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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function in both $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$. I want to show that there exists a sequence of functions $g_1,g_2,\ldots$ in the Schwartz class such that both $\|g_n-f\|_1\rightarrow 0$ and $\|g_n-f\|_2\rightarrow 0$ as $n\rightarrow \infty$.

Following the suggestion given in this post, I'm looking at the function $$g_m(x) = f(x)\cdot \chi_{[-m,m]}(x)\cdot \chi_{\{ \lvert f(y)\rvert \leqslant m\}}(x),$$ where $\chi$ denotes the characteristic function. I can see that the convergence follows from the dominated convergence theorem. But why would $g_m$ be in the Schwartz class? Since there is no assumption on the differentiability of $f$, the function $g_m$ might not even have derivatives of any order, right?

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PJ Miller
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  • No, $g_m$ isn't in the Schwartz class. But it is bounded and has compact support, so its convolution with any function from the Schwartz class belongs to the Schwartz class. Make it a convolution with a test function, and you have a test function. – Daniel Fischer Nov 13 '13 at 20:20
  • @DanielFischer Oh, I misread. So what do you mean by a "compactly supported approximation of the identity"? What is such a function? – PJ Miller Nov 13 '13 at 20:23
  • Actually, now I realize that this is much more complicated than I thought. Any easier examples of such functions $g_m$ would be welcome too. – PJ Miller Nov 13 '13 at 20:29

1 Answers1

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$g_m$ is in general not even continuous. But it is bounded and has compact support, and that makes it easy to check that for any $\varphi \in C_c^\infty(\mathbb{R})$ with $\int \varphi = 1$, you have

$$\begin{align}(\varphi_\varepsilon \ast g_m)(x) &= \int_\mathbb{R} \frac{1}{\varepsilon}\varphi\left(\frac{z}{\varepsilon}\right) g_m(x-z)\,dz\\ &=\int_\mathbb{R} \varphi(y)g_m(x-\varepsilon y)\,dy \xrightarrow{\varepsilon\to 0} g_m\tag{1} \end{align}$$

both in $L^1$ and in $L^2$. The family $\left\lbrace \varphi_\varepsilon : \varepsilon > 0 \right\rbrace$, where $\varphi_\varepsilon(x) := \varepsilon^{-1}(x/\varepsilon)$, is called an approximation of the identity (because $\lim\limits_{\varepsilon\to 0} \varphi_\varepsilon \ast h = h$ for $h \in L^p$ [and other reasonable function spaces], so convolution with $\varphi_\varepsilon$ is approximately the identity operation for small $\varepsilon > 0$).

Since $g_m$ and $\varphi$ have compact support, the support of $g_m$ is contained in $[-m,m]$, and the support of $\varphi$ say is contained in $[-K,K]$, the support of $\varphi_\varepsilon\ast g_m$ is contained in $[-(m+\varepsilon K), m+\varepsilon K]$, hence compact. Writing the convolution integral in the form

$$(\varphi_\varepsilon \ast g)(x) = \frac{1}{\varepsilon}\int_\mathbb{R} \varphi\left(\frac{x-z}{\varepsilon} \right)g_m(z)\,dz,$$

the dominated convergence theorem shows that $\varphi_\varepsilon \ast g_m$ is differentiable with

$$(\varphi_\varepsilon \ast g)'(x) = \frac{1}{\varepsilon^2}\int_\mathbb{R} \varphi'\left(\frac{x-z}{\varepsilon} \right)g_m(z)\,dz,$$

and iterating the argument shows that $\varphi_\varepsilon \ast g_m \in C^\infty(\mathbb{R})$.

And thus, to obtain a sequence in $\mathscr{S}(\mathbb{R})$ converging to $f$ in both $L^1$ and $L^2$, for every $n \in\mathbb{Z}^+$, choose an $m$ such that $\lVert g_m-g\rVert_p < \frac{1}{2n}$ for $p\in \{1,2\}$, and then choose $\varepsilon$ in $(1)$ small enough that you have $\lVert (\varphi_\varepsilon \ast g_m) - g_m\rVert_p < \frac{1}{2n}$ for $p\in\{1,2\}$.

Since for all $m$ and $\varepsilon > 0$ we have $\varphi_\varepsilon \ast g_m \in C_c^\infty(\mathbb{R}) \subset \mathscr{S}(\mathbb{R})$, the so-constructed sequence converges of Schwartz functions converges to $f$ simultaneously in $L^1$ and $L^2$.

Daniel Fischer
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  • Does your $\varphi_\varepsilon$ mean $\varphi/\varepsilon$ or something like that? Also, why is $\varphi_\varepsilon \ast g_m \in C_c^\infty(\mathbb{R})$? – PJ Miller Nov 13 '13 at 20:41
  • It actually means $\varphi_\varepsilon(x) = \dfrac{1}{\varepsilon}\varphi\left(\dfrac{x}{\varepsilon}\right)$. I have reparametrised the convolution integral directly, I think I shouldn't have. Let me edit that. As for $\varphi_\varepsilon\ast g_m \in C_c^\infty(\mathbb{R})$, it's pretty direct to show that the convolution has compact support, and the dominated convergence theorem yields differentiability with $(\varphi_\varepsilon \ast g_m)' = (\varphi_\varepsilon)' \ast g_m$. As you don't yet have much experience with mollifiers, let me say a bit more about it in an edit. – Daniel Fischer Nov 13 '13 at 20:49
  • Thanks Daniel. That will take some time for me to read, so let me accept your answer for now, and I'll come back later if I have anything to ask. – PJ Miller Nov 13 '13 at 21:10
  • Why is it that $\int_\mathbb{R} \varphi(y)g_m(x-\varepsilon y),dy \xrightarrow{\varepsilon\to 0} g_m$ both in $L^1$ and in $L^2$ – PJ Miller Nov 13 '13 at 22:01
  • Especially since $g_m$ is not necessarily continuous, I struggle to see why that's true. – PJ Miller Nov 13 '13 at 22:09
  • You're right, it's not that obvious, since $g_m$ isn't continuous, sorry. But by Lusin's theorem, for any $\delta>0$, there is a continuous function $h$ with $\lvert h\rvert\leqslant\lvert g_m\rvert$ such that $\mu({x:h(x)\neq g_m(x)})<\delta$. That gives $\lVert h-g_m\rVert_p\leqslant\delta\cdot(2m)^p$, and for the continuous $h$ (with compact support), you have a uniform approximation by the convolution, and since everything happens in a (fixed) compact set, that uniform approximation implies $L^p$ convergence. You can also argue with the denseness of continuous functions in $L^p$. – Daniel Fischer Nov 13 '13 at 22:23
  • But I think Lusin's theorem is the more elementary argument. – Daniel Fischer Nov 13 '13 at 22:24
  • Aha I've seen before that $C_c^\infty(\mathbb{R})$ is dense in $L^p(\mathbb{R})$. How would that yield this result? – PJ Miller Nov 13 '13 at 22:26
  • Actually that gives me another idea. This density statement gives us functions ${g_m}\in C_c^\infty(\mathbb{R})$ that converge to $f$ in $L^1$, and functions ${h_m}\in C_c^\infty(\mathbb{R})$ that converge to $f$ in $L^2$. Can we somehow combine $g_m$ and $h_m$ to get functions converging to $f$ in both $L^1$ and $L^2$? – PJ Miller Nov 13 '13 at 22:32
  • Convolution with $\varphi_\varepsilon$ is a linear operator with norm $\lVert \varphi_\varepsilon\rVert_1 = \lVert \varphi\rVert_1$. So $T_\varepsilon \colon h \mapsto(\varphi_\varepsilon \ast h)$ is an equicontinuous family of linear operators $L^p \to L^p$. We have $T_\varepsilon(h) \to h$ for $h \in C_c(\mathbb{R})$, and by Ascoli's theorem, that implies $T_\varepsilon(g) \to g$ for all $g\in L^p$. If you are familiar with equicontinuity arguments, that is probably more elegant than Lusin's theorem, but if not, maybe a bit obscure. – Daniel Fischer Nov 13 '13 at 22:32
  • That is the entire point, that we can do that (the problem statement asks for simultaneous approximation by Schwartz functions, but the construction yields even test functions). However, to show that requires a few gymnastics. – Daniel Fischer Nov 13 '13 at 22:36
  • I see. Wow, didn't think this would be so complicated, but thanks anyway. – PJ Miller Nov 13 '13 at 22:39
  • It is less complicated if one spends enough time thinking about how to present it beforehand. When writing a book or preparing a lecture, one does not produce the argument piecemeal because "oops, I didn't think of stating that, I took it for granted due to familiarity", one builds a coherent argument knowing what can be assumed because it was treated earlier. That tends to lead to more streamlined proofs. – Daniel Fischer Nov 13 '13 at 22:46