= Solution
Taking the <Fourier transform> of the distributional equation gives
$$
(1+|p|^{4044})\widehat u(p)=\widehat f(p),
$$
because $2022$ is even. The multiplier therefore gives the <regularity gain for one plus an even power of the Laplacian>
$$
u\in H^{r+4044}(\mathbb R^n).
$$
The <Sobolev embedding theorem> in its derivative form says that
$$
H^s(\mathbb R^n)\hookrightarrow C^k
\quad\text{when}\quad s>k+\frac n2.
$$
To make all derivatives through order $4044$ continuous, and simultaneously make $f\in H^r$ continuous, it is enough and in general necessary to have
$$
r>\frac n2.
$$
Under this condition the distributional identity is a pointwise identity, so $u$ is a classical solution. Thus
$$
\boxed{n<2r.}
$$
At the borderline $n=2r$, the standard Sobolev embedding does not in general give continuity.
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