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How do you show that the space $C_c (\mathbb{R})$ of continuous functions on $\mathbb{R}$ is dense in $L_1(\mu)$, where $\mu$ is a regular measure on the reals, without using Lusin's theorem directly?

More concretly, given an integrable function $f$, I'd like to know how to explicitely construct a sequence of functions in $C_c(\mathbb{R})$ that converges to it in $L_1$, without just using simple functions at all, or their density in $L_1(\mu)$.

I thought about defining $f_k:=f\chi_{E_k}$, where $E_k:=\{x\in\mathbb{R}: |f(x)|\geq \frac 1k\}$, but there's no guarantee that these would be continuous (there's no guarantee that $f$ is continuous). Perhaps there is a way to "smooth" them out enough to make them continuous?

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    Not directly related but you can look up *mollifier*. Really interesting stuff.2012-11-23

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What you do is convolution against "good" functions. If you do the convolution with a $C^\infty$ function, the result will be $C^\infty$.

You can find the details in Chapter 9 of Wheeden-Zygmund. There you can find proofs of the following facts (I'll write $1$-dimensional versions, but the results are $n$-dimensional):

1) If $f\in L^p(\mathbb R)$, $g\in C^k_0(\mathbb R)$, then $f*g\in C^k(\mathbb R)$ and $D^\alpha(f*g)(x)=(f*D^\alpha g)(x)$ for any $\alpha$.

2) Given $g\in C^\infty_0(\mathbb R)$ with $g\geq0$ and $\int_{\mathbb R}g=1$, $\varepsilon>0$, let $g_\varepsilon(x)=\varepsilon^{-1}g(\varepsilon^{-1}x)$. For any $f\in L^p(\mathbb R)$, $\|f*g_\varepsilon-f\|_p\to0$ as $\varepsilon\to0$.

3) Given $f\in L^p(\mathbb R)$, $\delta>0$, let $f=f_1+f_2$, where $f_1$ has compact support and $\|f_2\|_2<\delta$. Choose $g$ as in 2); then $\|f_1*g_\varepsilon-f_1\|_p\to0$ as $\varepsilon\to0$. Now $$ \|f_1*g_\varepsilon-f\|_p\leq \|f_1*g_\varepsilon-f_1\|_p+\|f_2\|_p<2\delta $$ if $\varepsilon$ is small enough. As $f_1*g_\varepsilon\in C_c(\mathbb R)$, we are done.

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    I don't have quick access to that Wheeden-Zygmund book. Could you please provide a quick sketch of how to prove that $\|f\ast g_\epsilon-f\|$ goes to 0 as $\epsilon \to 0$?2012-11-23
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    I think these three pages: http://ev03023.math.uregina.ca/Wheeden,%20Zygmund%20-%20Measure%20And%20Integral%20(Dekker,%201977).pdf contain all you need.2012-11-23
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    @MartinArgerami, how would you make this work for 0 < p < 1, though? I know FPP wasn't asking about that, but is there a way?2014-02-06
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    I'm afraid I have no idea. When $p<1$ one loses the triangle inequality, so proofs for most approximation techniques are likely fail (but maybe the techniques do work still).2014-02-06
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    @MartinArgerami, we do, however, have $ \|f_1*g_\varepsilon-f\|_{p}^{p}\leq \|f_1*g_\varepsilon-f_1\|_p^{p} +\|f_2\|_p^{p}<2\delta $, so wouldn't this do the trick due to the continuity of $f(x) = x^{p}$? Also, in your proof, how do you know there exists a decomposition $f = f_1 + f_2$ and a kernel $g_\epsilon$ with compact support?2014-02-06
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    You are right about that inequality. But you still need 2) above, and its proof uses Hölder's Inequality and maybe other (stuff) I haven't looked carefully lately, so it might still not work. The decomposition $f_1+f_2$ is a consequence of the fact that $\int_{\mathbb R}f=\lim_n\int_{B_n}f$, where $B_n$ is the ball of radius $n$ centered at the origin.2014-02-06
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    To construct the kernel $g$, you start with the function $h(x)=e^{-1/x^2}$ for positive $x$ and $0$ for negative $x$. Then $h(x-a)h(b-x)$ is $C^\infty$ with support $[a,b]$. And $h(|x|)h(r-|x|)$ is $C^\infty(\mathbb R^n)$ with support in the ball of radius $r$ around the origin.2014-02-06
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    @MartinArgerami, thanks for the feedback. As far as that decomposition goes, couldn't you have the $L^p$ norm of the set of discontinuities of $f$ to be strictly positive? I don't quite see why not at the moment, and I think this would prevent the "approximation" by $f_1$. And you're right in pointing out 2), although it seems that, using the p-th power inequality for $0 < p < 1$ and following the steps in the proof of that, you would just get $\|f*g_\varepsilon-f\|_p^{p}\to0$ instead, and that's all you'd need, wouldn't you?2014-02-06
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    There are no assumptions on the discontinuities of $f$. Only that it is in $L^p$. The proof that continuous functions with compact support are dense in $L^p$ is precisely what the answer is about. As for step 2), the proof uses Hölder, which I don't think has a version for $p<1$.2014-02-07
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    @MartinArgerami, the reason I mentioned discontinuities is that if that set indeed had positive measure, then how would the existence of $f_1$ follow from what you wrote above?2014-02-07
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    The convolution of an $L^p$ function with a $C^\infty$ function is $C^\infty$, for **any** $L^p$ function.2014-02-07
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    @MartinArgerami, ah, I see. I just realized that in your proof you did indeed state only that $f_1$ has compact support, not that $f_1 \in C_{c}(\mathbb R)$, which is what I was reading it as. So it makes more sense now, ugh. As for the proof of 2), I've seen it proven with Minkowski's inequality, as well, so it might not be a problem then. Plus, the version of Hölder's inequality I learned is actually applicable to all $p > 0$, not just $p>1$.2014-02-07
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    Hölder for $p<1$? For example with $p=1/2$ and $q=-1$?2014-02-07
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    @MartinArgerami, no, not that way, but in the sense that if $0 and $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$, then $\|fg\|_r \leq \|f\|_p\|g\|_q$.2014-02-07
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    I see. I'm not sure then, whether the proof of 2) can be done under those conditions.2014-02-07
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Another approach uses the representation of the dual $\left[ L^1(\mathbb{R})\right]^\star$ as $L^\infty(\mathbb{R})$ and the Hahn-Banach separation theorem. Namely, to prove that a vector subspace of a Banach space is dense we only need to show that the only continuous linear functional that vanishes on it is the null one.

To do this fix $\phi \in L^\infty(\mathbb{R})$ and suppose that $$\tag{1}\int_{-\infty}^\infty \phi(x)f(x)\, dx=0, \quad \forall f\in C_c(\mathbb{R}).$$ We claim that $\phi=0$ almost everywhere. Indeed, let $a be fixed numbers. Approximate the characteristic function $\chi_{[a,b]}$ with a family $\chi^{(\varepsilon)}_{[a, b]}$ of "trapezoidal-like" functions:

trapezoidal function

We have $$\int_{-\infty}^\infty \left\lvert \chi_{[a,b]}(x)-\chi_{[a,b]}^{(\varepsilon)}\right\rvert\, dx = 2\varepsilon$$ so \begin{align} \left\lvert \int_{-\infty}^\infty \phi(x)\chi_{[a,b]}(x)\, dx - \int_{-\infty}^\infty \phi(x)\chi_{[a, b]}^{(\varepsilon)}(x)\, dx \right\rvert & \le \lVert \phi\rVert_{\infty} \int_{-\infty}^\infty \left\lvert \chi_{[a,b]}(x)-\chi_{[a,b]}^{(\varepsilon)}(x)\right\rvert\, dx \\ &=2\varepsilon \lVert \phi\rVert_\infty. \end{align} In particular, $$\int_a^b \phi(x)\, dx=\lim_{\varepsilon \to 0} \int_{-\infty}^\infty \phi(x)\chi_{[a,b]}^{(\varepsilon)}\, dx,$$ and the last limit is $0$ due to our assumption (1): indeed, every $\chi_{[a, b]}^{(\varepsilon)}$ is a continuous function with compact support. We have thus shown that $$\tag{2} \int_a^b\phi(x)\, dx=0, \qquad \forall a It is intuitively clear that this can happen only if $\phi=0$ almost everywhere: for a rigorous proof of this you can apply the Lebesgue differentiation theorem or Lemma 1 of this post. This proves the claim.

To conclude we only need to recall that every continuous linear functional $\Lambda \in \left[ L^1(\mathbb{R})\right]^\star$ is of the form $$\Lambda f= \int_{-\infty}^\infty \phi(x)f(x)\, dx,\qquad f \in L^1(\mathbb{R}),$$ for a unique $\phi\in L^\infty(\mathbb{R})$, and then apply the Hahn-Banach separation theorem. $\square$

A final remark: Even if we did not mention convolutions explicitly, the present proof is not that different in nature from the ones presented above. Both rely on the possibility of approximating "rough" functions (like our $\chi_{[a, b]}$) with "smooth" ones.

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    @GiusseppeNegro Thank you for your detailed answer. But would you write the definition of the trapezoidal-like function?2016-08-09
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One standard way of doing this is with convolutions. Let $f\in L^1$.

First note that the sequence $f\chi_{[-n,n]}$ converges to $f$ in $L^1$ as $n\to \infty$, so it suffices to find compactly supported continuous functions converging to $f\chi_{[-n,n]}$. In other words, we may assume with no loss of generality that $f$ is compactly supported.

Let $\varphi\colon \mathbb{R}\to \mathbb{R}$ be a compactly supported smooth function that is nonnegative and has $\int \varphi\,d\mu = 1$. For each $\epsilon>0$, let $\varphi_\epsilon(x) = \epsilon^{-1}\varphi(x/\epsilon)$. Then $\varphi_\epsilon$ is also a nonnegative compactly supported smooth function with $\int\varphi_\epsilon\,d\mu = 1$. One can show that the convolution $f*\varphi_\epsilon$ is a compactly supported smooth function for each $\epsilon$, and that $f*\varphi_\epsilon\to f$ in $L^1$ as $\epsilon\to 0$.