This problem is about the space $V$ of real polynomials in the variables $x$ and $y$. If $f$ is a polynomial, $d_f$ will denote the operator $f(d/dx,d/dy)$ , and $d_f(g)$ will denote the result of applying this operator to a polynomial g.
(a) The rule $\langle f,g\rangle=d_f(g)_0$ defines a bilinear form on $V$, the subscript 0 denoting evaluation of a polynomial at the origin. Prove that this form is symmetric and positive definite, and that the monomials $x^iy^j$ form an orthogonal basis of $V$ (not an orthonormal basis).
(b) We also have the operator of multiplication by $f$ , which we write as $m_f$. So $m_f(g) = fg$. Prove that $d_f$ and $m_f$ are adjoint operators.
(c ) When $f = x^2 + y^2$, the operator $d_f$ is the Laplacian, which is often written as $\Delta$. A polynomial $h$ is harmonic if $\Delta h = 0$. Let $H$ denote the space of harmonic polynomials. Identify the space $H^\perp$ orthogonal to $H$ with respect to the given form.
\ \ \
I have solved part a, and am currently stuck on part b. Here is what I have so far:
Let us express the result of the operations in terms of the monomial basis given in the previous section.
$d_f(g)=((\sum_{i,j}a_{i,j}(\frac{d}{dx})^i(\frac{d}{dx})^j)*(\sum_{i,j}b_{i,j}x^iy^j))=\sum_{k,l} \frac{k!l!}{i!j!}a_{k-i,l-j}b_{k,l}x^iy^j$ thus we have $c_{i,j}=\sum_{k,l} \frac{k!l!}{i!j!}a_{k-i,l-j}b_{k,l}$, and if we order the monomials be increasing order of degree of each variable with the $y$ degrees having higher weight, we can write the transformation as the following:
$$\begin{pmatrix} a_{0,0}&&\frac{1!0!}{0!0!}a_{1,0}&&...&&i!j!a_{i,j}\\ 0&&a_{0,0}&&...&&i!j!a_{i-1,j}\\ .&&.&&...&&.\\ 0&&0&&...&&\frac{i!j!}{k!l!}a_{i-k,j-l}\\ .&&.&&...&&.\\ 0&&0&&...&&a_{0,0}\\ \end{pmatrix}* \begin{pmatrix} b_{0,0}\\ b_{1,0}\\ .\\ b_{0,1}\\ .\\ b_{i,j}\\ \end{pmatrix}= \begin{pmatrix} c_{0,0}\\ c_{1,0}\\ .\\ c_{0,1}\\ .\\ c_{i,j}\\ \end{pmatrix}$$
Note that all entries of the matrix are multiplied by the factorial of numbers denoting their column and divided by factorials representing their row.
$m_f(g)=(\sum_{i,j}a_{i,j}x^iy^j)*(\sum_{i,j}b_{i,j}x^iy^j)=\sum_{k,l}a_{i-k,j-l}b_{k,l}x^iy^j$ thus we have $c^*_{i,j}=\sum_{k,l}a_{i-k,j-l}b_{k,l}$, and if we order the monomials as before, we can write the transformation as the following:
$$\begin{pmatrix} a_{0,0}&&0&&...&&0\\ a_{1,0}&&a_{0,0}&&...&&0\\ .&&.&&...&&.\\ a_{k,l}&&a_{k-1,l}&&...&&0\\ .&&.&&...&&.\\ a_{i,j}&&a_{i-1,j}&&...&&a_{0,0}\\ \end{pmatrix}* \begin{pmatrix} b_{0,0}\\ b_{1,0}\\ .\\ b_{0,1}\\ .\\ b_{i,j}\\ \end{pmatrix}= \begin{pmatrix} c_{0,0}\\ c_{1,0}\\ .\\ c_{0,1}\\ .\\ c_{i,j}\\ \end{pmatrix}$$
Note that in this matrix there are no coefficients, but the $a$ entries are equal to the analogous transposed $a$ entries in the matrix of the $d_f$ operation.
Consider the basis of monomials $\frac{x^iy^j}{i!j!}$. The new transformation for $d_f$, where the $c$ values represent the new coefficients, is
$$\begin{pmatrix} a_{0,0}&&a_{1,0}&&...&&a_{i,j}\\ 0&&a_{0,0}&&...&&a_{i-1,j}\\ .&&.&&...&&.\\ 0&&0&&...&&a_{i-k,j-l}\\ .&&.&&...&&.\\ 0&&0&&...&&a_{0,0}\\ \end{pmatrix}* \begin{pmatrix} 0!0!b_{0,0}\\ 1!0!b_{1,0}\\ .\\ 0!1!b_{0,1}\\ .\\ i!j!b_{i,j}\\ \end{pmatrix}= \begin{pmatrix} c_{0,0}\\ c_{1,0}\\ .\\ c_{0,1}\\ .\\ c_{i,j}\\ \end{pmatrix}$$
This is where I am stuck. Clearly, the matrix of $d_f$ with respect to this orthonormal basis is equal to the adjoint of the matrix of $m_f$ with respect to the original orthogonal basis. However, I believe they need to be adjoint with respect to the same basis, and writing the transformation for $m_f$ in terms of the orthonormal basis, the matrix changes and is no longer the transpose. I feel like I must have messed up somewhere, but I've been at this for about 4 hours and haven't gotten anywhere.
Separately, looking at part c, I can see that for any function $g$, $\langle h,m_f(g)\rangle=\langle d_f(h)=0,g\rangle=0$, so I believe the answer would be all polynomials that have $x^2+y^2$ as a factor. This may be because I am very tired, but I have no idea how I might prove these are the only such polynomials; any hints?