TITLE: Infinite dimensional version of a simple fact on certain singular matrices
QUESTION [7 upvotes]: We consider the following simple fact about matrices. Then we try to generalize it in the context of smooth manifolds;
Let $L$ be the collection of all $n \times n$ real matrices $A=(a_{ij})$ with the following property:
$$\sum_{i=1}^{n} a_{ij}=0$$ for every fixed $j$.
Obviousely $L$ is a Lie algebra.(As I have already learned from Qiaochu Yuan in another MO post)
Moreover the linear map $X \mapsto AX$ has non trivial kernel.
These simple facts can be modelized in an infinite dimensional manner.(note that a vector $X\in\mathbb{R}^{n}$ can be considered as a function on a finite ($n$ pointed)set $M$ equiped  with discrete counting measure and a matrix is a function on $M\times M$. Now the product $AX$ has an integral representation if we replace $\sum$ by the integral sign. That is we read the expersion $\sum a_{ij}x_{j}$ in the integral form $\int_{M} a_{ij}x_{j}$ where the integration is based on the normalized counting measure.
Now we state our questions as generalization of the above simple fact about matrices.
Assume that $M$ is a compact orientable manifold or a Lie or topological group. So $M$ has a natural measure, correspond to volum form or the invariant metric or Haar measure.
Assume that $g: M \times M \to \mathbb{R}$ is a smooth function which satisfies $$\int_{M} g(x,y)dx=0\;\;\;\;\;(1) $$ for all $y \in M$.

Does the linear map $A$ on  $C^{\infty} (M)$ has nontrivial kernel?
  $$A(f)(x)=\int_{M} g(x,y)f(y)dy$$

Note that for topological groups we consider continuous functions, since smoothness is meaningles.
For our next question, we assume that $M$ is a symplectic manifold, so $M \times M$ has a natural symplectic structure. 
Let $L$ be the space of all smooth functions on $M\times M$ which satisfy the equation (1).

Is $L$ closed under Poisson bracket?

REPLY [4 votes]: Willie Wong answered the general case, and I'd like to give a counterexample for the symplectic case:
Let $M=S^2$, the standard 2-sphere embedded in $\mathbb{R}^3$ as the unit sphere, with the standard symplectic form corresponding to the Euclidean metric on $\mathbb{R}^3$. Take on $(M \times M, \Omega = \omega \oplus - \omega)$ the functions 
\begin{align}
f(z,w) &= z_3\cdot F(w),
\\
g(z,w) &= z_3 \cdot G(w)
\end{align}
both of which are in $L$. 
Now, $\{ f, g \} = \Omega(\mathrm{sgrad}(g), \mathrm{sgrad}(f)) =: \Omega'(\mathrm{d}f, \mathrm{d}g)$. Then,
\begin{align}
\{f,g\} 
&= (\omega' \oplus (-\omega')) \left( F(w)\mathrm{d} z_3 + z_3 \mathrm{d} F(w), G(w)\mathrm{d} z_3 + z_3 \mathrm{d} G(w) \right) \\
&= F(w)G(w) \omega'(\mathrm{d}z_3,\mathrm{d}z_3) - z_3^2 \omega'(\mathrm{d} F(w),\mathrm{d} G(w)) 
\\
&= z_3^2 \{G,F\}(w) =:h(z,w)
\end{align}
Now, unless $\{G,F\} \equiv 0$, $h$ will not be in $L$.