# 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!

• one thing to note is that translation is $L^p$ continuous so $\|f(x)-f(x-y)\|_2 \to 0$ as $|y| \to 0$ is in fact true! – guy3141 Mar 25 '20 at 0:12
• You're right! I just remembered that the way you prove this is by using that $C^0_c$ (compactly-supported continuous functions) is dense in $L^2$ so WLOG $f$ is compactly-supported and continuous, after which it is easy. Unfortunately we have the nasty $|\eta_\delta(y)|^2$ lying around, and $||\eta_\delta||^2_{L^2}\to \infty$ is possible... – Juan Carlos Ortiz Mar 25 '20 at 0:17

## 3 Answers

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.

• Ah, nice approach! I'd forgotten about Minkowski... Just to add a complete solution, I'll write out how to finish here: for any $\epsilon$ there exists $\delta'$ such that $||f - f(\bullet-y)||_2<\epsilon$ for $|y|\le\delta'$, by continuity of translations in $L^2$. So for $\delta<\delta'$, the above bound gives $$||f-f*\eta_\delta||_2 \le \int |\eta_\delta(y)| \cdot \epsilon dy = \epsilon \int \eta_\delta =\epsilon$$ and so $||f-f*\eta_\delta||_2$ is guaranteed to be long as soon as $\delta$ is small enough, as desired. – Juan Carlos Ortiz Mar 25 '20 at 15:36

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.

• +1, never seen this before :) Has the dawn ever seen your eyes? Have the days made you so unwise? – Calvin Khor Mar 28 '20 at 6:38
• Realize, you are! – Juan Carlos Ortiz Mar 29 '20 at 17:51

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.