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Let $f_1,f_2$ be Lebesgue-summable functions on the real line. I was wondering whether space $C_0^\infty(\mathbb{R})$ of infinitely differentiable compactly supported functions, intended as distributions, is "enough to distinguish" such $f_1,f_2\in L^1(\mathbb{R})$, i.e. whether it is true that $$\forall\varphi\in C_0^\infty(\mathbb{R})\quad\int_\mathbb{R}f_1\varphi d\mu=\int_\mathbb{R}f_2\varphi d\mu\quad\Rightarrow\quad f_1\sim f_2$$where I intend $f_1\sim f_2$ to mean that the functions are identical almost everywhere. I find no explicit statement of the implication, but I intuitively suspect that it may be true, although I cannot manage to prove it. Could anybody prove it, provided that it is true? I thank anybody very much for any answer!

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    $\begingroup$ What is the set $K$? $\endgroup$ – Wintermute Jan 3 '15 at 13:55
  • $\begingroup$ @mtiano Thank you for the remark! Sorry, I used Kolmogorov-Fomin's notation I'm accustomed to instead of (more usual, I guess) $C_0^\infty(\mathbb{R})$ $\endgroup$ – Self-teaching worker Jan 3 '15 at 14:06
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    $\begingroup$ For that matter, sometimes $C_o^\infty$ means smooth things vanishing at infinity, rather than compactly supported. The most unequivocal notation is $C_c^\infty$ for compactly supported. No one will be confused. $\endgroup$ – paul garrett Jan 3 '15 at 21:51
  • $\begingroup$ @paulgarrett Thank you for the information! I leave the notation I used in order not to create inconsistencies with the notation of the answers, but I'm grateful to you for letting me know this more unequivocal one. $\endgroup$ – Self-teaching worker Jan 8 '15 at 14:10
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This is just due to the regularity of measurable sets, and I think this is more or less the most elementary way of doing it. If you can show that $\displaystyle \int_E f=0$ for all measurable sets you would be done, since we can just take $E$ to be the set where $f$ takes positive or negative values. So we need to show $\displaystyle \int_E f=\int_\mathbb{R}f\,\chi_E=0\,.$ We can take compact sets $K_n\subset E$ and open sets $E\subset O_n$ such that $m(E-K_n)\to 0\,,m(O_n-E)\to 0$ as $n\to\infty\,.$ So take a sequence of $C^\infty_0$ functions $\phi_n$ such that $0\le\phi_n\le 1\,,\phi\equiv 1$ on $K_n\,,$ $\phi_n\equiv 0$ outside of $O_n\,.$ Then $\displaystyle \int_{\mathbb{R}}|\chi_E-\phi_n|=\int_{K_n}|\chi_{E}-\phi_n|+\int_{O_n-K_n}|\chi_E-\phi_n|=m(E-K_n)+\int_{O_n-K_n}|\chi_E-\phi_n|\to 0$ as $n\to\infty\,,$ so $\phi_n\to\chi_E$ in $L^1$ and so there exists a subseqeuence for which $\phi_{n_k}\to \chi_E$ almost everywhere. Then by dominated convergence theorem, $0=\int_{\mathbb{R}}f\phi_{n_k}\to\int_{\mathbb{R}}f\chi_E=\int_E f\,,$ and we are done.

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    $\begingroup$ ...where we choose $f:=(f_1-f_2)^\pm$. Thank you so much! $\endgroup$ – Self-teaching worker Jan 5 '15 at 16:58
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This is a special case of the du Bois-Reymond lemma / fundamental lemma of calculus of variations:

If $Ω \subset \mathbb R^n$ is open and $f ∈ L^1_{loc}(Ω)$ such that $$ \int_Ω f(x) \phi(x) dx = 0 \quad \forall \phi ∈ C^∞_0(Ω) $$ then $f$ is zero almost everywhere. Now just consider $Ω = \mathbb R$ and take $f$ as the difference of your two functions.

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  • $\begingroup$ Wow, very interesting! I don't know much of the calculus of variation, but I'm anxious to study it. Is there a "more elementary" method to prove it? Thank you so much again! $\endgroup$ – Self-teaching worker Jan 3 '15 at 14:22
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    $\begingroup$ I don't remember a proof for the general statement, but I don't think it's a very deep result, i.e. you should be fine with what you already know. Maybe I find the time later to look for an elementary proof. (For $f ∈ C(Ω)$ the proof, by contradiction, is quite elementary. Have you looked at that already?) $\endgroup$ – Three.OneFour Jan 3 '15 at 16:00
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    $\begingroup$ Maybe the references given here are of some use to you. $\endgroup$ – Three.OneFour Jan 3 '15 at 16:11
  • $\begingroup$ Forgive me if I have to choose another answer, which is a "more direct" proof assuming less previous knowledge. $\infty$ thanks again!!! $\endgroup$ – Self-teaching worker Jan 5 '15 at 16:56
  • $\begingroup$ Sure I'm okay with that. Feel free to accept the answer which is the most useful to you. :) $\endgroup$ – Three.OneFour Jan 6 '15 at 23:24
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I see that you have a satisfactory answer, and that's good.

Because I suggested you look into mollifiers, I wanted to mention a couple of results that are fundamental to the study of mollifiers. Mollifiers are convolution operators with positive $C^{\infty}_{0}(\mathbb{R}^{n})$ kernels $\varphi$ for which $\|\varphi\|_{L^{1}(\mathbb{R}^{n})}=1$. It turns out that $L^{1}$ convolution operators are well-behaved on every $L^{p}$ space. This is unexpected because the integral of an $L^{1}$ function multiplying an $L^{p}$ function may not be finite. However, their convolution is finite a.e., and the resulting operator is bounded on $L^{p}$.

Theorem: Let $\varphi\in L^{1}(\mathbb{R^{n}})$, and define $M_{\varphi}$ as the convolution operator $$ M_{\varphi}f = \int \varphi(x-y)f(y)dy = \varphi\star f. $$ If $f \in L^{p}$ for some $1 \le p \le \infty$, then $M_{\varphi}f \in L^{p}$ and $$ \|M_{\varphi}f\|_{p} \le \|\varphi\|_{1}\|f\|_{p}. $$

Proof: I'll let you work out the special cases where $p=1$ and $p=\infty$; both are fairly straightforward. Suppose $f \in L^{p}(\mathbb{R}^{n})$ for some $1 < p < \infty$. Let $q$ satisfy $1/p+1/q=1$, and use Holder's inequality to obtain $$ \begin{align} |(Mf)(x)|^{p} & \le \left(\int |\varphi(x-y)|^{1/q}|\varphi(x-y)|^{1/p}|f(y)|dy\right)^{p} \\ & \le \left(\int |\varphi(x-y)|dx\right)^{p/q}\int|\varphi(x-y)||f(y)|^{p}dy \\ & = \|\varphi\|_{1}^{p/q}\int|\varphi(x-y)||f(y)|^{p}dy \end{align} $$ Then integrating with respect to $x$ and applying Fubini's Theorem shows that $Mf \in L^{p}$ and $$ \|Mf\|_{p}^{p}\le \|\varphi\|_{1}^{p/q}\|\varphi\|_{1}\|f\|_{p}^{p} = \|\varphi\|_{1}^{p}\|f\|_{p}^{p}. $$ Hence, $\varphi(x-y)f(y)$ is absolutely integrable in $y$ for a.e. $x$, and $M\varphi \in L^{p}$ with $\|Mf\|_{p} \le \|\varphi\|_{1}\|f\|_{p}$. $\;\;\Box$

This last result makes approximation much easier. If the distance between two $L^{p}$ functions is small, then the same remains true of the mollified functions. Using Lusin's theorem to approximate on a finite interval with a continuous function, then the final approximation result is reduced to approximation of a continuous function by the mollified continuous function, which is known to be uniform and pointwise.

Theorem: Let $\varphi$ be a non-negative even $C^{\infty}_{0}(\mathbb{R})$ function which is supported in $[-1,1]$ such that $0 \le \varphi(x) \le \varphi(1)$ and $\int_{\mathbb{R}}\varphi(x) dx = 1$. Define $\varphi_{n}(x)=n\varphi(nx)$. Suppose that $f$ is a Lebesgue measurable function on $\mathbb{R}$ which is locally integrable. For each positive integer $n$, define $$ (M_{n}f)(x)= \int_{\mathbb{R}}\varphi_{n}(x-y)f(y)dy. $$ If $f \in L^{p}(\mathbb{R})$ for some $1 \le p \le \infty$, then $M_{n}f \in L^{p}(\mathbb{R})$ and $$ \|M_{n}f\|_{p} \le \|f\|_{p}. $$ Furthermore, if $f \in L^{p}(\mathbb{R})$ for some $1 \le p < \infty$, then $$ \lim_{n} \|M_{n}f - f\|_{p} = 0. $$ The above limit holds for $p=\infty$ if $f$ is uniformly continuous on $\mathbb{R}$.

Proof: The norm estimates follow from the previous theorem. Suppose $f \in L^{p}$ for some $1\le p < \infty$. Then $$ \lim_{a\rightarrow\infty}\|f-\chi_{[-a,a]}f\|_{p}=0, $$ which allows the argument to be reduced to the case where $f$ is $0$ outside a finite interval $[-a,a]$. Then $f \in L^{1}\cap L^{p}$. So there is a continuous function $g$ on $[-a,a]$ such that $\|f-g\|_{p} < \epsilon$ and, without loss of generality $g(-a)=g(a)=0$. Extend $g$ to be $0$ outside $[-a,a]$. Then $M_{n}g$ converges uniformly to $g$ on every compact subset of $\mathbb{R}$, which means $M_{n}g$ also converges in $L^{p}$ to $g$. Then $$ \begin{align} f-M_{n}f & = (f-\chi_{[-a,a]}f) \\ & + (\chi_{[-a,a]}f-g) \\ & + (g-M_{n}g) \\ & + M_{n}(g-\chi_{[-a,a]}f) \\ & + M_{n}(\chi_{[-a,a]}f-f). \end{align} $$ The first term is made $L^{p}$ small by choosing '$a$' large enough. The next term is made $L^{p}$ small by choosing $g$ correctly. The third term is made $L^{p}$ small by choosing $n$ large enough. The fourth term is already small because $\|M_{n}(g-\chi_{[-a,a]}f)\|_{p} \le \|g-\chi_{[-a,a]}f\|_{p}$, and similarly for the fifth term. $\;\;\Box$

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    $\begingroup$ @Self-teachingDavide : You're welcome. Sometimes it's hard to sift out the important results when you're first learning a subject. These are important results: (1) $L^{1}$ convolutions are continuous on every $L^{p}$, $1 \le p \le \infty$ and (2) Mollifiers convergence pointwise to the identity for continuous functions and converge in $L^{p}$, $1\le p < \infty$ to the identity as well. That gets you started with Sobolev spaces, too, where weak derivatives are in $L^{p}$ spaces. Mollified Sobolev elements converge along with derivatives in $L^{p}$, which gives you a reduction to $C^{\infty}$. $\endgroup$ – DisintegratingByParts Jan 5 '15 at 18:38

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