Let $V$ be the vector space of the polynomials over $R$ of degree less than or
equal to $3$ with the inner product space $(f|g)=\int_{0} ^{1}f(t)g(t) dt$, and let $D$ be the differentiation operator on V. Find $D^*$
Attempt
As I did the calculation was very large, so I will explain what I did step by step so that you say if I made a mistake or I hit
STEP 1: First I considered a canonical basis of $R _{\leq 2} [X]$ that is $ \{ 1,x,x^2,x^3\}$. I did the Gram-Schmidt Process to orthogonalize this base. Resulted in
$$ \left\{ 1, x-\frac{1}{2}, x^2-x+ \frac{1}{6}, x^3-\frac{3}{2}x^2+\frac{3}{5}x- \frac{1}{2} \right\}$$
STEP 2:
Orthonormalize this base, dividing each element by its norm (in this part I may have been wrong because they gave many calculations). Resulted in
$$ \{ 1, \sqrt{12} \left(x-\frac{1}{2} \right) , \sqrt{180} \left( x^2-x+ \frac{1}{6} \right) , \sqrt{\frac{33600}{18772}} \left( x^3-\frac{3}{2}x^2+\frac{3}{5}x- \frac{1}{2} \right) \}$$
STEP 3: Write the matrix of $D$
$$\begin{bmatrix}0&\sqrt{12}&0& \frac{1}{10} \sqrt{\frac{33600}{18772}}\\0&0&2\sqrt{15}& 0\\ 0&0&0& \sqrt{\frac{33600}{33378960}} \\ 0&0&0& 0 \end{bmatrix}$$
STEP4: The matrix of $D^*$ is the conjugate transpose of the matrix of $D$,(in this case only transpose) according to the corollary:
Let $V$ be a finite-dimensional inner product space, and let
$T$ be a linear operator on $V$. In any orthonormal basis for $V$, the matrix of $T^*$ is the conjugate transpose of the matrix of $T$.
At least the idea is correct? Is there any easier way to do this exercise?
Best Answer
Yes, the idea is correct (I didn't verify the coefficients in the orthonormalization).
Here's an alternative method: Integrating by parts gives $$\langle f, Dg \rangle = \int_0^1 f g' dx = f(1) g(1) - f(0) g(0) + \int_0^1 - f' g \,dx = f(1) g(1) - f(0) g(0) + \langle -Df, g \rangle.$$ Thus, if we can find an operator $S$ such that $\langle S f, g \rangle = f(1) g(1) - f(0) g(0)$, we will have $$\langle f, D g \rangle = \langle (S - D) f, g \rangle$$ and thus $$D^* = S - D .$$
Remark Up to this point, the formulation works for all functions in $L^2([0, 1])$, and if we enlarge our space to include the Dirac delta function, by definition we can write $S = \delta(t - 1) - \delta(t)$ and thus $D^* = -D + \delta(t - 1) - \delta(t) .$)
To find an explicit formula for $S$, we look for polynomials $r, s \in V$ such that $$\langle S f , g \rangle = \langle f(1) r - f(0) s , g \rangle$$ for all $f , g \in V$. On the one hand, $\langle S f, g \rangle = \langle D f, g \rangle + \langle f, D g \rangle$, and except for $k = l = 0$, this gives $\langle S(x^k), x^l \rangle = 1$. So, writing $r = \sum_{i = 0}^3 r_i x^i$ and $s = \sum_{i = 0}^3 s_i x^i$ and integrating against monomials $f = x^k, g = x^l$ gives (for $k > 0$) $$1 = \langle S (x^k), x^l \rangle = \langle r, x^l \rangle .$$ Writing this in terms of $r_0, r_1, r_2, r_3$ for $l = 0, 1, 2, 3$ and integrating gives rise to the system $$\pmatrix{1&\frac{1}{2}&\frac{1}{3}&\frac{1}{4}\\\frac{1}{2}&\frac{1}{3}&\frac{1}{4}&\frac{1}{5}\\\frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\frac{1}{6}\\\frac{1}{4}&\frac{1}{5}&\frac{1}{6}&\frac{1}{7}} \pmatrix{r_0\\r_1\\r_2\\r_3} = \pmatrix{1\\1\\1\\1} .$$ Solving gives $$r_0 = -4, r_1 = 60 , r_2 = -180, r_3 = 140 ,$$ so $r(x) = 4 (35 x^3 - 45 x^2 + 15 x - 1)$. A similar argument (and using that $\langle S(1), 1 \rangle = 0$) gives that $s(x) = -4 (35 x^3 - 60 x^2 + 30 x - 4)$. Putting everything together gives $$\boxed{D^* f = -D f + 4 [f(1) (35 x^3 - 45 x^2 + 15 x - 1) - f(0) (35 x^3 - 60 x^2 + 30 x - 4)]} .$$
The matrix on the left-hand side is the $4 \times 4$ Hilbert matrix; there is a general formula for the inverse of the analogous $n \times n$ matrix, which means we can write down an explicit formula for the adjoint for the differentiation operator on the space of polynomials of degree $\leq n$ for general $n$.