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Finding Total number of positive continuous

function $g(x)$ in $[0,1]$ which satisfy $\displaystyle \int^1_{0}g(x)dx=1,\;\int^{1}_{0}xg(x)dx=2\,\ \int^1_0x^2g(x)dx=4.$

What I try :

I am Trying to solve using Cauchy schwarz Inequality

Using Cauchy Schwartz Inequality

$\displaystyle \bigg(\int^{1}_{0}(g(x))^2dx\bigg)\bigg(\int^1_0 x^2dx\bigg)\geq \bigg(\int^1_0 xg(x)dx\bigg)^2$

Equality Hold when $g(x)=x$

But I could not fit the above data

Please have a look on that how i solve it

Thanks

jacky
  • 5,194

2 Answers2

2

Notice that:

$$\begin{eqnarray}\int_0^1 (x - 2)^2 g(x) dx & = & \int_0^1 (x^2 - 4x + 4)g(x) dx \\ & = & \int_0^1 x^2 g(x) dx - 4 \int_0^1 x g(x) dx + 4 \int_0^1 g(x) dx \\ & = & 4-8+4 \\ & = & 0 \end{eqnarray}$$

But $(x-2)^2 g(x) > 0$ on $[0, 1]$, so the integral should also be strictly positive, which is impossible.

ConMan
  • 24,300
1

If such a $g$ would exist, then acording to the mean value theorem for definite integrals there is a $\xi\in[0,1]$ s.t. $$2=\int_0^1xg(x)dx=\xi\int_0^1g(x)dx=\xi\times 1=\xi$$ But as $\xi\in[0,1]$ it cannot be equal to 2. Thus there is not such $g$.

Maksim
  • 1,706
  • Thanks Maskim could you please explain me. According to M.V.T for integral , If $g(x)$ is continuous in $[0,1].$ Then $\displaystyle \int^1_0g(x)dx=g(c)(1-0),$ Where $c\in[0,1]$ – jacky Aug 27 '23 at 08:34
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    You are right, there a numerous versions of the MVT for definite integrals. The version which is used above is different from the one you have cited. The version used above can be found here: https://proofwiki.org/wiki/Mean_Value_Theorem_for_Integrals/Generalization – Maksim Aug 27 '23 at 08:50