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!

I wanted to add an answer using a different estimate using Plancherel's identity:
$$||f-f*\eta_\delta||_{L^2} = ||\hat{f}-\widehat{f*\eta_\delta}||_{L^2} = ||\hat{f}(1-\hat{\eta_\delta})||_{L^2} = \left(\int |\hat{f}(\xi)|^2|1-\hat{\eta_\delta}(\xi)|^2d\xi\right)^{1/2} $$
converges to zero because $|\hat{f}|^2|1-\hat{\eta_\delta}|^2$ is dominated by $|\hat{f}|^2\in L^1$ and pointwise we claim we have $|\hat{f}(\xi)|^2|1-\hat{\eta_\delta}(\xi)|^2\to 0$ as $\delta\to 0$ (keeping $\xi$ fixed); hence the Dominated Convergence Theorem applies and gives us $||f-f*\eta_\delta||_{L^2}\to 0$ as desired. The only thing left to check is $\hat{f}(\xi)(1-\hat{\eta_\delta}(\xi))\to 0$ if $\xi$ is fixed, i.e. that $|1-\hat{\eta_\delta}(\xi)|\to 0$. We compute this is equal to 
$$\bigg|1-\int \eta_\delta(x)e^{-ix\xi}dx\bigg| \le \int |\eta_\delta(x)(1-e^{-ix\xi})dx| \\ \le \left(\int \eta_\delta(x) dx\right) \cdot \sup_{|x|\le\delta}|1-e^{-ix\xi}| = \sup_{|x|\le\delta}|1-e^{-ix\xi}|$$
goes to $0$ because $\xi$ is fixed and $x\mapsto 1-e^{-ix\xi}$ is a continuous function which at $x=0$ evaluates to $0$, QED.
