Approximate $L^2$ function by convolving with mollifiers

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

While we're adding extra answers, here's a fix of your original attempt (although I suspect that if you squint hard enough, its equivalent to my other answer). The issue is that its not always true that $$ \left(\int_X f d\mu\right)^2 \le \int_X f^2 d\mu$$ (For instance, try $f=\frac{1}{1+|x|}\in L^2(\mathbb R)\setminus L^1(\mathbb R)$.) A sufficient condition is that $\mu$ be a probability measure, and then it is a special case of Jensen's inequality. So if the $\delta$ ball around the origin $B_\delta(0)$ has Lebesgue measure $C\delta^n$ in $\mathbb R^n$, we can write the $y$ integral in terms of the uniform probability measure on $B_\delta(0)$, $$d\mu = \frac{dy}{C\delta^n}$$ to get (for $p=2$ as well as any $p\in[1,\infty)$) \begin{align} I_\delta&:=\int_{\mathbb{R}^2}\bigg| \int_{\mathbb{R}^2} \eta_\delta(y)(f(x)-f(x-y)) dy\bigg|^p dx \\ &= \int_{\mathbb{R}^2}C^p\delta^{np}\bigg| \frac{1}{C\delta^n}\int_{|y|<\delta} \eta_\delta(y)(f(x)-f(x-y)) dy\bigg|^p dx \\ &\le \int_{|y|<\delta}C^{p-1}\delta^{n(p-1)}\eta_\delta(y)^p\left(\int_{\mathbb{R}^2}|f(x)-f(x-y)|^p dx\right) dy \\ &\le \sup_{|z|<\delta}\|f-f(\bullet - z)\|_{L^p}^p \int_{|y|<\delta}C^{p-1}\delta^{n(p-1)}\eta_\delta(y)^{p-1} \eta_\delta(y) dy \end{align} Recalling that $\eta_\delta(y) = \delta^{-n}\eta(\frac y\delta)$, setting $w = y/\delta$, we have $$\eta_\delta(y)dy = \eta(w)dw, \quad \eta_\delta(x)^{p-1}=\delta^{-n(p-1)}\eta(w)^{p-1}$$ giving perfect cancellation of all powers of $\delta$: $$ I_\delta \le C^{p-1}\sup_{|z|<\delta}\|f-f(\bullet - z)\|_{L^p}^p\int_{|w|<1}\eta(w)^p dw $$ Since one usually takes $\eta$ to be $C^\infty_c\subset L^p_{\text{loc}}(\mathbb R^n)$, this final integral is finite, and independent of $\delta$. Therefore, the continuity of translations in $L^p$ again gives the conclusion.