Big Rudin: Chapter 4, Exercise 15

As promised, here is the solution to exercise 15 of chapter 4 in Walter Rudin’s Real and Complex Analysis. So the reader does not have to flip through the text, I have reproduced the statement of problem below.

Compute

\displaystyle\min_{a,b,c\in\mathbb{R}}\int_{0}^{\infty}\left|x^{3}-a-bx-cx^{2}\right|^{2}e^{-x}dx

State and solve the corresponding maximum problem, as in Exercise 14.

Define a weight function w:[0,\infty)\rightarrow[0,\infty) by w(x):=e^{-x}. Note that the expression

\displaystyle\langle{f,g}\rangle_{w}=\int_{0}^{\infty}f(x)\overline{g(x)}w(x)dx

is well-defined for measurable complex-valued functions f and g satisfying

\displaystyle\max\left\{\int_{0}^{\infty}\left|f(x)\right|^{2}w(x)dx,\int_{0}^{\infty}\left|g(x)\right|^{2}w(x)\right\}<\infty

since, by Cauchy-Schwarz,

\displaystyle\int_{0}^{\infty}\left|f(x)\overline{g(x)}w(x)\right|dx=\int_{0}^{\infty}\left|f(x)w(x)^{\frac{1}{2}}\right|\left|\overline{g(x)}w(x)^{\frac{1}{2}}\right|dx\leq\left\|f\right\|_{w}\left\|g\right\|_{w}

I leave it to the reader that \langle{,}\rangle_{w} defines an inner product.

Define polynomials p_{0}:=1, p_{1}:=x, and p_{2}:=x^{2}. We use the Gram-Schmidt algorithm to obtain an orthonormal basis for M. First, an elementary integral calculus lemma.

Lemma 1. Let n be a nonnegative natural number. Then

\displaystyle\int_{0}^{\infty}x^{n}e^{-x}dx=n!

Proof. We prove this result by induction on n. From basic integral calculus, we see that

\displaystyle\int_{0}^{\infty}e^{-x}dx=\left[-e^{-x}\right]_{x=0}^{x=\infty}=1

Now suppose that the result is true for some n>1. Integrating by parts, we obtain

\displaystyle\int_{0}^{\infty}x^{n+1}e^{-x}dx=\left[-x^{n+1}e^{-x}\right]_{x=0}^{x=\infty}+(n+1)\int_{0}^{\infty}x^{n}e^{-x}dx=(n+1)n!=(n+1)!

\Box

An easy consequence of the lemma and the linearity of the integral is that, for any polynomial p=a_{0}+a_{1}x+\cdots+a_{n}x^{n} of degree n,

\displaystyle\int_{0}^{\infty}p(x)e^{-x}dx=\sum_{i=0}^{n}a_{i}\int_{0}^{\infty}x^{i}e^{-x}dx=\sum_{i=0}^{n}a_{i}i!

Using the lemma, we see that

\left\|p_{0}\right\|=\left(\int_{0}^{\infty}e^{-x}dx\right)^{\frac{1}{2}}=\left(\left[-e^{-x}\right]_{x=0}^{x=\infty}\right)^{\frac{1}{2}}=1

So we define e_{0}:=p_{0}=1. Applying the Gram-Schmidt algorithm, we obtain that

\begin{array}{lcl}\displaystyle e_{1}:=\dfrac{x-\langle{x,1}\rangle_{w}}{\left\|x-\langle{x,1}\rangle_{w}\right\|}=\dfrac{x-\int_{0}^{\infty}xe^{-x}dx}{\left\|x-\langle{x,1}\rangle_{w}\right\|}&=&\displaystyle\dfrac{x-\int_{0}^{\infty}e^{-x}dx}{\left\|x-\langle{x,1}\rangle_{w}\right\|}\\[2 em]&=&\displaystyle\dfrac{x-1}{\left(\int_{0}^{\infty}(x-1)^{2}e^{-x}dx\right)^{\frac{1}{2}}}\\[2 em]&=&\displaystyle x-1\end{array}

Lastly,

\begin{array}{lcl}\displaystyle e_{2}&=&\displaystyle\dfrac{x^{2}-\langle{x^{2},e_{1}}\rangle_{w}e_{1}-\langle{x^{2},e_{0}}\rangle_{w}e_{0}}{\left\|x^{2}-\langle{x^{2},e_{1}}\rangle_{w}e_{1}-\langle{x^{2},e_{0}}\rangle_{w}\right\|_{w}}\\[2 em]&=&\displaystyle\dfrac{x^{2}-\left(\int_{0}^{\infty}x^{2}(x-1)e^{-x}dx\right)(x-1)-\left(\int_{0}^{\infty}x^{2}e^{-x}dx\right)}{\left\|x^{2}-\langle{x^{2},e_{1}}\rangle_{w}e_{1}-\langle{x^{2},e_{0}}\rangle_{w}\right\|_{w}}\\[2 em]&=&\displaystyle\dfrac{x^{2}-4(x-1)-2}{\left\|x^{2}-\langle{x^{2},e_{1}}\rangle_{w}e_{1}-\langle{x^{2},e_{0}}\rangle_{w}\right\|_{w}}\\[2 em]&=&\displaystyle\dfrac{x^{2}-4x+2}{\left(\int_{0}^{\infty}(x^{2}-4x+2)^{2}e^{-x}dx\right)^{\frac{1}{2}}}\\[2 em]&=&\displaystyle\dfrac{x^{2}-4x+2}{\left(\int_{0}^{\infty}[x^{4}-8x^{3}+20x^{2}-16x+4]e^{-x}dx\right)^{\frac{1}{2}}}\\[2 em]&=&\displaystyle\dfrac{1}{2}x^{2}-2x+1\end{array}

The element of M which minimizes the weighted L^{2}-distance to x^{3} is the polynomial

\displaystyle\langle{x^{3},e_{2}}\rangle_{w}e_{2}+\langle{x^{3},e_{1}}\rangle_{w}e_{1}+\langle{x^{3},e_{0}}\rangle_{w}e_{0}

\displaystyle\langle{x^{3},e_{2}}\rangle_{w}=\int_{0}^{1}x^{3}\left(\dfrac{1}{2}x^{2}-2x+1\right)e^{-x}dx=\dfrac{5!}{2}-2(4!)+3!=18

Next,

\displaystyle\langle{x^{3},e_{1}}\rangle_{w}=\int_{0}^{1}x^{3}(x-1)e^{-x}dx=4!-3=18

Lastly,

\displaystyle\langle{x^{3},e_{0}}\rangle_{w}=\int_{0}^{1}x^{3}e^{-x}dx=6

Substituting in the computed coefficients, we see that the solution to the minimization problem is the polynomial

\displaystyle3\left(6x^{2}+6x-1\right)

I am not going to prove the second part of the exercise; the reader has seen enough to solve it by him- or herself. However, I will prove a more general result, which may be of some help.

Proposition 2. If x_{0}\in H and M is a closed subspace of H, then

\displaystyle\min\left\{\left\|x-x_{0}\right\|:x\in M\right\}=\max\left\{\left|\langle{x_{0},y}\rangle\right|:y\in M^{\perp},\left\|y\right\|=1\right\}

Proof. Fix x_{0}\in M. Let \delta_{1} and \delta_{2} denote the quantities on the left- and right-hand sides, respectively. Since M is a closed and a fortiori convex, there exist unique elements Px_{0}\in M and Qx_{0}\in M^{\perp} such that x_{0}=Px_{0}+Qx_{0}. Observe that

\displaystyle\left\|Px_{0}-x_{0}\right\|=\delta_{1}\text{ and }\left\|Qx_{0}-x_{0}\right\|=\min\left\{\left\|y-x_{0}\right\|: y \in M^{\perp}\right\}

For any element y\in M^{\perp} with \left\|y\right\|=1, we have that

\displaystyle\left|\langle{x_{0},y}\rangle\right|=\left|\langle{Qx_{0}+Px_{0},y}\rangle\right|=\left|\langle{Qx_{0},y}\rangle\right|\leq\left\|Qx_{0}\right\|\left\|y\right\|=\left\|Qx_{0}\right\|=\delta_{1},

where we use the orthogonality Px_{0}\perp M^{\perp} and the Cauchy-Schwarz inequality. Taking the supremum of the left-hand side of the inequality over \left\{y \in M^{\perp}:\left\|y\right\|=1\right\}, we see that \delta_{2}\leq\delta_{1}. But the element \frac{Qx_{0}}{\left\|Qx_{0}\right\|} has norm 1, belongs to M^{\perp}, and attains the upper bound, which shows that \delta_{2}=\delta_{1}. \Box

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