Is the convolution operator with kernel $(1+|x|^2)^\frac{n}{2}$ bounded between Lebesgue spaces?

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Suppose we try to define a convolution operator with kernel $K(x) = (1+|x|^2)^\frac{n}{2}$.

For any measurable function $f$ we define

$T[f](x) = \int_{\mathbb R^n} f(x-y) (1+|x|^2)^\frac{n}{2} dy$

It is clear that this a bounded mapping $T : L^1(\mathbb R^n) \rightarrow L^\infty(\mathbb R^n)$. The kernel also happens to be a tempered distribution.

For which exponents $1 \leq p, q \leq \infty$ is this a bounded mapping $T : L^p(\mathbb R^n) \rightarrow L^q(\mathbb R^n)$.