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Is the subset consisting of all integrable (or square integrable) smooth functions of the set of all integrable (or square integrable) functions, dense under the usual Euclidean or integral of absolute difference metric?

By smooth I mean derivatives of all orders exist.

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exit -> exist. (Also, no need for a signature as you can see your name is added automatically at the bottom of the question) –  Tobias Kienzler Nov 1 '10 at 9:49

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up vote 12 down vote accepted

Yes. In fact, by the Stone-Weierstrass theorem and the existence of smooth bump functions, smooth functions with compact support are uniformly dense in the space of continuous functions with compact support. Uniform density implies $L^2$ and $L^1$ density for functions with compact (and therefore finite measure) support, and since continuous functions with compact support are dense in $L^2$ and $L^1$, the result follows.

If you wanted to see this more directly, you can go through the iterations of approximating an arbitrary ($L^1$ or $L^2$) function with a bounded function with bounded support, then with a simple function, then with a step function, and finally approximate the step function with a smooth function using bump functions.

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What about noncompact smooth functions ?...compactness is not imposed –  Rajesh D Nov 1 '10 at 7:11
    
@Rajesh D: There are of course smooth functions without compact support in $L^1$ and $L^2$, but what I am saying is that if you consider the subset of smooth functions with compact support, that is already dense. A set containing a dense subset is dense. –  Jonas Meyer Nov 1 '10 at 7:15
    
got it....thanks –  Rajesh D Nov 1 '10 at 7:17
    
@Jonas Meyer: how could you approximate a noncompact function in $L^2$ with a smooth compact function ? Please explain. Your answer seems to me that it is applicable to space of compact functions and i am not able to uderstand how it is applicable to noncompact functions. –  Rajesh D Dec 7 '10 at 3:15
    
@Rajesh: Let $f$ be an arbitrary function in $L^2$, with compact support or not. Consider the sequence $(f_n)$ defined by $f_n=f$ on $B(0,n)$ (the ball of radius $n$ centered at the origin) and $f_n=0$ elsewhere. Show that $f_n\to f$ in $L^2$. Each $f_n$ has compact support, so if you are satisfied with approximating functions with compact support with smooth functions, then you are done. –  Jonas Meyer Dec 7 '10 at 3:45

Jonas's argument is good. Another proof is: given $f \in L^p$ (here $p=1,2$), take the convolution of $f$ with a sequence of mollifiers $\eta_\epsilon$. Using properties of convolutions, it's easy to check that $f * \eta_\epsilon$ is a smooth function, and that $f * \eta_\epsilon \to f$ in $L^p$ as $\epsilon \to 0$. This has the advantage of being a little more direct.

Edit: For a reference, see Folland's Real Analysis, section 8.2.

The smoothness of $f * \eta_\epsilon$ is Proposition 8.10 and comes from differentiating under the integral sign in the convolution (with justification!), and choosing to put the derivative on $\eta_\epsilon$. Intuitively, it comes from the idea that convolution is an "averaging" operation and tends to smooth, smear, or blur rough areas of $f$ together, and so should be a smoothing operation. (The wikipedia article has a nice animation illustrating this.)

The fact that $f * \eta_\epsilon \to f$ in $L^p$ is Folland's Theorem 8.14 (a), and it's pretty elementary. He also has Proposition 8.17 which proves that $C^\infty_c$ is dense in $L^p$, but it sort of inexplicably starts by using the fact that $C_c$ is dense in $L^p$. I suppose this is used to get a compactly supported function, so that you can approximate $f \in L^p$ by functions which are not only smooth (which $f * \eta_\epsilon$ is) but also compactly supported (which $f * \eta_\epsilon$ need not be, although $\eta_\epsilon$ is). But an easier argument would be to first approximate $f$ in $L^p$ norm by a function $g$ which is compactly supported but not necessarily continuous; for example, $g = f 1_{[-N,N]}$ for large $N$ (this works by dominated convergence), and then apply mollifiers to $g$. Unless, of course, there is some subtlety that I've missed.

Edit 2: Indeed there is. Folland's 8.14 (a) relies upon the fact that translation is strongly continuous in $L^p$, which uses the density of $C_c$. So apparently it is not so easy to bypass this step, and that destroys a lot of the "directness" of my argument.

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Let 'f' be a continous but nowhere differentiable function. Is f convolved with mollifier, a smooth function ? –  Rajesh D Nov 2 '10 at 3:48
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@Rajesh D: Yes. Indeed, $f$ can be much worse: if $f$ is any distribution and $\phi$ is a smooth function with compact support, then $f * \phi$ is a smooth function. –  Nate Eldredge Nov 2 '10 at 5:52
    
could you please suggest a reference for the proof. –  Rajesh D Nov 2 '10 at 7:31
    
the direct method of starting with convolution integral and differentiating doesnt work in this special case. –  Rajesh D Nov 2 '10 at 7:31
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@Rajesh, @Nate: In the proof of Theorem 8.14, Folland uses Proposition 8.5. In the proof of 8.5 he has to start from continuous functions with compact support, and then he uses the denseness of those functions. This is exactly what I meant in my above comment. –  Hendrik Vogt Nov 3 '10 at 16:01

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