1

What are the number of possible continuous functions $f(x)$ defined on $[0,1]$ for which $$ I_1 = \int_0^1 f(x)dx=1,\\ I_2 =\int_0^1 xf(x)dx =a,\\ I_3 = \int_0^1 x^2f(x)dx=a^2.$$ I have no idea how to to solve it. Can anyone help me?

SvanN
  • 2,307
Girish Kumar Chandora
  • 931

2 Answers2

6

Without a condition on the sign of the function, you can find a continuous $f$ satisfying all conditions. Namely, it is enough to choose $$ f(x) = \alpha x^2 + \beta x + \gamma $$ and to find $\alpha, \beta, \gamma$ by solving the (uniquely solvable) linear system obtained imposing the three conditions. (The solution should be $\alpha = (6a^2-6a+1)/30$, $\beta = -180 a^2 +192 a -36$, $\gamma = 30 a^2 - 36 a + 9$.)

On the other hand, there are no continuous functions satisfying $f\geq 0$ or $f\leq 0$ and the three given conditions. Indeed, one has $$ \int_0^1 (x-a)^2 f(x) \, dx = \int_0^1 x^2 f(x)\, dx - 2a \int_0^1 x f(x)\, dx + a^2 \int_0^1 f(x) \, dx = 0, $$ but the unique continuous function $f\geq 0$ (or $f\leq 0$) satisfying this condition is $f\equiv 0$, that clearly does not satisfy the first requirement.

Rigel
  • 14,434
1

One may exploit the fact that shifted Legendre polynomials provide an orthogonal base of $L^2(0,1)$ with respect to the standard inner product. The given constraints lead to $$ \int_{0}^{1}f(x)P_0(2x-1)\,dx = 1,\quad \int_{0}^{1}P_1(2x-1)\,dx=2a-1$$ $$\int_{0}^{1}f(x)P_2(2x-1)\,dx=6a^2-6a+1 $$ hence any function of the form $$ 1+(6a-3)P_1(2x-1)+(30a^2-30a+5)P_2(2x-1)+\sum_{m\geq 3}c_m P_m(2x-1) $$ does the job. However, if we assume $a\geq 0$ and further impose $f(x)\geq 0$ on $[0,1]$ we get the unique solution $f(x)\equiv 0$ associated to $a=0$, since the constraints ensure we are in the equality case of the Cauchy-Schwarz inequality $$ \left(\int_{0}^{1} x f(x)\,dx\right)^2\leq \int_{0}^{1}f(x)\,dx \int_{0}^{1}x^2 f(x)\,dx.$$

Jack D'Aurizio
  • 353,855
  • I didnt got your Cauchy schwartz inequality – Girish Kumar Chandora Aug 11 '17 at 19:23
  • @GirishKumarChandora: write $x,f(x)$ as $x\sqrt{f(x)}\cdot\sqrt{f(x)}$. Is it clear now? – Jack D'Aurizio Aug 11 '17 at 19:25
  • I got cauchy schwartz inequlaity. But i am ug student so i am unable to understand your answer. It will be helpful for me if u explain by other method or in detail – Girish Kumar Chandora Aug 11 '17 at 19:28
  • 1
    @GirishKumarChandora: let $a,b$ non-negative, square-integrable functions over $(0,1)$. If $$\left(\int_{0}^{1}a(x)b(x),dx\right)^2 = \int_{0}^{1}a(x)^2,dx \int_{0}^{1}b(x)^2,dx $$ holds, then $\frac{a}{b}$ is almost-everywhere constant on $(0,1)$. In your case, $x\sqrt{f(x)}=\lambda\sqrt{f(x)}$ implies $f\equiv 0$ a.e. on $(0,1)$, hence $f\equiv 0$ since $f$ is continuous. – Jack D'Aurizio Aug 11 '17 at 19:31
  • In our question how are u sure that f(x) >0 for x belongs to [0,1]? – Girish Kumar Chandora Aug 11 '17 at 19:35
  • The first part of my answer does not assume that. The second part of my answer states that if we assume $f(x)\geq 0$ on $[0,1]$, then... – Jack D'Aurizio Aug 11 '17 at 19:38
  • What would happen if f(x)<0 for given interval, – Girish Kumar Chandora Aug 11 '17 at 19:40
  • Then $a$ would be negative. – Jack D'Aurizio Aug 11 '17 at 19:41
  • Uptil now i only have concept of cauch schwartz inequality and i got only your last step...thanks for explaining . If u explain by other method except legendre polynomial then i will definitly got it. – Girish Kumar Chandora Aug 11 '17 at 19:45
  • . In our case, $x\sqrt{f(x)}=\lambda\sqrt{f(x)}$ implies either $\sqrt{f(x)}=0 or \x-\lambda=0$. Why we cant take $x= \lambda$ – Girish Kumar Chandora Aug 11 '17 at 19:52
  • @GirishKumarChandora: the only non negative function that for some $\lambda$ ensures $x\sqrt{f(x)}=\lambda\sqrt{f(x)}$ for any $x\in(0,1)$ is the zero function. – Jack D'Aurizio Aug 11 '17 at 19:54
  • Thnks a lot for explaining. But i am still stuck by f(x)<0 or f(x) >0 . As we assumes f(x) >0 for given interval so we can apply schwartz equlality. If f(x)<0 for some sub interval then how we can do – Girish Kumar Chandora Aug 11 '17 at 20:01
  • If you have $f(x)\leq 0$ on $[0,1]$ and $a\leq 0$ then by renaming $f(x)$ as $-f(x)$ and $a$ as $-a$ you are in the same situation outlined by the second part of my answer. So, for short, if $a\neq 0$ and $f(x)$ is $\geq 0$ or $\leq 0$ on $[0,1]$, then $f\equiv 0$. If we allow $f(x)$ to change its sign on $[0,1]$, there are plenty of solutions for any value of $a$, see the first part of my answer. – Jack D'Aurizio Aug 11 '17 at 20:11
  • Thank you. I think i will get it only after reading the concept of legendre. – Girish Kumar Chandora Aug 11 '17 at 20:15