The first item is a direct calculation. An convenient way to see this is to use the following two facts

1. (1)

   A homogeneous polynomial degree $k$ polynomial restricts to an eigenfunction of the Hodge-Laplacian $\Delta_{S^{2}}$ on the unit sphere, with eigenvalue $k(k+1)$.

2. (2)

   Given an eigenfunction $h$ of $\Delta_{S^{2}}$ on the unit sphere with eigenvalue $k$, for any $l$, we can extend $h$ to a homogeneous function $h_{l}$ on $\mathbb{R}^{3}\setminus\{0\}$ of degree $l$, and

   (3.162)

   $$\Delta_{0}h_{l}=r^{-2}(k-l(l+1))h_{l}$$

For the second item it is possible to write down an explicit inverse to $\Delta_{0}$. Here we provide a quick abstract proof. First we notice $\square:\mathcal{P}_{4}\to\mathcal{P}_{4}$ is a well-defined linear map. This follows from the standard computations

Since each term in the above formula is a polynomial in $\mathcal{P}_{4}$, so $\square f\in\mathcal{P}_{4}$.

Now to prove $\square$ is an isomorphism it suffices to prove it has a trivial kernel in $\mathcal{P}_{4}$. Let $u\equiv r^{-3}f$, then $u=O(r)$ for both $r\to 0$ and $r\to\infty$. If $\Delta_{0}(u)=0$, then $u$ is harmonic on $\mathbb{R}^{3}\setminus\{0\}$. The removable singularity theorem implies that $u$ extends smoothly on $\mathbb{R}^{3}$. Since $u=O(r)$ as $r\to\infty$, applying the standard derivative estimate for harmonic functions, we conclude $\nabla^{2}u\equiv 0$. Therefore, $u$ must be a linear function. Noticing $f\in\mathcal{P}_{4}$, we conclude $f\equiv 0$. The proof is done.

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