= Solution
Fix $t,x$. The <Cauchy-Schwarz inequality> in the integration variable $v_*$ gives
$$
|K_tg(x,v)|^2
\leq\left(\int_{\mathbb R^d}|k(t,x,v,v_*)|^2\,dv_*\right)
\left(\int_{\mathbb R^d}|g(x,v_*)|^2\,dv_*\right).
$$
Integrate first in $v$, then in $x$. The uniform kernel hypothesis bounds the first factor after velocity integration by $C^2$, independently of $t,x$. Hence
$$
\boxed{\|K_tg\|_{L^2_{x,v}}^2\leq C^2\|g\|_{L^2_{x,v}}^2,\qquad \|K_t\|\leq C.}
$$
The <linear Boltzmann collision operator> is linear by its integral definition, so this proves it is a <bounded linear operator> on $L^2_{x,v}$. The estimate is the <Hilbert-Schmidt kernel bound> applied at each spatial point. The printed real-kernel square is $|k|^2$ for a complex kernel. Measurability of the coefficients is understood so that the displayed integrals are defined.
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