# 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 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