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I am very beginner in the theory of weak derivative. I am trying to fix the following problem:

Suppose that $f\in{L_{loc}^{1}}$ and $\int_{a}^{b}f(x)\phi({x})dx=0$ for all $\phi\in{C_{0}^{\infty}}$ then $f(x)=0$ $a.e.$ on $(a,b)\subset{\mathbb{R}}$.

I will appreciate your ideas.

  • No apologies for trivial things! People here on this site are fond of it, since it gives them the opportunity to provide an answer. Continuously they are eagerly searching for that. – drhab Feb 28 '14 at 08:31
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    Do you believe that a $\phi$ can have small, measurable support and be positive? (Typically called "bump functions".) If so, what must the value of $f(x)$ be on that support? (and on every measurable subset of that support for which we can construct a suitable alternative bump function) – Eric Towers Feb 28 '14 at 08:42
  • @drhab This one is actually far from being trivial. – Etienne Feb 28 '14 at 08:49
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    @Algebra See for example here: http://math.stackexchange.com/questions/78142/the-constant-distribution – Etienne Feb 28 '14 at 08:51
  • @Etienne In the original question (there was an edit) the OP apologized for asking maybe something trivial. Only on that I wrote my comment. The question itself is beyond my scope. – drhab Feb 28 '14 at 08:52
  • @ Etienne: Thanks for providing a link for the similar problem. –  Feb 28 '14 at 08:53
  • @drhab I understand. Sorry if my comment sounded "agressive"; this was not at all my intention. – Etienne Feb 28 '14 at 09:37

1 Answers1

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One useful technique you can use here is regularization by convolution. For instance, a typical regularization result (that is relevant here) is the following:

Regularization: Let $\phi\in C^\infty_0$ be nonnegative, with $\int \phi\,dx = 1$. For each $t>0$, define $\phi_t = t^{-1}\phi(x/t)$. Then for any $f\in L^1_{loc}$, the convolutions $f*\phi_t$ are smooth functions and $f*\phi_t\to f$ in $L^1_{loc}$ as $t\to 0$.

Now let's apply this result to your problem. Choose a $\phi$ as in the theorem. By definition, $$(f*\phi_t)(x) = \int f(y)\phi_t(x - y)\,dy.$$ But the function $y\mapsto \varphi_t(x - y)$ is a compactly supported smooth function, so by your assumption on $f$, the integral is $0$, i.e., $(f*\phi_t)(x) = 0$ for all $x$. That is to say $f*\phi_t \equiv 0$. As $t$ was arbitrary, if you let $t\to 0$ the regularization theorem says that $0 \equiv (f*\phi_t)\to f$ in $L^1_{loc}$. Thus $f = 0$ in $L^1_{loc}$.

froggie
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  • But our case is $\phi:(a,b)\rightarrow{\mathbb{R}}$ then $x/t$ may not be in $(a,b)$ and then $\phi(x/t)$ is not defined. How do we control this issue? –  Mar 07 '14 at 23:35
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    @Algebra: you can always extend $\phi$ to a function defined on all of $\mathbb{R}$ simply by setting $\phi\equiv 0$ outside of $(a,b)$. This function remains compactly supported and smooth. – froggie Mar 08 '14 at 09:34