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Suppose I am given $F \subset [0,T]$ of non-zero measure (where $T \neq 0$), and a measurable function $f(x)$ on $F$ that is not zero on a set of non-zero measure. We can define, for measurable functions $\beta$ on $F$, the following quantity: $$ I_{F} (\beta) = \int_{F} f(x)\beta(x) \operatorname{dx} $$

My question is the following: does there exist $\beta$ ($\neq 0$ on set of positive measure) such that $I_F(\beta) = 0$?

This seems to be solvable by contradiction, rather than construction (due to the generality of the setting). Suppose there is no such $\beta$. Then, for all $\alpha$, $I_F(\alpha) \neq 0$. By taking $-\alpha$ instead of $\alpha$, we see that $I_F(\alpha)$ can also take negative values.

However, I am not able to see why if the function can take negative and positive values, then it can take the zero value. Furthermore, even if it can take such a value, I cannot see why $\beta$ can't be zero in that case.

If I am mistaken, and the proof actually proceeds through construction, then I would happy to know. Moreover, in the book I am reading, this lemma was classed as having "trivial" proof, so that baffles me even more.

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    You are just looking for a nonzero element of the orthogonal complement of the span of $f$ in $L^2(F,dx)$. – 2017-02-10
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    @Ian That's a new perspective. This would exist if the span of $f$ were not dense in $L^2 (F,dx)$, right? – 2017-02-10
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    That's correct. You can also construct such a thing if you are given an orthonormal basis of $L^2(F,dx)$. – 2017-02-10
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    @Ian All right, so in my case, it turns out the $f$ I have is just a measurable function. When can I assert a statement of non-density? Suppose I go even further and let $f$ be continuous, then can I assert this? – 2017-02-10
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    I'm fairly sure $L^2(F,dx)$ is isometrically isomorphic to $L^2([0,1],dx)$ for any measurable $F \subset [0,1]$ with positive measure. And in that case the desired result is clear (and quite explicit). – 2017-02-10
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    @Ian I do not know anything about $F$ (It could even be a fat Cantor set, for example). Nevertheless, if this isomorphism exists, then I think I can see the result you are hinting at. Can you try to describe this isomorphism? – 2017-02-10
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    I think you could begin with an orthonormal basis of $L^2([0,1],dx)$, restrict them to $F$, and then use Gram-Schmidt to re-orthogonalize upon restriction to $F$. The linear independence is retained upon restriction to $F$, so I think everything works out. (You could also do this with abstract Hilbert space theory.) – 2017-02-10
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    Ok, I see how this goes. Great, thanks for the input. – 2017-02-10
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    You can also see $I$ as a linear operator from $L^2$ to $R$. It is clear now that its kernel is non trivial. – 2017-02-10
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    @Michael Thank you that is also a nice way of looking at it, but why must the kernel be non-trivial (i.e. if the kernel were trivial, what can we conclude about $f$?) – 2017-02-11
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    @астонвіллаолофмэллбэрг $L^2$ is infinite dimensional vector space and $R$ has dimension 1. – 2017-02-11
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    @Michael Oh, of course! If at least one-non-zero value is taken, then we have that the map is surjective. Fine, that's excellent. Thank you. – 2017-02-13

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