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Find the maximum of $\int\limits_{-1}^1x^3f(x)~dx$ for real-valued measurable functions $f(x)$ satisfying $$\int\limits_{-1}^1 |f(x)|^2~dx=1$$ $$\int\limits_{-1}^1f(x)~dx=\int\limits_{-1}^1 x f(x)~dx=\int\limits_{-1}^1 x^2 f(x)~dx=0.$$

My guess is to use Lusin's theorem and Stone-Weierstrass theorem, but then I do not know how to maximize the integral over all polynomials satisfying these conditions. I may have overlooked the properties of some orthogonal polynomials.

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  • $\begingroup$ Study simple functions satisfying the conditions. $\endgroup$ – md2perpe Jul 17 '17 at 19:59
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Legendre polynomials given an orthogonal base of $L^2(-1,1)$ with respect to the standard inner product. Assuming $$ f(x)\stackrel{L^2}{=}\sum_{n\geq 0}c_n P_n(x) \tag{1}$$ the given constraints can be written as $$ c_0=c_1=c_2=0,\qquad \sum_{n\geq 0}\frac{2c_n^2}{2n+1}=1 \tag{2} $$ hence the solution is given by a multiple of $P_3(x)=\frac{1}{2}\left(5x^3-3x\right)$, namely $\pm\sqrt{\frac{7}{2}}\,P_3(x)$.

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Using Lagrange multipliers, let $$ F(f, \lambda_0, \lambda_1, \lambda_2, \mu) = \int_{-1}^{1} f(x) (x^3 + \mu f(x) + \lambda_0 + \lambda_1 x + \lambda_2 x^2) \; dx $$ Then by taking the variation with respect to $f(x)$, we should have $$ x^3 + 2 \mu f(x) + \lambda_0 + \lambda_1 x + \lambda_2 x^2 = 0$$ i.e. $f(x)$ is a cubic polynomial that satisfies the constraints.

That polynomial turns out to be $\pm \frac{\sqrt{14}}{4} (5 x^3 - 3 x)$. For a maximum, you want the $\pm$ to be $-$.

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