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My question is this : given $f \in L^\infty(\mathbb{R}^2)$, can we find a sequence $\phi_n$ of smooth, compactly supported functions (test functions) such that for any $g \in L^1(\mathbb{R}^2)$,

$$\int g \phi_n \rightarrow_n \int g f$$

i.e. $\phi_n$ converges weakly to $f$ in the weak * topology of $L^\infty$ ?

I know that a strong convergence is true for $L^p$, $p < \infty$ and wrong for $p=\infty$. However, it seems that if you only ask weak convergence it should be true even in $L^\infty$...

I have not been able to find a reference for this, either in Rudin's Functional Analysis or Brezis's book.

Is this true ? If so, can anyone provide me with a reference ?

Thanks in advance.

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Why do you work in $\Bbb R^2$ rather than in the real line? –  Davide Giraudo Oct 29 '12 at 17:50
    
The title and the body of the question are different. The answer to the question in the title is no. Weak closure of $C_c^\infty$ = norm closure of $C_c^\infty$ = continuous functions vanishing at infinity. –  commenter Oct 29 '12 at 18:16
    
@DavideGiraudo : because the problem I encountered which prompted me to ask myself this question is in the complex plane. I don't think the dimension matters here though. –  Glougloubarbaki Oct 29 '12 at 19:16
    
@commenter I don't see how the question in the title and the question asked are different. Also your answer is clearly wrong, since the constant function 1 does not tend to 0 at infinity (obviously...) and it is approximable by plateau function in the weak-* topology. –  Glougloubarbaki Oct 29 '12 at 19:19
    
In the title you ask for weakly dense (implying duality with $(L^\infty)'$) as opposed to weak*-dense (duality with $L^1$). That's all I'm saying. The constant function $1$ is not in the weak closure because any Hahn-Banach extension of the limit functional separates it from $C_{c}^\infty$. More of a nitpick than anything deep but I thought it was worth pointing it out. –  commenter Oct 29 '12 at 19:23

1 Answer 1

up vote 1 down vote accepted

It is true. Firstly, it is easy to see that there exists $\{\phi_n\}$, such that for each $n\ge 1$, $\phi_n$ satisfies the following conditions: (i) $\phi_n$ is smooth and compactly supported; (ii) $\|\phi_n\|_\infty\le \|f\|_\infty$; and (iii) $\lim_{n\to\infty} \phi_n=f$ a.e. on $\mathbb{R}^2$. Then the conclusion follows from dominated convergence theorem directly.

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could you please explain briefly how you would construct such $\phi_n$ ? I'm thinking it's a sort of diagonal argument where you approach $f$ on disks $D(0,n)$ covering the plane, but I don't see exactly how to approach $f$ –  Glougloubarbaki Oct 29 '12 at 19:15
1  
I think I got it. You take the convolution of $f_{|D(0,n)}$ with a sequence of mollifiers. this converges $L^1$ to $f_{|D(0,n)}$ so there is a subsequence which converges almost everywhere. then you use a diagonal extraction. did you have something simpler in mind ? –  Glougloubarbaki Oct 29 '12 at 19:56
    
Essentially my construction is the same as yours. If you want to avoid the diagonal argument, you may construct $\phi_n$ convergent to $f$ in $L^1(\mathbb{R}^2)$ first, and then take an a.e. convergent subsequence. –  23rd Oct 29 '12 at 20:24
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hum... except that $f$ is not in $L^1$ a priori, so there is no hope to find $\phi_n$ with compact supports converging $L^1$ to $f$... –  Glougloubarbaki Oct 29 '12 at 20:41
    
Ah, thank you for reminding me of this. –  23rd Oct 29 '12 at 20:53

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