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I want to use it for an alternative proof of What locally integrable function $f$ satisfies $\int_a^ b f(x) \phi'(x)dx=0 $ for each $\phi \in C_0^\infty(a,b)$

I need to pick a $\phi_n$ sufficiently close to $1_A$ but such that $||\phi_n+1_A||_\infty$ is not too big.

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Guillermo Mosse
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  • Small and bounded are different adjectives! In general you cannot have a uniform approximation by means of compactly supported functions. However, boundedness should be easier to get. – Siminore Jul 06 '16 at 14:14
  • I edited my question so it's more clear. Bounding an only $\phi$ should be quite easy. I want to bound all of them uniformly after certain $n$. I guess I can't. – Guillermo Mosse Jul 06 '16 at 14:45
  • What is $A$? what does "not so big" mean? Do you want to pick a sequence (as in the body of the question), or are you given such a sequence and you want to know if $|\cdot|_\infty$ is bounded (as in the title)? –  Jul 06 '16 at 15:05
  • I'm happy if I can do this with $A$ a closed, bounded interval of $\mathbb{R}$. My title question was wrong [again]. I want what's in the body of it. – Guillermo Mosse Jul 06 '16 at 15:43
  • Still, what does not so big mean. And please edit your question. –  Jul 06 '16 at 16:49
  • @ArcticChar, I edited my question 4 hours ago, when I answered you. – Guillermo Mosse Jul 06 '16 at 19:50

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$\phi$ is a Schwartz function (because it is in $C_0^\infty$). This implies there exists a constant $c>0$ such that $$ |\phi(x)|<\frac{c}{|x|^2} $$ for all x.

Maybe that helps?

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