$\mathbb C$$, use the pattern that is easily discerned in the coefficient matrices in the first matrix equation above, and replace the $$c_k$$ and $$b_k$$ with $$\overline{c_k}$$ and $$\overline{b_k}$$.