$$\int_0^1xf(x) dx=\int_0^1f(x) dx = 1$$

prove that

$$\int_0^1(f(x))^2 dx \geq 4$$

EDIT My attempt is as follows - I can use only the $\int xf(x)$dx part to get a bound $\int f^2(x) dx \geq 3$ from cauchy schwarz. I can't think of ways how to incorporate the other given condition.

  • 2
    $\begingroup$ I'm sure you would get some help if you showed some effort in solving the problem. Most people here are not interested in doing your homework for you. $\endgroup$
    – Winther
    Aug 10 '14 at 18:16
  • $\begingroup$ @Winther I added my attempt. $\endgroup$
    – Soham
    Aug 10 '14 at 18:31
  • 1
    $\begingroup$ Thats what I too tried at first:) As you say, CS is too weak here as it doesn't incorporate the second condition. Expanding $f$ in a basis of some orthogonal polynomials is the best way to go (see Jack's answer)! $\endgroup$
    – Winther
    Aug 10 '14 at 18:40
  • $\begingroup$ This is a very nice problem, indeed. Equality holds only for $f(x)=6x-2$, as shown below. $\endgroup$ Aug 10 '14 at 18:40

Of course once the inequality condition is obtained (or guessed by trying out a general linear polynomial), Cauchy-Schwarz is a breeze:

$$\int_0^1 f^2 dx \ge \frac{\left(\int_0^1 (3xf-f)dx\right)^2}{\int_0^1(3x-1)^2 dx } = \frac{4}{1}=4$$

Equality is when $f(x)$ is proportional to $3x-1$.

In general we can have $$\int_0^1 f^2 dx \ge \frac{\left(\int_0^1 (axf+bf)dx\right)^2}{\int_0^1(ax+b)^2 dx } = \frac{(a+b)^2}{a^2/3+ab+b^2}$$ where the maximum is when $(a, b)$ is proportional to $(3, -1)$, so $4$ is indeed the best possible.

  • $\begingroup$ I think you're missing an $x$ in the numerator of the middle term in the last expression. Nice derivation based only on the Cauchy-Schwarz inequality, with optimal case following from equality case of the Cauchy-Schwarz inequality. $\endgroup$ Aug 11 '14 at 8:28
  • $\begingroup$ @user161825 You're right, I missed it and edited it in now.. thanks. $\endgroup$
    – Macavity
    Aug 11 '14 at 8:29

Write your function in terms of the shifted Legendre polynomials $\tilde{L}_n(x)=L_n(2x-1)$, that are an orthogonal base of $L^2((0,1))$ with respect to the usual inner product. Assuming: $$ f(x) = \sum_{n=0}^{+\infty} a_n\,\tilde{L}_n(x), $$ the constraints give: $$ a_0 = 1,\qquad \frac{a_0}{2}+\frac{a_1}{6} = 1, $$ hence $a_0=1$ and $a_1=3$. This implies:

$$\int_{0}^{1}f(x)^2 dx = a_0^2+\sum_{n=1}^{+\infty}\frac{a_n^2}{2n+1}\geq 1+\frac{9}{3}=4.$$

Moreover, you have that equality holds only for $f(x)=6x-2$.

  • $\begingroup$ What made you think of using Legendre polynomials? As far as I can tell, this is much simpler than using the Fourier system for instance. $\endgroup$ Aug 10 '14 at 19:35
  • 4
    $\begingroup$ You have the need to write $\int x\,f(x)\,dx =1$ in terms of a finite number of coefficients. I would have used the Fourier system if the constraints were something like $\int_{0}^{1}f(x)\sin(2\pi x)\,dx = 1$, for instance. $\endgroup$ Aug 10 '14 at 19:40

We have for each $a,b$: $$ \int_0^1 \Big(f(x)-ax-b\Big)^2dx \geq 0$$ So we have \begin{eqnarray*} \int_0^1 f(x)^2 dx &\geq &2\int _0^1f(x)(ax+b)dx -\int _0^1 (ax+b)^2dx \\ &=&2(a+b) -{(a+b)^3-b^3\over 3a}\\ &=&2(a+b) -{a^2+3ab+3b^2\over 3} =:E \end{eqnarray*} This inequality is valid for all $a,b$, so even when $E$ achieves maximum: \begin{eqnarray*} E &=&\underbrace{-b^2 +b(2-a) - {(2-a)^2\over 4}} + \underbrace{{(2-a)^2\over 4} -{a^2\over 3}}\\ &=&-\Big(b - {2-a\over 2}\Big)^2+{-a^2-12a+12\over 12}\\ &\leq & {-a^2-12a+12\over 12} \\ &=& {-(a-6)^2+48\over 12}\\ &\leq & 4 \end{eqnarray*} Equality is achieved at $a=6$ and $b={2-a\over 2}= -2$, that is when $f(x)=6x-2$.


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