I am in high school and have been trying to solve this problem for a few days. There should be no advanced calculus. I believe that there are no functions that work, but I can’t figure out how to prove that using math for every type of function. Any help would be appreciated. Here is the qustion:

Find all continuous positive functions, $0\le x \le 1$ That satisfy: $$\int_0^1f(x)dx=1$$ $$\int_0^1f(x)xdx=\alpha$$ $$\int_0^1f(x)x^2dx=\alpha^2$$ where $\alpha$ is an arbitrary real constant

Thanks for any advice.

  • 3
    $\begingroup$ Welcome. Strange to title “easy” when you in fact find it hard. You should provide a description of what the problem is in the title itself $\endgroup$
    – FShrike
    Jun 19, 2022 at 20:34
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    $\begingroup$ hint: since $f$ is positive and normalized, it is a probability distribution. Use the problem information to calculate the variance of $f$ $\endgroup$
    – user619894
    Jun 19, 2022 at 20:46
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    $\begingroup$ Equivalent to the intuition about probability, $\int_0^1f(x)(x-\alpha)^2\,\mathrm{d}x=0$ and the positive continuous criterion yields $f(x)(x-\alpha)^2\equiv0,f\equiv0$ which contradicts the hypotheses $\endgroup$
    – FShrike
    Jun 19, 2022 at 20:52
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    $\begingroup$ It's here and I remember it has been posted on this site as well but couldn't find the duplicate. $\endgroup$
    – WhatsUp
    Jun 19, 2022 at 21:29
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    $\begingroup$ I don't suppose that this is the intended answer, but if we allow the Dirac delta function and $0<\alpha<1$ then $f(x)=\delta (x-\alpha)$ does the trick. $\endgroup$
    – Blitzer
    Jun 20, 2022 at 10:32

1 Answer 1


Call the integrals $I_1$, $I_2$, $I_3$ for convenience.

$$I_1 = \int_0^1 f(x) \text{dx} = 1$$

$$I_2 = \int_0^1 xf(x) \text{dx} = \alpha$$

$$I_3 = \int_0^1 x^2f(x) \text{dx} = \alpha^2$$

Then $\alpha^2I_1 - 2\alpha I_2 + I_3 = \alpha^2 - 2\alpha^2 + \alpha^2 = 0$.

But also, $\alpha^2 I_1 - 2\alpha I_2 + I_3 = \int_0^1 \alpha^2f(x) - 2axf(x) + x^2f(x) \text{dx} = \int^1_0(\alpha-x)^2f(x)\text{dx}$.

Therefore $\int_0^1 (\alpha-x)^2 f(x) \text{dx} = 0$, which forces $f(x) = 0$. Which means that $I_1 \neq 1$ so no function $f$ exists.


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