# Old graduate analysis qualifying exam question

I am working through an old real analysis qualifying exam. Most of the problems are measure theory related except perhaps one, which I am having some trouble (probably because I have not seen this type of question before). I would be very appreciated if someone can help me. Here is the question:

Compute $$\underset{a,b,c}{\min}{\textstyle\int_{-1}^{1}} \left\vert x^{3}-a-bx-cx^{2}\right\vert dx$$

and find $$\max{\int_{-1}^{1}}x^{3}g(x)dx,$$ where $g$ is subject to the restrictions $$\begin{gather*} \int_{-1}^{1}g(x)dx={\int_{-1}^{1}x}g(x)dx={\int_{-1}^{1}}x^{2}g(x)dx=0;\\ {\int_{-1}^{1}}\left\vert g(x)\right\vert ^{2}dx=1 \end{gather*}$$

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I was initially thinking this was very similar to math.stackexchange.com/questions/231109/…, but I misread, and it might not be quite so similar. It would be easier if the integrand first were squared. The second part makes it look like inner product space methods should be used. –  Jonas Meyer Dec 13 '12 at 6:05
I would hazard a guess that the first involves showing that $a=c=0$ and then minimizing over $b$. The second can be relaxed to have $\|g\|_2 \leq 1$. –  copper.hat Dec 13 '12 at 6:06
First part is a special case of math.stackexchange.com/questions/97628 –  David Speyer Dec 13 '12 at 14:37

For the first, let $\phi(a,b,c) = \int_{-1}^{1} | x^{3}-a-bx-cx^{2}| dx$. Notice that $\phi(a,b,c) = \phi(-a,b,-c)$ and since $\phi$ is convex, we have $\phi(0,b,0) \leq \frac{1}{2} (\phi(a,b,c) +\phi(-a,b,-c))$. Hence we can assume that $a=c=0$ and the problem reduces to minimizing $\xi(b) = \int_{-1}^{1} | x^{3}-bx| dx = 2\int_{0}^{1} x| x^{2}-b| dx$. If $b<0$, we have $| x^{2}-b| = x^2-b>x^2$, hence $\xi(b) \geq \xi(0)$. So we can assume that $b \geq 0$. Similarly, if $b>1$, we have $| x^{2}-b| =b-x^2 > 1-x^2 = |x^2-1|$ over the range of integration. Hence $\xi(b) \geq \xi(1)$. Hence we may assume that $b \in [0,1]$. Evaluating the integral gives $\xi(b) = 2\frac{1}{4}(2 b^2-2b+1)$, and minimizing with respect to $b$ over $[0,1]$ gives $b = \frac{1}{2}$, and evaluating gives $\xi(\frac{1}{2}) = \frac{1}{4}$.

For the second, letting $p_k(x) = x^k$, and choosing the space as $L_2[-1,1]$ lets us write the problem as $\sup \{ \langle p_3, g \rangle | \langle p_k, g \rangle = 0 , k=0,1,2, \ \ \|g\| = 1\}$. If we let $S = \text{sp} \{p_k\}_{k=0}^2$ (which is closed, since it is finite dimensional), the problem can be written as $\sup_{g \in S^\bot, \|g\| = 1} \langle p_3, g \rangle$. Since the function $g \mapsto\langle p_3, g \rangle$ is linear,this problem is the same as $\sup_{g \in S^\bot, \|g\| \leq 1} \langle p_3, g \rangle$. Now let $\Pi$ be the orthogonal projection onto $S^\bot$, then we can write the problem as $\sup_{\|\Pi g\| \leq 1} \langle p_3, \Pi g \rangle$, and since $\|\Pi g\| \leq \|g\|$, we see that this is equal to $\sup_{\| g\| \leq 1} \langle p_3, \Pi g \rangle = \sup_{\| g\| \leq 1} \langle \Pi^* p_3, g \rangle = \|\Pi p_3\|$, since $\Pi$ is self-adjoint. Let $P_n$ be the Legendre polynomials, then we see that $p_3 = \sum_{k=0}^3 \alpha_k P_k$, and $S = \text{sp} \{P_k\}_{k=0}^2$. Since the $P_n$ are orthogonal, we see that $\Pi p_3 = \alpha_3 P_3$, and the solution to the problem is $|\alpha_3\|\|P_3\|$. Since $P_3(x) = \frac{1}{2}(5 x^3-3x)$, we see that we must have $\alpha_3 = \frac{2}{5}$. Since $\|P_3\| = \sqrt{\frac{2}{7}}$, we see that the answer is $|\alpha_3\|\|P_3\| = \frac{2}{5} \sqrt{\frac{2}{7}} = \sqrt{\frac{8}{175}}$.

(Addendum: Note that a minimizing $g$ is $g = \frac{1}{\|\Pi p_3\|} \Pi p_3 = \frac{1}{\|P_3\|} P_3 = \sqrt{\frac{7}{2}}\frac{1}{2}(5 p_3-3p_1)$, or $g(x) = \sqrt{\frac{7}{8}}(5 x^3-3x)$.)

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Note that I had made a mistake in the computation of $b$ for the first answer (I miscalculated the integral). The minimum value of $xi$ remains unchanged. –  copper.hat Dec 13 '12 at 16:10

In this answer, I'll give some relevant literature references. I will not prove any results.

1. Answer: $\mathbf{a=c=0,\ b=1/2}$.

The solution to your first problem is given by a Chebyshev polynomial of the second kind (properly scaled). By definition, the Chebyshev polynomial $U_k(t)$ is the polynomial $p(t)$ of degree $k$ with the leading coefficient $2^k$ that minimizes $\int_0^1 |p(t)| dt$. See e.g. http://en.wikipedia.org/wiki/Chebyshev_polynomials . It is known that $$U_3(t) = 8x^3 - 4x.$$ Accordingly, the polynomial of degree 3 with the leading coefficient $1$ that minimizes $\int_0^1 |p(t)| dt$ is $x^3 - x/2$. So the answer to your first question is $a=c=0$, $b=1/2$.

2. Answer: g(x) = $\mathbf{\sqrt{7/8}\, (5x^3 -3x)}$, $\mathbf{\int_{-1}^1 x^3 g(x) dx = \frac{2\sqrt{14}}{35}}$.

Your second question asks to find a function $g(x)\in L_2[-1,1]$ that

• has norm $1$,

• is orthogonal to the linear space of polynomials of degree at most $2$,

• maximizes the inner product with $x^3$.

Let us prove that $g(x)$ is a polynomial of degree $3$. Indeed, let $g(x)$ be the optimal solution. Suppose that $g(x)$ is not a polynomial of degree 3. Then let $f(x)$ be the orthogonal projection of $g(x)$ on the set of polynomials of degree $3$. We have,

• $f(x)$ is orthogonal to all polynomials of degree at most 2.

• $\int_{-1}^1 x^3 \cdot f(x) dx = \int_{-1}^1 x^3\cdot g(x) dx$

• and $\|f\| < \|g\| =1$.

Thus $f(x)/\|f\|$ is a better solution than $g(x)$: $$\int_{-1}^1 x^3 \cdot \frac{f(x)}{\|f\|} dx = \int_{-1}^1 x^3 \cdot \frac{g(x)}{\|f\|} dx > \int_{-1}^1 x^3 \cdot g(x) dx.$$ This contradicts to our assumption that $g(x)$ is the optimal solution.

That is, the solution to this problem is a polynomial $g(x)$ of degree 3 that is orthogonal to all polynomials of degree at most 2 and has norm 1.

A polynomial of degree $k$ that is orthogonal to all polynomials of degree less than $k$ is known as the Legendre polynomial (there is only one such polynomial up to normalization). See http://en.wikipedia.org/wiki/Legendre_polynomials .

The answer to your question is the Legendre polynomial of degree 3, properly scaled: $$\sqrt{7/8}\, (5x^3 -3x).$$

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$\int_{-1}^1 x^3 g(x) dx = \sqrt{\frac{8}{175}}$. –  copper.hat Dec 13 '12 at 17:20
@copper.hat: $\sqrt{8/175} = \frac{2\sqrt{14}}{35}$ ;-) –  Yury Dec 13 '12 at 20:46
My apologies, I completely tripped on that! –  copper.hat Dec 13 '12 at 20:52