# Understanding denseness of $C^\infty$ in $L^p$ space.

I am not able to comprehend the proof " $C^\infty$ is dense in $L^p$ " given (here, page number 4). I would like to understand it properly . Any help would be appreciated. I am not able to understand the discussion from the second line onwards . Thank you !

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You take an $L^p$-function and approximate it by smooth functions which are constructed as convolutions of the original function with certain nice smooth functions. Which details are unclear to you? –  Rasmus May 21 '12 at 22:11
@ Rasmus first question is how can i show that its closure is the whole $L^p$ space ? and i don't know what $f_X(k)$ means in the proof. –  Theorem May 21 '12 at 22:15
@Ananda: There is no $f_X(k)$ in the proof - there is, however, the product $f\chi_{K}$, which is $f$ multiplied by the characteristic function of the set $K$. –  Martin Wanvik May 21 '12 at 22:18
You show that the closure is the entire space by showing that any $L^p$ function can be approximated arbitrarily well by a function in $C_0^\infty$, i.e. for every $f \in L^p$ and $\varepsilon > 0$, there exists a $g \in C_0^\infty$ such that $\| f - g \|_p < \varepsilon$. –  Martin Wanvik May 21 '12 at 22:20
BTW, you are not using comment replies properly. If you actually want the person you're replying to to be notified about your comment, you need to remove the space between the @-character and the username (see meta.stackexchange.com/questions/43019/…) –  Martin Wanvik May 21 '12 at 22:45

## 1 Answer

By looking at the proof, I guess there are two points that can be confusing. Feel free to ask more questions if this does not answer you original one.

The idea, as Martin and Rasmus suggested, is to take an arbitrary function $f\in L^p$, and for all $\epsilon>0$, find a smooth function $g$ with compact support such that $\|f-g\|_p<\epsilon$. In this proof, we have $g=(f\chi_K)*\phi_t$, where $\phi$ is a $C^\infty$-function with compact support and integral equal to $1$, and $K$ is a suitably chosen compact set. Now, in the proof, we have the following statement:

Let $K$ be a compact set chosen in such a way that $\|f\chi_{K^c}\|_p<\epsilon/2$.

This can be done, because $f\in L^p$ (so finite $L^p$-norm) and because $\mathbb{R}^n$ is a countable union of compact sets.

Finally, when choosing $t_0$, we use Theorem 1.4, and the fact that $(f\chi_K)*\phi_t$ is a $C^\infty$-function with compact support for all $t>0$ follows from Proposition 1.2 and Theorem 1.3.

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@M Turgeon i didn't understand how u actually choose $t_0$ , can you explain me ? –  Theorem May 21 '12 at 22:52
@Ananda Well, Theorem 1.4 (i) tells us that $(f\chi_K)*\phi_t$ converges to $f\chi_K$ in the $L^p$-norm (as $t$ goes to 0); hence, for any $\epsilon>0$, there exists $t_0$ such that, whenever $0<t<t_0$, we have $$\|f\chi_K-(f\chi_K)*\phi_t\|_p<\epsilon/2.$$ –  M Turgeon May 21 '12 at 23:00
@M Turgeon Thanks a lot :) I will go through it , and when i have doubt i will come again. –  Theorem May 21 '12 at 23:06