# Easy Putnam Question [closed]

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.

• 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 Jun 19, 2022 at 20:34
• hint: since $f$ is positive and normalized, it is a probability distribution. Use the problem information to calculate the variance of $f$ Jun 19, 2022 at 20:46
• 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 Jun 19, 2022 at 20:52
• It's here and I remember it has been posted on this site as well but couldn't find the duplicate. Jun 19, 2022 at 21:29
• 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. Jun 20, 2022 at 10:32

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.