# $f(c)\int_{1}^{\infty} f(x) \phi(x) dx =g(c)\int_{1}^{\infty} g(x) \phi(x) dx$

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.

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

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 $$f(2.5) = g(2.5) = 0$$.

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