Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Calculate the maximal value of $\int_{-1}^1g(x)x^3 \, \mathrm{d}x$, where $g$ is subject to the conditions

$\int_{-1}^1g(x)\, \mathrm{d}x = 0;\;\;\;\;\;\;\;$ $\int_{-1}^1g(x)x^2\, \mathrm{d}x = 0;\;\;\;\;\;\;\;\;\;$ $\int_{-1}^1|g(x)|^2\, \mathrm{d}x = 1.$


I should mention that $g\in C[-1,1]$ with real scalars, and usual inner product. Just looking for the actual answer!

share|improve this question

3 Answers 3

up vote 2 down vote accepted

We denote with $$I= \int_{-1}^1\!dx\,g(x)x^3$$ the value of the integral. The constraints, we include via Lagrange multipliers. So we would like to maximize the integral $$J= I + \lambda \int_{-1}^1\!dx\,g(x) + \mu \int_{-1}^1\!dx\,g(x) x^2 + \nu \int_{-1}^1\!dx\,g(x)^2.$$

At the maximum of $J$, the first variation has to vanish: $$ \delta J =x^3 + \lambda + \mu x^2 + 2 \nu g(x)=0.$$

Thus we have $$g(x) = c_0 + c_2 x^2 + c_3 x^3.$$

  • From $\int_{-1}^1\!dx\,g(x)=0$ it follows $3 c_0 + c_2 =0$.

  • From $\int_{-1}^1\!dx\,g(x) x^2=0$ it follows $5 c_0 + 3c_2 =0\quad$ ($\Rightarrow c_0 = c_2 =0$)

  • From $\int_{-1}^1\!dx\,g(x)^2$ it follows $c_3 =\sqrt{7/2}$.

Thus, we have $$g(x) = \sqrt{\frac72} x^3$$ and the maximum value of $I$ reads $$ I_\text{max} = \sqrt{\frac27}.$$

share|improve this answer
    
The maximum value in this case if also the lower bound when one applies Hölder's inequality. But for Hölder only the third condition is necessary. The other two become interestingly redundant. –  johnny May 12 '12 at 20:54

Using some Hilbert space theory with the inner product $<f,g> = \int_{-1}^1 f(x)g(x)\,dx$, maximizing $\int_{-1}^1 x^3 g(x)$ with $\int_{-1}^1 g(x)^2\,dx = 1$ is the same as maximizing $<x^3,g>$ with the constraint that $||g|| = 1$. Thus $g(x) = cx^3$ for the constant $c$ giving $\int_{-1}^1 (cx^3)^2\,dx = 1$. The other two constraints are satisfied for such $g(x)$, so this will be your answer. Solving for $c$ you have $$c^2 \int_{-1}^1 x^6 = 1$$ This gives $c = \sqrt{7 \over 2}$ and $g(x) = \sqrt{7 \over 2}x^3$. In this case $\int_{-1}^1 x^3 g(x) = \sqrt{2 \over 7}$.

share|improve this answer

One solution to this problem would be to use Hilbert space methods. There is a well-known orthogonal basis to $L^2([-1,1])$ given by the Legendre polynmials. Let $P_0,P_1,\ldots$ be the Legendre polynomials, as defined in the Wikipedia link. Your conditions $$\tag{$*$}\int_{-1}^1 g(x)\,dx = 0\mbox{ and }\int_{-1}^1g(x)x^2\,dx = 0$$ are equivalent to the inner product equations $\langle g, P_0\rangle = 0$ and $\langle g, P_2\rangle = 0$. Moreover, it is easy to derive that $x^3 = \frac{2}{5}P_3 + \frac{3}{5}P_1$. If $\hat{P}_1$ and $\hat{P}_3$ denote the normalizations of $P_1$ and $P_2$, then $$\hat{P}_1 = \sqrt{\frac{3}{2}}P_1\mbox{ and }\hat{P}_3 = \sqrt{\frac{7}{2}}P_3,$$ so that $$x^3 = \frac{2}{5}\sqrt{\frac{2}{7}}\hat{P}_3 + \frac{3}{5}\sqrt{\frac{2}{3}}\hat{P}_1.$$ In order for $\langle g, x^3\rangle$ to be maximal, $g$ can therefore only have components in the $P_1$ and $P_3$ directions, i.e., $g = a\hat{P}_1 + b\hat{P}_3$ with $a^2 + b^2 = 1$. We have therefore reduced the problem to maximizing $$\langle g, x^3\rangle = \frac{2}{5}\sqrt{\frac{2}{7}}b + \frac{3}{5}\sqrt{\frac{2}{3}}a$$ under the constraint $a^2 + b^2 = 1$. Standard Lagrange multipliers techniques give that this quantity is maximized when $$a = \frac{3}{5}\sqrt{\frac{7}{3}}\mbox{ and }b = \frac{2}{5},$$ and that for these values of $a$ and $b$ one has $\langle g, x^3\rangle = \sqrt{2/7}.$ Moreover, $g$ is explicitly computed as $$g = a\hat{P}_1 + b\hat{P}_3 = \sqrt{\frac{7}{2}}x^3.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.