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Let $f:[0,1] \to \mathbb R$ be a continuous function s.t. $$\int_0^1f(x)g(x) dx=\int_0^1f(x)dx \int_0^1g(x)dx, $$ for all $g:[0,1]\to \mathbb R$ continuous and not differentiable. Prove that $f$ is constant.

I've found that choosing $h(x)=f(x)-\int_0^1f(t)dt$ gives $$\int_0^1h(x)g(x)dx=0. $$ I don't know how to proceed. Can somebody give me some tips, please?

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  • $\begingroup$ Proof by contradiction may be easier. $\endgroup$ – Yadati Kiran Mar 10 at 17:46
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    $\begingroup$ What happens if you plug in $g:= h$ ? $\endgroup$ – Max Mar 10 at 17:46
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    $\begingroup$ Are we sure about this formulation? Maybe you mean $g$ continuous but not per se differentiable? $\endgroup$ – Shashi Mar 10 at 17:50
  • $\begingroup$ if you plug g=h then integral of $h^2$ =0 $\endgroup$ – Gaboru Mar 10 at 17:52
  • $\begingroup$ And that means that $h(x)=0$ $\endgroup$ – Andrei Mar 10 at 17:54
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As Max et al have noted, the special case $g=h$, if valid, gives $\int_0^1 h^2(x) dx=0$. But $h$ is real-valued, so $h^2(x)\ge 0$ with equality iff $h(x)=0$. For $h^2$ to integrate to $0$, the fact that $h$ is continuous implies $h$ is identically zero, so $f(x)=\int_0^1 f(t) dt$ is constant.

You seem insistent all we know about $f$ is that each non-differentiable continuous choice of $g$ obtains $\int_0^1f(x)g(x)dx=\int_0^1f(x)dx\int_0^1g(x)dx$. But it doesn't matter: we can choose such $g$ arbitrarily close to $h$, by adding to $h$ a non-differentiable continuous noise $\eta$ times an arbitrarily small coefficient, viz. $g=h+\epsilon\eta$. Then $$\int_0^1 h^2(x)dx=lim_{\epsilon\to 0}\int_0^1 h(x)(h(x)+\epsilon\eta(x))dx=\lim_{\epsilon\to 0}0=0.$$

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  • $\begingroup$ why is the last $lim 0$? $\endgroup$ – Gaboru Mar 10 at 20:25
  • $\begingroup$ @Gaboru Because we're taking $g=h+\epsilon\eta$. $\endgroup$ – J.G. Mar 10 at 20:26
  • $\begingroup$ and why h(x)(h(x)+...)=0? $\endgroup$ – Gaboru Mar 10 at 20:46

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