Kernel of certain differential operators Let $U \subset \mathbb C^2$ be a domain and $f : U \to \mathbb C$ smooth (not necessarily holomorphic). Let $f$ further be in the kernel of the following differential operators.
$$\frac{\partial^2}{\partial x_i \partial x_j} + \frac{\partial^2}{\partial y_i \partial y_j}$$
and
$$\frac{\partial^2}{\partial x_i \partial y_j} - \frac{\partial^2}{\partial x_j \partial y_i}$$
for $i,j \in \{1,2\}$. What is the solution space of $f$? Is it only the polynomials of degree less than 2?.
Motivation: This should be just the kernel of $\partial \overline{\partial}$ sending $0$-forms to $(1,1)$-forms.
 A: Your guess is correct. A complex-valued function is in the kernel of all these operators if and only if it's in the kernel of $\partial\bar\partial$ (which is equivalent to its real and imaginary parts being pluriharmonic, meaning their restrictions to every complex line are harmonic, or equivalently each is locally the real part of a holomorphic function).
Here's a proof. (This works in $\mathbb C^n$ for any $n$, not just $\mathbb C^2$.)
First let's note that all of your differential operators take real functions to real functions and imaginary ones to imaginary ones, so a complex-valued function is in their common kernel if and only if its real and imaginary parts are. Similarly, the operator $i\partial\bar\partial$ takes real functions to real $2$-forms and imaginary ones to imaginary ones (this is an easy consequence of the fact that $\bar\partial\partial = - \partial\bar\partial$), so $f$ is in the kernel of $\partial\bar\partial$ if and only if its real and imaginary parts are. Thus it suffices to assume $f$ is real and show that $\partial\bar\partial f = 0$ if and only if $f$ is in the kernel of all your operators.
In holomorphic coordinates $(z^1,\dots,z^n)$, we have
$$
\partial\bar\partial f = \sum_{j,k=1}^n \frac{\partial^2 f}{\partial z^j \partial \overline{z^k}} \, dz^j\wedge d\overline {z^k}.
$$
Since the differential forms $dz^j\wedge d\overline{z^k}$ are all linearly independent over $\mathbb C$,
$$
\partial\bar\partial f = 0 \quad \iff \quad 
\frac{\partial^2 f}{\partial z^j \partial \overline{z^k}} =0 \text{ for all $j,k$}.
$$
Now expand these operators thus:
$$
\frac{\partial }{\partial z^j } = \frac 1 2 \left( \frac{\partial}{\partial x^j} - i \frac{\partial}{\partial y^j}\right),
\qquad 
\frac{\partial }{\partial \overline{z^k} } = \frac 1 2 \left( \frac{\partial}{\partial x^k} + i \frac{\partial}{\partial y^k}\right).
$$
Some simple algebra then yields
$$
\partial\bar\partial f = 0 
\quad \iff \quad
\forall j,k,\ 
\left(
\frac{\partial^2 f}{\partial x^j \partial x^k} + 
\frac{\partial^2 f}{\partial y^j \partial y^k} \right)
+ i \left(
\frac{\partial^2 f}{\partial x^j \partial y^k} -
\frac{\partial^2 f}{\partial y^j \partial x^k} \right)=0.
$$
Since we're assuming $f$ is real, this is true if and only if $f$ is in the
common kernel of your operators.
