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1) Apply best approximation theorem to compute min$\int_{-1}^1| x^3-a-bx-cx^2|^2 dx$.

2) Find max $\int_{-1}^1 x^3 g(x) dx$, where $g$ is subject to the constraints $$\int_{-1}^1 g(x)dx = \int_{-1}^1xg(x)dx=\int_{-1}^1 x^2 g(x) dx=0,\quad\text{and}\quad \int_{-1}^1|g(x)|^2 dx=1.$$

I don't know how to start to solve this problem, anyone can help me.

Robert Z
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J. Klins
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  • Robert Z They aren't unrelated. You can use the result of the first question to answer the second. –  Nov 27 '17 at 12:38
  • Use the theorem of projection on the $span{1,x,x^2}$ in $L^2(-1,1)$ – Guy Fsone Nov 27 '17 at 12:44
  • See https://math.stackexchange.com/questions/336182/a-proof-with-legendre-polynomials-and-an-integral-minimum-value – Robert Z Nov 27 '17 at 12:45
  • @Professor Vector Thanks for your enlightening comment! – Robert Z Nov 27 '17 at 13:57
  • @Robert Z You're welcome. In fact, my comment was a bit obscure, due to lack of time, but I had in mind Cauchy-Schwartz, indeed. –  Nov 27 '17 at 20:42

1 Answers1

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Hint for 1). Since $\int_{-1}^1x^d dx=0$ when $d$ is odd, and $\int_{-1}^1x^d dx=\frac{2}{d+1}$ when $d$ is even, by expanding the square we get $$\int_{-1}^1(x^3-a-bx-cx^2)^2 dx =\frac{2}{3}b^2-\frac{4}{5}b+\frac{2}{7}+2a^2+\frac{4}{3}ac+\frac{2}{5}c^2\\ =\frac{2}{3}\left(b-\frac{3}{5}\right)^2+\frac{8}{175}+2\left(a^2+\frac{2}{3}ac+\frac{1}{5}c^2\right). $$ Moreover $(1/3)^2-1/5<0$ implies that $\left(a^2+\frac{2}{3}ac+\frac{1}{5}c^2\right)\geq 0$.

Hint for 2) (see Professor Vector's comment). By the given constraints and the Cauchy-Schwarz inequality, $$\begin{align} \int_{-1}^1 x^3 g(x) dx&=\int_{-1}^1 (x^3-a-bx-cx^2)g(x) dx\\ &\leq \left(\int_{-1}^1 (x^3-a-bx-cx^2)^2dx\right)^{1/2} \left(\int_{-1}^1 (g(x))^2dx\right)^{1/2}\\ &\leq \left(\int_{-1}^1 (x^3-a-bx-cx^2)^2dx\right)^{1/2}. \end{align}$$ Hence the maximum of the LHS is less or equal to the minimum of the RHS (see 1)).

Robert Z
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