I have a question which asks:

Let $g\in C[-1,1]$ and the usual inner product $\langle f,g\rangle = \int_{-1}^{1} f(x)g(x)dx$.

Find the max value of $\int_{-1}^{1}g(x)x^3dx$ where $g$ is subject to the restrictions:

$\int_{-1}^{1} g(x)dx=0$, $ \ \ \ $ $\int_{-1}^{1} g(x)x^2dx=0$, $ \ \ \ $ $\int_{-1}^{1} |g(x)|^2dx=1$

I'm not really sure how I am supposed to proceed with this question. It is in relation to Hilbert spaces and the orthogonal projection and this question Finding the min of an integral

Please Help.

  • $\begingroup$ Don't you feel $g(x)=x$ ! :) $\endgroup$
    – ABC
    Apr 8 '13 at 12:28
  • 1
    $\begingroup$ Did you try rewriting everything as a Hilbert space question? $\endgroup$ Apr 8 '13 at 12:34
  • $\begingroup$ @HaraldHanche-Olsen Your comment is perfect. Mine was silly, sorry. $\endgroup$
    – Julien
    Apr 8 '13 at 12:50
  • $\begingroup$ @HaraldHanche-Olsen I'm sorry I don't quite understand what you mean by re-writing everything as a Hilbert spaces question. I can see that my restrictions are requirements for an orthonormal basis but I am a bit confused as to what you mean? Thanks for the help $\endgroup$
    – hmmmm
    Apr 8 '13 at 13:00
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    $\begingroup$ I mean like this: “Find the maximum of $\langle g,x^3\rangle$ subject to $\langle g,1\rangle=\langle g,x^2\rangle=0$ and $\lVert g\rVert=1$.” $\endgroup$ Apr 8 '13 at 13:45

Here is a hint: Let $P$ be the projection on the span of $1$ and $x^2$, so $Q=I-P$ is the projection on the orthogonal complement of that space. By two of the constraints $g=Qg$, so that $\langle g,x^3\rangle=\langle Qg,x^3\rangle=\langle g,Qx^3\rangle$. So you want to compute $Qx^3$. $g$ is the unit vector with the largest possible inner product with that vector.

  • $\begingroup$ Isn't $Qx^3$ simply $x^3$? In that case, I am not sure how this helps. $\endgroup$
    – Julien
    Apr 8 '13 at 14:33
  • $\begingroup$ @julien Yes, so we are doing some unnecessary work, which is fine with me. 8-) $\endgroup$ Apr 8 '13 at 15:20

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