Let $K\in L^1(\mathbb{R}^d)$ with Lebesgue measure. Suppose that $\psi_n\in L^2(\mathbb{R}^d)$ is a sequence of functions such that $\psi_n\to \psi$ weakly in $L^2$ and $\psi_n\equiv 0$ when $|x|>1$
Show that $f_n(x) = \int_{\mathbb{R}^d}K(x-y)\psi_n(y)\ dy$ converges to $f(x) = \int_{\mathbb{R}^d}K(x-y)\psi(y)\ dy$ strongly in $L^2(\mathbb{R}^d)$
My Attempt: $$\int |f_n(x) - f(x)|^2\ dx \leq \int_{\mathbb{R}^d}(\int_{\mathbb{R}^d}|K(x-y)|^{1/2+1/2}|\psi_n(y)-\psi(y)|\ dy)^2\ dx \leq \int_{\mathbb{R}^d}(\int_{\mathbb{R}^d}|K(x-y)|\ dy)(\int_{\mathbb{R}^d}|K(x-y)||\psi_n(y)-\psi(y)|^2\ dy)\ dx\leq\\ ||K||_{L^1}^2\int_{\mathbb{R}^d}|\psi_n(y)-\psi(y)|^2\ dy$$
I feel like this is close, but I'm missing something to show that $\int_{\mathbb{R}^d}|\psi_n(y)-\psi(y)|^2\ dy \to 0$
Weak $L^2$ convergence does not imply strong $L^2$ convergence. This means that you cannot solve this problem by reducing it to showing $\|\psi_n - \psi\|_{L^2} \to 0$. I would look again at the definition of weak convergence and look to apply it more directly.