In the frequency domain, $G(f) H(f) = G(f)$ if $H$ is a brickwall filter with a higher bandwidth than $G$.
I want to show this in the time domain.
My attempt: The time domain representation would be
$$ \int g(\tau) \operatorname{sinc}\left(\frac{t-\tau}{T}\right) d\tau = g(t) , $$
where the sinc is the ideal brickwall filter with bandwidth $1/(2T)$.
By definition, if $g$ is bandlimited to $1/(2 T_g)$, it can be expanded using the WKS interpolation formula, having the samples as coefficients:
$$ g(t) = \sum_{n=-\infty}^{\infty} g(n T_g) \operatorname{sinc}\left(\frac{t-n T_g}{T_g}\right) = \sum_{n-\infty}^{\infty} g[n] \operatorname{sinc}\left(\frac{t-n T_g}{T_g}\right) $$
Plugging this into the original equation:
$$ \int \sum_{n=-\infty}^{\infty} g(n T_g) \operatorname{sinc}\left(\frac{\tau-n T_g}{T_g}\right) \operatorname{sinc}\left(\frac{t-\tau}{T}\right) d\tau \\ = \sum_{n=-\infty}^{\infty} g(n T_g) \left[ \int \operatorname{sinc}\left(\frac{\tau-n T_g}{T_g}\right) \operatorname{sinc}\left(\frac{t-\tau}{T}\right) d\tau \right] $$
From this, the term in bracket would need to be equivalent to $\operatorname{sinc}\left(\frac{t-n T_g}{T_g}\right)$ but only if $T \leq T_g$!
I tried calculating this integral using sines and exponentials but the expressions just get messier. I don't see this working out.