Approximate $L^2$ function by convolving with mollifiers Let $\eta_\delta$ be a mollifier (i.e. positive real-valued function on $\mathbb{R}^2$, supported on the ball of radius $\delta$ centered at the origin, whose integral is 1), and $f$ is a compactly-supported $L^2$-function. How can we prove that
$$ || f - f*\eta_\delta||^2_{L^2} \rightarrow 0 $$
as $\delta\to 0$? (This is standard in the proof that we can approximate $L^2$-functions via smooth functions, by the use of mollifiers). The computation leads to bounding 
$$ \int_{\mathbb{R}^2}\bigg| \int_{\mathbb{R}^2}  \eta_\delta(y)(f(x)-f(x-y)) dy\bigg|^2 dx \le \int_{|y|<\delta}|\eta_\delta(y)|^2\left(\int_{\mathbb{R}^2}|f(x)-f(x-y)|^2 dx\right) dy,$$
at which point I get stuck. Is it true that $|| f(x)-f(x-y)||^2_{L^2}\to 0$ as $|y|\to 0$? This could be used above but it wouldn't even finish, I think. Thank you for your help!

Minkowski’s inequality should be used to get $$\|f-f*\eta_\delta\|_2 = \left\| \int \eta_\delta (y)(f- f(\bullet -y)) dy \right\|_2 \le \int \| \eta_\delta (y)(f- f(\bullet -y)) \|_2 dy $$
Then you use the continuity of translations in $L^p$: only one factor of $\eta_\delta$ appears. 
