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We know that the set of continuous functions and the set of smooth functions in $\mathcal{L}^1$ or $\mathcal{L}^2$ are dense in $\mathcal{L}^1$ and $\mathcal{L}^2$. Is the set of $\mathcal{C}^k$ functions dense in $\mathcal{L}^1$ or $\mathcal{L}^2$ for any $k\in\mathbb{N}$?

EDIT

Is the set $\mathcal{C}^k\setminus \mathcal{C}^{\infty}$ in $\mathcal{L}^1$ or $\mathcal{L}^2$ dense in $\mathcal{L}^1$ or $\mathcal{L}^2$. How would one prove this one way or the other?

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Every smooth function is $C^k$, right? –  Carl Mummert Dec 6 '10 at 13:49
    
At least to me, the edit does not clarify the question. Do you mean to ask whether the set of functions which are $C^k$ but not $C^{\infty}$ are dense in $L^p$? The answer is yes. –  Pete L. Clark Dec 6 '10 at 13:59
    
@Pete L. Clark : Could you please give a hint for the proof. –  Rajesh D Dec 6 '10 at 14:13
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What space? Not all continuous functions are in $L^1(\mathbb R)$. For example $f(x) = 1$. –  Jonas Teuwen Dec 6 '10 at 14:20
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Why not just take a smooth approximation, and then add a small multiple of a function which is $C^k$ but not smooth? –  Akhil Mathew Dec 6 '10 at 14:50
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Here is the idea behind the proof.

1) Step functions are dense in $L^p$ ($p=1,2$).

2) Given a step function $T$ and $\varepsilon>0$ there is a $g\in C^k$ $$\|T-g\|_{L^p}\lt\varepsilon$$ ($p=1,2$).

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@AD. : how to prove the second statement ...is it through non uniform convergence ? please suggest . –  Rajesh D Dec 6 '10 at 14:46
    
@AD. :Also suggest a nice reference for proof of statement 1. –  Rajesh D Dec 6 '10 at 14:47
    
Statement 1 is probably in most real analysis texts that talk about $L^p$. Rudin (Real and Complex Analysis) or Folland (Real Analysis: Modern Techniques and their Applications) should have it. –  Matt Dec 6 '10 at 16:34
    
The second can be shown with Urysohn or something like that (maybe even easier). –  Jonas Teuwen Dec 6 '10 at 16:44
    
@Jonas T: Right, maybe even easier: 2.1) consider a characteristic function and $C^k$-smooth it around the edges, 2.2) a general step function is a linear combination of characteristic functions. –  AD. Dec 6 '10 at 17:23
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One basic way to see that smooth functions are dense in any reasonable function space is to use "approximations to the identity." The point is that convolving with the delta function is the identity operator. However, the delta function can be approximated by smooth functions (take a smooth function supported in a small ball of total integral one). If $\phi$ is such a smooth function (which approximates the $\delta$ function), and $f$ is any function, then $\phi \ast f$ is probably close to $f$ in the metric you are interested in (this is always true in $L^p$, at least). But $\phi \ast f$ is smooth since to differentiate this, just differentiate $\phi$. (It is a nice property of convolution that convolving anything with a smooth function gives a smooth function---if the integrals all converge nicely.)

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Also if $\phi_n$ is an approximate identity then for $f \in L^p(G)$ with $1 \leq p < \infty$ where $G$ is a locally compact group with a left Haar measure then $\phi_n * f \to f$ in $L^p$. A $C^\infty$ continuous approximate identity that works is easy to construct here, take for example a Gaussian $g$ and set $g_\epsilon(x) = \epsilon^{-n} g(x/\epsilon)$. –  Jonas Teuwen Dec 6 '10 at 17:16
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Work in a normal (resp. regular) borel measure space $(X,S;\mu)$, where $X$ is a normal topological space (resp. a locally compact hausdorff space).

Fix $p\in[1,\infty]$. Let $f\in L^{p}(\mu)$. First note that we can approximate $f$ by simple functions, where, by approximate, I mean topologically with respect to the norm $\|\cdot\|_{p}$ (even for the case that $p=\infty$). Next we can approximate simple functions (i.e. measurable functions with countable image), by simple functions with finite image (even for the case that $p=\infty$). Since simple functions are simply a sum of weighted characteristic functions, it suffices to approximate characteristic functions by continuous functions, for then the weighted sum of these continuous functions will be an approximation of the finite simple function.

So consider $A\subseteq X$ measurable so that $\chi_{A}\in L^{p}(\mu)$. Now assume $p<\infty$. Let $\varepsilon\in\mathbb{R}^{+}$. Let $C$ be closed (resp. compact) and $U\supseteq C$ open (resp. $\sigma$-bounded open), so that $C\setminus A, A\setminus U$ are null, and $\mu(U\setminus C)\leq\varepsilon^{p}$. Let $f:X\to[0,1]$ be continuous and separate $C$ from $X\setminus U$ (the topological conditions on $X$ ensure this is possible). Clearly $f\in L^{p}(\mu)$, since $\int |f|^{p}~\operatorname{d}\mu\leq\mu(U)\leq\mu(U\setminus C)+\mu(A)\leq\varepsilon+\|\chi_{A}\|_{p}^{p}<\infty$. Also $\int |f-\chi_{A}|^{p}~\operatorname{d}\mu\leq\varepsilon^{p}$, so $\|f-\chi_{A}\|_{p}\leq\varepsilon$.

This demonstrates that we can always approximate measurable functions from $L^{p}(\mu)$ by continuous functions in $L^{p}(\mu)$ for $1\leq p <\infty$.

(For $p=\infty$, this fails as follows. Consider e.g. $X=[0,1]$ and $\mu=\mathcal{L}|[0,1]$ the lebesgue measure. Let $A=[0,1/2]$ say. Suppose $f_{n}:X\to[0,1]$ were continuous and $f_{n}\rightarrow \chi_{A}$ in $L^{\infty}(\mu)$. Then $(f_{n})_{n}$ would be cauchy in the polish space $C(X,[0,1])$, and thus would converge within that space. But then the limit of the $f_{n}$ would be both continuous and not continuous. A contradiction.)

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Ah, I just realised I misread the question. Hmm, yes I would consider the case of (locally compact hausdorff) topological groups. –  user8873 Mar 30 '11 at 6:35
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