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Number of all positive continuous function $f(x)$ in $\left[0,1\right]$ which satisfy $\displaystyle \int^{1}_{0}f(x)dx=1$ and $\displaystyle \int^{1}_{0}xf(x)dx=\alpha$ and $\displaystyle \int^{1}_{0}x^2f(x)dx=\alpha^2$

Where $\alpha$ is a given real numbers.

$\bf{My\; Try::}$ :: Adding $(1)$ and $(3)$ and subtracting $2\times (2),$ we. Get $$\displaystyle \int^{1}_{0}(x-1)^2f(x)dx=(\alpha-1)^2$$ now how can I solve it after that, Thanks

Community
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juantheron
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3 Answers3

7

One way is to think of $f(x)$ as a probability measure. Then you are seeking a continuous random variable $Y$ such that $E(Y)^2=E(Y^2)$. Since $E(Y^2)-E(Y)^2$ is the variance of $Y$, this value is zero if and only if $Y$ is a constant, and thus $f$ would be a delta function, not a continuous real-valued function.

The standard proof of this is to note:

$$\begin{align} 0<\int_{0}^1 (x-a)^2 f(x)\,dx &= \int_0^1 x^2f(x)\,dx-2a\int_0^1 xf(x)\,dx + a^2\int_0^1 f(x)\,dx\\ &= \int_0^1 x^2f(x)\,dx - 2a^2+a^2\\ &=\int_0^1 x^2f(x)\,dx -a^2 \end{align}$$

So $$\int_0^1 x^2f(x)\,dx >a^2$$

Thomas Andrews
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5

By the Cauchy-Schwarz inequality,

$$ \alpha^2 = \bigg( \int_{0}^{1} x f(x) \, dx \bigg)^2 \leq \bigg( \int_{0}^{1} f(x) \, dx \bigg)\bigg( \int_{0}^{1} x^2 f(x) \, dx \bigg) = \alpha^2. $$

Since the inequality is saturated, it reduces to an equality. This implies that $f(x)$ is a constant multiple of $x^2 f(x)$, which is impossible. Therefore no such function $f$ exist.

Remark. Positivity of $f$ is essential in proving non-existence of such $f$. If the positivity of $f$ is dropped off, then there are infinitely many solutions. Even you can find a quadratic solution!

Sangchul Lee
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  • Thanks sos440, But i did not understand Why no such positive function for $f(x) = \lambda x^2f(x),$ bcz When $\displaystyle x=\pm \frac{1}{\sqrt{\lambda}},$ Then these two are equal, plz clearfy me – juantheron Sep 15 '16 at 03:00
  • @juantheron, The point is that $f(x) = \lambda x^2 f(x)$ is an equality between functions on $[0, 1]$. In particular, this should be true for all $x \in [0, 1]$ simultaneously for your $\lambda$. Now your computation shows that this cannot happen. – Sangchul Lee Sep 15 '16 at 03:02
  • Thanks sos, But i did not understand Last line. – juantheron Sep 15 '16 at 03:10
  • @juantheron Your computation shows that $x$ must reduce to $1/\sqrt{\lambda}$. Is it possible that this holds for all $x \in [0, 1]$? – Sangchul Lee Sep 15 '16 at 03:13
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I believe there are no such functions. Indeed, by combining the equations, we get that for every polynomial $P$ of degree $\le2$, we have

$$\int_0^1 P(x) f(x) dx = P(\alpha) $$

In particular, for $P=(x-\alpha)^2 $, we get

$$\int_0^1 (x-\alpha)^2 f(x) dx = 0$$

The integrand $x \mapsto (x-\alpha)^2 f(x)$ is a non negative continuous function whose integral is zero, so it's zero, and $f=0$, which contradicts the fact that it's integral over $[0,1] $ is $1$.

Joel Cohen
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