# Inequality of integrals $\int_0^1(f(x))^2 dx \geq 4$ if $\int_0^1xf(x) dx=\int_0^1f(x) dx = 1$

If

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

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.

• 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. Aug 10 '14 at 18:16
• @Winther I added my attempt. Aug 10 '14 at 18:31
• 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)! Aug 10 '14 at 18:40
• This is a very nice problem, indeed. Equality holds only for $f(x)=6x-2$, as shown below. 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.

• 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. Aug 11 '14 at 8:28
• @user161825 You're right, I missed it and edited it in now.. thanks. 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 $f(x)=6x-2$.

• 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. Aug 10 '14 at 19:35
• 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. 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 $f(x)=6x-2$.