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If $f$ and $g$ are continuous on $[1,\infty)$ and $\phi(x)\geq0$ and be integrable in $[1,\infty) $. Also $ \int_{1}^{\infty}\phi(x) dx \neq 0$ and $ \int_{1}^{\infty} g(x) \phi(x) dx \neq 0$ then there exists some $c\in(1,\infty)$ such that $$ g(c)\int_1^\infty f(x) \phi(x) dx = f(c)\int_1^\infty g(x) \phi(x) dx $$

MY TRY - Let $$ \frac{\int_1^\infty f(x) \phi(x) dx} {\int_1^\infty g(x) \phi(x) dx} = \lambda $$ so $$ \int_{1}^{\infty} (f(x)-\lambda g(x)) \phi(x)dx =0 $$ After this i am stuck.

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  • $\begingroup$ You only want one pair of dollar signs around an entire expression. I've fixed that for you. $\endgroup$
    – J.G.
    Jun 18, 2020 at 15:35
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    $\begingroup$ @J.G. thanks. Any hints for the problem?? $\endgroup$
    – user775902
    Jun 18, 2020 at 15:36
  • $\begingroup$ @gt6989b Thanks $\endgroup$
    – user775902
    Jun 18, 2020 at 15:38
  • $\begingroup$ In your attempt, don't forget the parentheses: the last line should be $\int^\infty_1 \big( f(x) - \lambda g(x)\big) \phi(x) dx = 0$. $\endgroup$
    – ccroth
    Jun 18, 2020 at 15:38
  • $\begingroup$ @ccroth Thanks. Any ideas to attack the problem? $\endgroup$
    – user775902
    Jun 18, 2020 at 15:42

1 Answer 1

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If $$\int_1^\infty (f(x) - \lambda g(x))\phi(x)dx =0$$ With $\phi \geq 0$ (and $\phi \not\equiv 0$), then by continuity, there must be some $c\in(1,\infty)$ such that $f(c) - \lambda g(c) = 0$, otherwise if $f(x) -\lambda g(x) >0$ for all $x\in(1,\infty)$, the integral would be positive. If $f(x) - \lambda g(x) < 0$ for all $x\in(1,\infty)$ the integral would be negative.

Edit: I will discuss the case when $g(c) =0$ as this implies $f(c)=0$ and $\lambda =1$. I have shown $$\exists c\in (1,\infty), \text{ such that } g(c)\int_1^\infty f(x)\phi(x)dx = f(c)\int_1^\infty g(x)\phi(x)dx, \quad (1)$$

As a counter example to the original claim $$\exists c\in (1,\infty), \text{ such that } \frac{\int_1^\infty f(x)\phi(x)dx}{\int_1^\infty g(x) \phi(x)dx} = \frac{f(c)}{g(c)}, \quad (2)$$

Consider taking $\phi = 1_{[1,4]}$, and $f$ to be a bump function with support in $[1,2]$ and $g$ a shifted bump function of $f$ with support in $[3,4]$. This implies $\lambda =1$ and \begin{align} \int_1^\infty(f(x) - \lambda g(x))\phi(x)dx &= \int_1^4(f(x) - g(x))dx \\ &= \int_1^2 f(x)dx - \int_3^4g(x)dx =0 \end{align} But there is no point $c\in [1,\infty)$ such that $$\frac{f(c)}{g(c)} = 1$$ since $f \equiv 0$ on the support of $g$. Therefor you cannot guarantee (2), but you Can guarentee the exitence of $c\in (1,\infty)$ for the equality (1). In this example $f(2.5) = g(2.5) = 0$.

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  • $\begingroup$ Integral also contains $\phi$. How is the integral positive if f(x)-$\lambda$g(x)>0?? $\endgroup$
    – user775902
    Jun 18, 2020 at 15:51
  • $\begingroup$ @user775902 Because $\phi \geq 0$ $\endgroup$
    – Dayton
    Jun 18, 2020 at 15:52
  • $\begingroup$ so integral can be equal to 0? $\endgroup$
    – user775902
    Jun 18, 2020 at 15:52
  • $\begingroup$ I want c in open interval (1,$\infty$). How does this solution provide me that? $\endgroup$
    – user775902
    Jun 18, 2020 at 15:57
  • $\begingroup$ @ I got a c $\in$[1,$\infty$). How to exclude c=1.? $\endgroup$
    – user775902
    Jun 18, 2020 at 15:59

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