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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 $langle f,grangle=d_f(g)_0$ 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 $g$, $langle h,m_f(g)rangle=langle d_f(h)=0,grangle=0$, 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?