2
$\begingroup$

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!

$\endgroup$
2
  • 1
    $\begingroup$ 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! $\endgroup$ – guy3141 Mar 25 '20 at 0:12
  • $\begingroup$ 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... $\endgroup$ – Juan Carlos Ortiz Mar 25 '20 at 0:17
1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ – Juan Carlos Ortiz Mar 25 '20 at 15:36
1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ +1, never seen this before :) Has the dawn ever seen your eyes? Have the days made you so unwise? $\endgroup$ – Calvin Khor Mar 28 '20 at 6:38
  • $\begingroup$ Realize, you are! $\endgroup$ – Juan Carlos Ortiz Mar 29 '20 at 17:51
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.