Maximum of a definite integral

Question

I am required to find

$max\int_{-1}^{1}x^3g(x)dx$

given that

$\int_{-1}^{1}x^ng(x)dx=0$ for $n=0,1,2$ and $\int_{-1}^{1}|g(x)|^2dx=1$.

The hint is to find $\underset{a,b,c}{min}\int_{-1}^{1}|x^3-a-bx-cx^2|^2dx$ first.

Below is what I have found.

$\underset{a,b,c}{min}\int_{-1}^{1}|x^3-a-bx-cx^2|^2dx$ is quite easy.

Let $f(x)=x^3-a-bx-cx^2$, then by Holder's Inequality,

$\int_{-1}^{1}x^{3}g(x)dx\leq \int_{-1}^{1}|f(x)g(x)|dx\leq (\int_{-1}^{1}|f(x)|^2dx)^{\frac{1}{2}}(\int_{-1}^{1}|g(x)|^2dx)^{\frac{1}{2}}=(\int_{-1}^{1}|f(x)|^2dx)^{\frac{1}{2}}$ which has a minimum value from above.

But I am not sure if I am working in the right direction since it seems like I am only able to find an upper bound rather than the maximum value.

Any advice?

Answer 1

Have you tried verifying that $f$ satisfies the orthogonality properties verified by $g$ (which should be the case to satisfy the minimum) and replace $g$ by $f$ to have a case of equality in your inequality ?

Answer 2

I suppose you started with $$f(x)=\frac25P_3(x)+\frac35P_1(x)+c_0P_0(x)+c_1P_1(x)+c_2P_2(x)$$ And found that $$\int_{-1}^1|f(x)|^2dx=\frac27\left(\frac25\right)^2+\frac25|c_2|^2+\frac23\left|\frac35+c_1\right|^2+\frac21|c_0|^2$$ And so found the minimum value of $\frac8{175}$ then $f(x)=\frac25P_3(x)$. From that point I am assuming that $g(x)$ is also a real function of $x$, else how could you find the maximum of $\int_{-1}^1x^3g(x)dx$?

Let $$\begin{align}I(\alpha)&=\int_{-1}^1\left(g(x)-\alpha f(x)\right)^2dx\\ &=\int_{-1}^1(g(x))^2dx-2\alpha\int_{-1}^1g(x)f(x)dx+\alpha^2\int_{-1}^1(f(x))^2dx\\ &\ge0\end{align}$$ Then $I(\alpha)$ has a minimum when $$I^{\prime}(\alpha)=0=-2\int_{-1}^1g(x)f(x)dx+2\alpha\int_{-1}^1(f(x))^2dx$$ So applying this value of $\alpha$, $$1-\frac{\left[\int_{-1}^1g(x)f(x)dx\right]^2}{\int_{-1}^1(f(x))^2dx}\ge0$$ So that $$\begin{align}\left[\int_{-1}^1g(x)f(x)dx\right]^2&=\left[\int_{-1}^1g(x)(x^3-a-bx-cx^2)dx\right]^2\\ &=\left[\int_{-1}^1x^3g(x)dx\right]^2\le\int_{-1}^1(f(x))^2dx\end{align}$$ For any real $f(x)=x^3-a-bx-cx^2$, therefore it can be no bigger than the minimum possible value, $$\left[\int_{-1}^1x^3g(x)dx\right]^2\le\int_{-1}^1\left(\frac25P_3(x)\right)^2dx=\frac4{25}\frac27=\frac8{175}$$ Of course for $g(x)=\sqrt{\frac72}P_3(x)$, $$\int_{-1}^1(g(x))^2dx=1$$ and $$\int_{-1}^11\cdot g(x)dx=\int_{-1}^1xg(x)dx=\int_{-1}^1x^2g(x)dx=0$$ And $$\int_{-1}^1x^3g(x)dx=\int_{-1}^1\frac25P_3(x)\sqrt{\frac72}P_3(x)dx=\frac25\sqrt{\frac27}$$ So the bound is actually attainable.