\documentclass{article} \begin{document} \LARGE{Assignment IV} \large{Luenberger Problems from Section 3.13} \normalsize 1. "if": Suppose $\alpha x + \beta y = 0$ for some scalars $\alpha$, $\beta$ not both equal to $0$. If $x = 0$, then $ = 0 = 0 ||y|| = ||x|| ||y||$; similarly, if $y = 0$, then $ = 0 = ||x|| ||y||$. Therefore, we can assume that $x \neq 0$ and $y \neq 0$, so $\beta \neq 0$ because $x \neq 0$ (otherwise $\alpha x = 0$, which implies that $\alpha = 0$ as well) and similarly, $\alpha \neq 0$ because $y \neq 0$. Therefore, if $x \neq 0$ then $x = \beta * (\alpha)^{-1} y$ and $y = \alpha (\beta)^{-1}x$. Therefore, $ = <\beta * (\alpha)^{-1}y|y> = \beta * \alpha^{-1} $ and $ = = $conj$(\alpha * \beta^{-1}) $ so $ = (\beta * \alpha^{-1})$conj$(\alpha * \beta^{-1}$); taking absolute values yields $||^2 = ||x||^2 ||y||^2$ and taking square roots gives $ = ||x|| ||y||$ in this case. "only if": Suppose that there are no scalars $\alpha$ and $\beta$ which are not both $0$ such that $\alpha x + \beta y = 0$. (Observe that $y \neq 0$; otherwise setting $\alpha = 0$ and $\beta = 1$ would yield $\alpha x + \beta y = 0$.) Then, for all scalars $\lambda$ we have $0 <$ $$ $= - \lambda - conj(\lambda) $. In particular, for $\lambda = /$ (we can do this as $y$ is nonzero), we have $0 $ $< - \frac{||^2}{}$ so $|| < \sqrt{} = ||x|| ||y||$; therefore, equality in the Cauchy-Schwartz inequality cannot hold if there are no scalars $\alpha$ and $\beta$ which are not both $0$ such that $\alpha x + \beta y = 0$. Therefore, the Cauchy-Schwartz inequality becomes an equality precisely when there are scalars $\alpha$ and $\beta$ which are not both $0$ such that $\alpha x + \beta y = 0$. 3. First observe that from linear algebra, if $A$ is a real-valued $n$ x $n$ matrix then the trace of $A$ is equal to the trace of $A'$ (and both are real). Therefore, $ = $Trace$ [B'QA] = $Trace$[(A' Q' B'')'] = $Trace$[A' Q B]$ (by the symmetry of $Q$; further, observe that as $A$, $Q$, and $B$ are real-valued matrices, so is $A' Q B$), so $ = $. As $ = $ and both are real, $ = $conj$()$. Further, suppose $A$, $B$, and $C$ are elements of $H$ and $\lambda$ is a scalar; then $ = $Trace$[(A + B)'QC] = $Trace$[(A'QB + B'QC)] = $Trace$[A'QC] + $Trace$[B'QC] = + $ and $<\lambda A|B> = $Trace$[(\lambda A)' Q B = \lambda $Trace$[(A' Q B)] = \lambda$. Finally, because $Q$ is positive definite, $ = $Trace$[(A' Q A)]$, which is always nonnegative and equal to zero iff $A$ is equal to zero. As $< | >$ satisfies the four axioms of an inner product, it follows that $H$ is an inner product space. Now suppose $\{x_1, \dots, x_k, \dots\}$ is a Cauchy sequence in $H$ and let $x$ be its componentwise limit. However, observe that because $||x - x_k||$ (defining the norm based on the inner product) $= = $Trace$[(x - x_k)' Q (x - x_k)] \leq (Z_k)^2 m^2 Y$ where $Z_k$ is the maximal entry of $x - x_k$ and $Y$ is the maximal eigenvalue of $Q$, which implies that $||x - x_k||$ approaches $0$ as $x - x_k$ approaches 0 componentwise; therefore, $\{x_1, \dots, x_k, \dots\}$ converges to $x$ which implies that $H$ is complete. 5. Observe that if $x(t) = a + bt$, $\int_{-1}^1 ([t^2 - x(t)]^2)$ $= \int_{-1}^1 ([t^2 - (a + bt)]^2)$ $= \int_{-1}^1 (t^4 - 2bt^3 + b^2 t^2 - 2at^2 + 2abt + a^2)$ $= (1/5 * 1^5 - 2b * 1/4 * 1^4 + b^2 * 1/3 * 1^3 - 2a * 1/3 * 1^3 + 2ab * 1/2 * 1^2 + a^2 * 1)$ $- (1/5 * (-1)^5 - 2b * 1/4 * (-1)^4 + b^2 * 1/3 * (-1)^3 - 2a * 1/3 * (-1)^3 + 2ab * 1/2 * (-1)^2 + a^2 * 1)$ $= 1/5 * 2 + 2b^2 * 1/3 - 4a * 1/3 + a^2 * 2 = 2a^2 - 4/3 a + 2/3 b^2 + 2/5$. Therefore, finding $x(t)$ to minimize the integral is the same as finding real $a$ and $b$ to minimize the expression $2a^2 - (4/3) a + (2/3) b^2 + 2/5 = 2(a^2 - (2/3) a + 1/9) + (2/3) b^2 + 2/5 - 2/9$ $= 2(a - 1/3)^2 + (2/3) b^2 + 8/45$, which is clearly minimized at $a = 1/3$ and $b = 0$. Therefore, the integral is minimized at $x(t) = 1/3 + 0t = 1/3$. 9. Suppose $T$ is a closed linear subspace containing $S$. Because $T$ is closed and linear, $T^{\bot \bot} = T$ by Theorem 1 of Section 3.4. However, as $S \subset T$, it follows that $T^{\bot} \subset S^{\bot}$ by Part 3 of Proposition 1 of Section 3.4; therefore, $S^{\bot \bot} \subset T^{\bot \bot} = T$ so any closed linear subspace containing $S$ must also contain $S^{\bot \bot}$. As $S^{\bot \bot}$ is a closed subspace by Part 1 of Proposition 1 of Section 3.4 and contains $S$ by Part 2 of Proposition 1 of Section 3.4, it follows that $S^{\bot \bot}$ is the smallest closed subspace containing $S$. 24. Consider the Banach space $l^\infty$ (as a real vector space). Let $x = \{1, 0, \dots, 0, \dots\}$ be a point in this vector space and let $K$ be the subspace of $l^\infty$ consisting of all points in $l^\infty$ which have a zero in their first coordinate; being a subspace implies convexity and it is closed because it is the preimage of the point $0$ under the (continuous) projection map from $l^\infty$ to $\textbf{R}$ sending each sequence in $l^\infty$ to its first coordinate. Now observe that if $k \in K$, $||k - x|| \geq |1 - 0| = 1$ (as $k$ and $x$ differ by $1$ in their first coordinate). Therefore, if $j$ is a vector in $K$ with $||x - j|| = 1$ then $||x - j|| \leq ||x - k|| \forall k \in K$. However, the vectors $j_1 = \{0, \dots, 0, \dots\}$ and $j_2 = \{0, 1, 0, \dots, 0, \dots\}$ ($j_2$ has $1$ in its 2nd coordinate and $0$ in all others) both are at a distance $1$ from $x$; therefore, although there exists a vector $k_0 \in K$ such that $||x - k_0|| \leq ||x - k|| \forall k \in K$ (pick $j_1$ or $j_2$), the choice is not unique. Writing: 1. First observe that if $y_1$ and $y_2$ are elements of $Y$ then $y_1 + y_2$ is a sequence such that for each positive integer $n$, the nth coordinate of $y_1 + y_2$ is equal to zero ONLY IF the nth coordinate of either $y_1$ or $y_2$ is equal to 0; as this can only happen in finitely many places (as the finite union of finite sets is finite) it follows that $y_1 + y_2$ has all but finitely many coordinates equal to $0$; therefore, $y_1 + y_2 \in Y$. This implies that $Y$ is closed under vector addition. Similarly, if $\alpha$ is a scalar, $\alpha y_1$ is a sequence such that for each positive integer $n$, the nth coordinate of $\alpha y_1$ is equal to zero ONLY IF the nth coordinate of $y_1$ is equal to zero; therefore, it has all but finitely many coordinates equal to ero. This implies that $Y$ is closed under scalar multiplication and is therefore a subspace of $l_\infty$. However, for each positive integer $n$ we can let $z_n \in Y$ be the sequence whose kth term is $1/k$ if $k \leq n$ and whose kth term is $0$ otherwise. The sequence $\{z_n\}$ is Cauchy and converges to the sequence $\{1, 1/2, 1/3, \dots\}$; however, this limit point of the sequence, despite being in $l_\infty$, is not in $Y$ which implies that $Y$ is not closed under the usual norm on $l_\infty$. Now let $Z$ be the set of all sequences in $l_\infty$ which converge to $0$. Suppose $w = \{w_1, \dots, w_k, \dots \}$ is a sequence in $Z$. For each positive integer $n$ let $v_n$ be the sequence whose kth term is equal to $w_k$ if $k \leq n$ and whose kth term is equal to $0$ otherwise; each $v_n$ is in $Y$ and the sequence $\{v_1, \dots, v_k, \dots \}$ converges to $w$ (because $\{w_1, \dots, w_k, \dots \}$ converges to $0$), so $w \in $clos$(Y)$. As $w$ is an arbitrary sequence in $Z$, this implies that $Z \subset $clos$(Y)$. Further, suppose $m \in l_\infty$ is NOT in $Z$. Then there exists $\epsilon > 0$ such that there are infinitely many terms in $m$ of magnitude greater than $\epsilon$. Therefore, if $z$ is a sequence in $Z$ (and there exists $N$ such that whenever $n > N$, the nth term of $z$ has magnitude less than $\epsilon/2$; also, by the last sentence there exists $k > m$ such that the kth term of $m$ has magnitude of at least $\epsilon$), $||m - z|| \geq \epsilon/2 > \epsilon/3$ so the open ball centered at $m$ of radius $\epsilon/3$ is disjoint from $Z$. Therefore, $Z$ is closed in $l_\infty$ so $Z = $clos$(Z)$, which implies that clos$(Y) \subset $clos$(Z) = Z \subset $clos$(Y)$, so $Z$ is the closure of $Y$. As $Z$ is the closure of $Y$, it follows that $Z$ is the smallest complete subset of $l_\infty$ containing $Y$; as $Y$ is a dense subset of $Z$, it follows that $Z$ is isomorphic to a completion of $Y$ with respect to the usual norm on $l_\infty$. By the discussion on Example 4, $l_p$ is a Banach space and therefore is a complete set containing $Y$. However, if $w = \{w_1, \dots, w_k, \dots \}$ is a sequence in $Z$, then for each positive integer $n$ we can let $v_n$ be the sequence whose kth term is equal to $w_k$ if $k \leq n$ and whose kth term is equal to $0$ otherwise; as $\{v_n\}$ is a sequence in $Y$ which converges to $w$, $w$ is in the closure of $Y$ so as $w$ is an arbitrary element of $l_p$, $l_p$ is contained in the closure of $Y$ and therefore IS the closure of $Y$ (being complete, $l_p$ is closed). Therefore, the completion of $Y$ with respect to the p-norm is isometrically isomorphic to $l_p$. 2. Observe that the harmonic sequence $\{1/k\}$ is not in $l_1$ because the harmonic series diverges; however, as the series $1/(1^2) + \dots + 1/(k^2) + \dots$ converges to $(\pi^2)/6$, it follows that the harmonic sequence is in $l_2$; therefore, $l_2$ is not a subset of $l_1$. On the last homework we showed that $l_1 \subset l_2$; therefore, $l_1$ is a proper subset of $l_2$. I would expect that $l_1$ is a proper subset of $l_2$ by the $p$-test for series, which says that while the harmonic series diverges, the series $1/(1^p) + \dots + 1/(k^p) + \dots$ converges whenever $p > 1$. The generalization to that (using the sequence $\{1/(k^{1/p})\}$ as a counterexample) is that whenever $p < q$, $l_p$ is a proper subset of $l_q$ (this holds for $q = \infty$ as well because the constant sequence $\{1, \dots, 1, \dots \}$ is clearly in $l_\infty$ as it is bounded but not in $l_p$ for any other $p$. However, the statement $L_1[0, 1] \subset L_2[0, 1]$ is NOT true because while the function defined to equal log $x$ on $(0, 1]$ and equal $0$ at $0$ is in $L_1[0, 1]$ (with a norm of $1 - 1 $log$ 1 = 1$), it is not in $L_2[0, 1]$ because it gets too large as $x$ approaches $0$. \end{document}