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Let $f$ be a (Riemann) integrable function such that $\displaystyle \int_a^x f\,dx=0$ for all $x\in [a,b]$.
Prove that $\displaystyle \int_a^b fg\,dx=0$ for any integrable $g$.

The question will be easy if we assume that $f$ is continuous, since the given condition will imply $f=0$ everywhere.
However, when $f$ is only required to be integrable, it can be some strange function, like the popcorn function. In this case, I have no idea how to start with, and I can only hope that the given condition will imply $\displaystyle \int_a^b f^2\,dx = 0$.
Then we can use the Cauchy inequality $$\left(\int_a^b fg \,dx\right)^2\leq \left(\int_a^b f^2 \,dx \right) \left( \int_a^b g^2 \,dx\right)=0$$

StubbornAtom
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Howardli621
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  • Are you aware that a Riemann integrable function is continuous almost everywhere? – copper.hat May 11 '20 at 06:15
  • Yes, but I cannot figure out how it helps me to prove the proposition. Perhaps the set where $f \neq 0$ has measure 0 but I want a simpler proof. – Howardli621 May 11 '20 at 06:23
  • Perhaps you could approximate $g$ be a sequence of step functions? – copper.hat May 11 '20 at 06:27
  • I think it should be possible to prove that if $[c, d] $ is any subinterval of $[a, b] $ then $f$ vanishes at some point of $[a, b] $. This will imply the desired conclusion. However I am not able to show the property of $f$. – Paramanand Singh May 11 '20 at 10:06

2 Answers2

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If $g$ is integrable in the Riemannian sense, then $g$ is bounded. Say $$ m \leq g(x) \leq M $$ for all $x \in [a,b]$. Then by monotonicity of the integral we have that

$$ m \int_a^b f(x) dx \leq \int_{a}^b f(x) g(x) dx \leq M \int_{a}^{b} f(x) dx. $$

Yet $$ \int_a^b f(x) dx = 0 $$ therefore $$ 0 \leq \int_a^b f(x) g(x) dx \leq 0. $$

As was pointed out in the comment below, this works for $f \geq 0$. To fix this write observe that $$ \int_a^b f(x) dx = 0 \Rightarrow \int_a^b f^2(x) dx = 0. $$ A proof of this fact is done below. Then run the above argument with $f^2(x)$ instead of just $f$.

Moon Bears-C-
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    To have this inequality, you need to know that $f \geq 0$. And we don't know this. – Paul May 11 '20 at 06:36
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    Decompose $f = f^+ - f^-$ – Moon Bears-C- May 11 '20 at 06:37
  • Can we just take the absolute value? Since $\int_a^b |f|dx = 0$ is probably true. – Howardli621 May 11 '20 at 06:57
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    You can always decompose an integrable function as the difference of two non-negative functions. Take $f^+ = max(f(x), 0)$ and $f^- = max(-f(x),0)$. Then write $f = f^+ - f^-$. Since Integration is linear you can break it into two separate integrals, with each integrand being non-negative – Moon Bears-C- May 11 '20 at 07:01
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    To make your argument with the decomposition work you would have to show that $\int_a^b f^+(x) dx = 0$ and $\int_a^b f^-(x) dx = 0$. – Martin R May 11 '20 at 07:26
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    Btw is $f^-(x)=\max(-f(x),0)$ instead? – Howardli621 May 11 '20 at 07:38
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    @MoonBears-C-: Will you address my above concern? Why does $ \int_a^x f,dx=0$ for all $x$ imply that the integral for $f^+$ and $f^-$ is zero? – Martin R May 11 '20 at 07:59
  • Yes, we seem to have assumed that the set ${x: f(x)\geq 0}$ can be express as a union of disjoint intervals $(a_i,b_i)$ – Howardli621 May 11 '20 at 10:09
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    You should update your answer to address the concern raised by @MartinR. More so because your answer is in accepted state. – Paramanand Singh May 11 '20 at 10:25
  • Yes sorry, I went to bed last night. Thank you for your patience in pointing out the cases I didn't consider. I deeply appreciate it. – Moon Bears-C- May 11 '20 at 16:35
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    @MoonBears-C-: Sorry for bothering you again, but if you run your argument with $f^2(x)$ then you'll get $\int_a^b f(x)^2 g(x) dx = 0$, not $\int_a^b f(x) g(x) dx $. – Martin R May 11 '20 at 17:41
  • Fair point. Lemme play around with square roots, convexity, and triangle inequality and see if I can get something. – Moon Bears-C- May 11 '20 at 17:55
  • @MartinR I was able to get it with this approach. Unfortunately I'm in the middle of finals for the next few days. I'll update it again once this is all over. Thank you for pointing out the mistakes. This is essentially how grad school went for me: I do something that's attempting to be cute and clever, and my friends just point out cases until I have something that resembles being correct – Moon Bears-C- May 12 '20 at 03:00
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Let $F(x) =\int_{a} ^{x} f(t) \, dt$ so that $F(x) =0$ on whole of $[a, b] $. Next we observe that if $[c, d] $ is a sub-interval of $[a, b] $ then $f$ is continuous at some point $\xi\in[c, d] $ and thus $f(\xi) =F'(\xi) =0$ via Fundamental Theorem of Calculus.

Next let us assume that $\int_{a} ^{b} f(x) g(x)\, dx>0$ (the case of $<0$ can be handled by replacing $g$ with $-g$). Then there is sub-interval $[c, d] $ of $[a, b] $ of positive length on which $f(x) g(x) >0$. But this contradicts the fact that $f$ vanishes somewhere on this sub-interval. The contradiction proves the desired result.

Paramanand Singh
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