Mathematics Prelims

Analysis Prelim Passed!

August 1, 2008

The analysis prelim was one week ago today, and I just received an email a little bit ago saying that I passed it, so I’m happy. Classes don’t start until August 22 (or so), so this blog will probably not have much of any posts until then, as I plan on just kind of taking it easy for the next couple of weeks.

I’ll plan start studying for stochastics much earlier this time around though, and so stochastics posts will probably start sometime around the start of the semester. See you then.

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The Banach Fixed Point Theorem

July 21, 2008

A mapping $T : X \to X$ from a metric space $X$ to itself is called a contraction if there exists an $\alpha \in (0, 1)$ such that for every $x, y \in X$ we have $d(Tx, Ty) \leq \alpha d(x, y)$ . The Banach fixed point theorem (aka the contraction theorem) says that every contraction on a non-empty complete metric space has exactly one fixed point.

Proof: Suppose $X$ is a complete metric space and that $T : X \to X$ is a contraction. Let $x_0 \in X$ be any point. Define $x_1 = T x_0$ , $x_2 = T x_1 = T^2 x_0$ , $x_3 = T x_2 = T^3 x_0$ and so forth. If we write $d(x_{n+1}, x_n)$ in terms of $Tx_n$ and $Tx_{n-1}$ , we see that $d(x_{n+1}, x_n) \leq \alpha^n d(x_1, x_0)$ . Now consider $d(x_n, x_m)$ where we assume, without loss of generality, that $n > m$ . By the triangle inequality we have

$\displaystyle d(x_n, x_m) \leq (\alpha^{n-1} + \alpha^{n-2} + ... + \alpha^m) d(x_1, x_0)$

$\displaystyle \qquad = \frac{\alpha^m (1 - \alpha^{n - m})}{1 - \alpha} d(x_1, x_0)$

$\displaystyle \qquad \leq \frac{\alpha^m}{1 - \alpha} d(x_0, x_1)$

We can obviously make this value as small as we’d like by picking a large enough $m$ , so the sequence $(x_n)_{n \in \mathbb{N}}$ is Cauchy and must converge. Call the limit of this sequence $x$ . Now consider the distance between $x$ and $Tx$ .

$\displaystyle d(x, Tx) \leq d(x, x_m) + d(x_m, Tx) \leq d(x, x_m) + \alpha d(x_{m-1}, x)$

By picking a large enough $m$ , we can make this as small as we’d like as well, so we have $d(x, Tx) = 0$ so $T$ has a fixed point. For the uniqueness of $x$ , suppose $y$ is also a fixed point.

$\displaystyle d(x, y) = d(Tx, Ty) \leq \alpha d(x, y)$

The only way this holds is if $d(x, y) = 0$ , so $x = y$ . Note that the $x$ we constructed didn’t depend on on our initial value of $x_0$ , so every sequence we construct by picking a point in $X$ and iterating $T$ will result in the same limit.

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The Continuous Dual Space

July 20, 2008

An important subspace of the algebraic dual space of a normed space $X$ is the space which consists only of bounded linear functionals on $X$ . This space is called the continuous space (or just dual space, and the conjugate space in some older texts, such as Kolmogorov), and is denoted $X'$ . Note that we can define the algebraic dual space for any vector space, but require a norm for the (continuous) dual. This space is itself a normed space where functionals are given the operator norm. In fact, regardless of the normed space $X$ , the dual $X'$ is a Banach space.

Let $(f_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $X'$ . Then for each $\epsilon > 0$ there exists a $N \in \mathbb{N}$ such that $\| f_n - f_m \| < \epsilon$ for all $m, n > N$ . This means that

$\displaystyle \| f_n - f_m \| = \sup_{x \in X, \, \|x\| = 1} |f_n(x) - f_m(x)| < \epsilon$

So for a given $x \in X$ with $\|x \| = 1$ , $(f_n(x))$ forms a Cauchy sequence in $\mathbb{R}$ , and so converges. We will then define a functional $f$ pointwise as follows.

$\displaystyle f(x) = \lim_{n \to \infty} f_n(x)$ .

We must show that this functional is linear and bounded. Linearity follows easily from the linearity of each $f_n$ and properties of limits.

$\displaystyle f(\alpha x + \beta y) = \lim_{n \to \infty} f_n(\alpha x + \beta y) = \alpha \lim_{n \to \infty} f_n(x) + \beta \lim_{n \to \infty} f_n(y)$ .

For boundedness,

$\displaystyle |f(x)| = \lim_{n \to \infty} |f_n(x)| \leq \lim_{n \to \infty} \| f_n \| \|x\| \leq M \|x\|$

for some $M$ (this is because every Cauchy sequence is bounded).

So $f$ is a bounded linear operator on $X$ , and we just need to show $f_n \to f$ .

Let $\epsilon > 0$ be given. As $(f_n)$ is Cauchy, there is an $N \in \mathbb{N}$ such that for all $m, n > N$ we have $\|f_m - f_n\| < \epsilon$ . Letting $m \to \infty$ , we have $\|f - f_n\| \leq \epsilon$ for all $n > N$ , so $f_n \to f$ .

And so the dual of any normed space is Banach, regardless of whether the original space was or not.

Note that for every finite dimesional normed space, the continuous and algebraic duals are in fact the same as all linear operators are bounded in finite dimensions.

(This proof is a modified version of a proof in Eidelman.)

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The Monotone Convergence Theorem

July 20, 2008

If $(f_n : E \to \mathbb{R})_{n \in \mathbb{N}}$ is a sequence of non-negative measurable functions and $(f_n(x))_{n \in \mathbb{N}}$ increases monotonically to $f(x)$ for each $x \in E$ , $f_n \nearrow f$ pointwise (this can actually be relaxed to just having $f_n \nearrow f$ a.e.), then

$\displaystyle \lim_{n \to \infty} \int_E f_n(x) \, dm = \int_E f \, dm$ .

Proof:

As $f_n \nearrow f$ a.e., we have that $f_n \leq f$ a.e. for each $n \in \mathbb{N}$ , and so $\int_E f_n \, dm \leq \int_E f \, dm$ for each $n$ . Now also notice that since $f_n \leq f_{n + 1}$ , that $\int_E f_n \, dm$ forms a bounded monotonic sequence sequence of real numbers, so it must converge. So we have

$\displaystyle \lim_{n \to \infty} \int_E f_n \, dm \leq \int_E f \, dm = \int_E \lim_{n \to \infty} f \, dm$

By Fatou’s lemma we have that

$\displaystyle \int_E f \, dm = \int_E \lim_{n \to \infty} f_n \, dm = \int_E \liminf_{n \to \infty} f_n \, dm$

$\displaystyle \qquad \leq \liminf_{n \to \infty} \int_E f_n \, dm = \lim_{n \to \infty} \int_E f_n \, dm$

Since we have inequalities both ways, we must have equality.

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Fatou’s Lemma, Part I

July 20, 2008

If $E \subset \mathbb{R}$ is a Lebesgue measurable set and $(f_n : E \to \mathbb{R})_{n \in \mathbb{N}}$ is a sequence of non-negative measurable functions, then

$\displaystyle \liminf_{n \to \infty} \int_E f_n dm \geq \int_e \liminf_{n \to \infty} f_n dm$

Proof: Let $g_n = \inf_{k \geq n} f_k$ and $f = \liminf_{n \to \infty} f_n = \lim_{n \to \infty} g_n$ . Let $\phi$ be a simple function with $\phi \leq f$ . Assume $f > 0$ on $E$ , as the case where $f = 0$ is trivial. Now define $\overline{\phi}$ as follows.

$\overline{\phi}(x) = \left\{ \begin{array}{lr} \phi(x) - \epsilon &: \phi(x) > 0 \\ 0 &: \phi(x) = 0 \end{array} \right.$

Where $\epsilon > 0$ is a value that ensures $\overline{\phi} \geq 0$ . Now note that $\overline{\phi} < f$ and $g_n \nearrow f$ , so there must exist an $N \in \mathbb{N}$ such that $g_n \geq \overline{\phi}$ for all $n > N$ .

Let $A_k = \{ x \in E : g_k(x) \geq \overline{\phi}(x) \}$ and notice that $A_k \subseteq A_{k+1}$ . Also, $\bigcup_{k=1}^\infty A_k = E$ . For $k \geq n$ we have the following.

$\displaystyle \int_{A_n \cap E} \overline{\phi} \, dm \leq \int_{A_n \cap E} g_n \, dm$

$\displaystyle \qquad \leq \int_{A_n \cap E} f_k \, dm$

$\displaystyle \qquad \leq \int_E f_k \, dm$

So we have

$\displaystyle \int_{A_n \cap E} \overline{\phi} \, dm \leq \liminf_{k \to \infty} \int_E f_k \, dm$

Now notice that $\overline{\phi}$ is a simple function, and so we can write it as

$\displaystyle \overline{\phi} = \sum_{i=1}^\ell c_i 1_{B_i}$

Now we have

$\displaystyle \int_{A_n \cap E} \overline{\phi} \, dm = \sum_{i=1}^\ell c_i m(A_n \cap E \cap B_i)$

Letting $n \to \infty$ this gives us

$\displaystyle \int_E \overline{\phi} \, dm = \sum_{i=1}^\ell c_i m(E \cap B_i)$

This leaves us with

$\displaystyle \int_E \overline{\phi} \, dm \leq \liminf_{k \to \infty} \int_E f_k \, dm$

If $m(\{ x \in E : \phi(x) > 0 \}) < \infty$ , then we have

$\int_E \overline{\phi} \, dm = \int_E \phi \, dm - \epsilon \, m(\{ x \in E : \phi (x) > 0 \})$

This gives the desired result if we let $\epsilon \to 0$ as $\phi$ is an arbitrary simple function with $\phi \leq f$ .

Note: This is an awesomely useful lemma, but I hate this proof.

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The Algebraic Dual Space

July 18, 2008

Suppose $X$ is a vector space over a field $K$ . We say a function $f : X \to K$ is a linear functional if for every $\alpha, \beta \in K$ and every $x, y \in X$ , we have $f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$ . We will always assume that $K$ is either $\mathbb{R}$ or $\mathbb{C}$ . Properties and theorems associated with “traditional” linear operators apply since $\mathbb{R}$ and $\mathbb{C}$ can be thought of as normed spaces with the “traditional” norms (absolute values). Note that the set of all linear functionals on $X$ , which is denoted $X^*$ , is itself a vector space if we allow scalar multiplication and addition of functions in the traditional way ( $(\alpha f)(x) = \alpha f(x)$ and $(f + g)(x) = f(x) + g(x)$ ) .

The vector space $X^*$ is referred to as the algebraic dual of $X$ . Since $X^*$ is itself a vector space, we can define its algebraic dual, $(X^*)^* = X^{**}$ , which is called the second algebraic dual of $X$ . An important property of $X^{**}$ is that there exists an injective mapping $C : X \to X^{**}$ called the canonical mapping of $X$ into $X^{**}$ . This mapping is given by taking an $x \in X$ and considering the functional $g_x : X^* \to \mathbb{R}$ such that for each $f \in X^*$ we map $g_x(f) = f(x)$ .

Since $C$ is a linear map from $X$ to $X^{**}$ (this follows from the fact that each $f \in X^*$ is linear), we have $X$ is isomorphic to a subspace of of $X^{**}$ (recall that the range of a linear operator is a subspace of the operator’s codomain). For this reason, $C$ is sometimes called the canonical embedding of $X$ into $X^{**}$ . (A space $A$ is said to be embeddable in $B$ if $A$ is isomorphic to a subspace of $B$ .) In the event that $C$ is also surjective, so we have $X$ is isomorphic to all of $X^{**}$ , we say that $X$ is algebraically reflexive.

Every finite dimensional vector space is algebraically reflexive.

Proof: Suppose $X$ is an n-dimensional vector space with basis $\{ e_1, ..., e_n \}$ . Let $f \in X^*$ and $x \in X$ with $x = \alpha_1 e_1 + ... + \alpha_n e_n$ . As $f$ is linear, we have $f(x) = f(\alpha_1 e_1 + ... + \alpha_n e_n) = \alpha_1 f(e_1) + ... + \alpha_n f(e_n)$ . This implies that $f$ is uniquely determined by the values of $f(e_1), ..., f(e_n)$ , which means we can view $f$ as an n-tuple of scalars. That in turn means that $\dim X^* = n$ and that the set of n-tuples which have a single one and then n-1 zeroes form a basis for $X^*$ . This is called the dual basis of $X$ . Applying the same procedure to $X^*$ , we see that $\dim X^{**} = n$ . Now, since $C$ is an injective linear map from $X$ into $X^{**}$ we have that $X$ is isomorphic to an n-dimensional subspace of $X^{**}$ , and since the only n-dimensional subspace of $X^{**}$ is $X^{**}$ itself, we have that $X$ is isomorphic to $X^{**}$ , and so every finite dimensional vector space is algebraically reflexive.

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Linear Operators

July 18, 2008

If $X$ and $Y$ are normed spaced and $\mathcal{D}_T \subseteq X$ , then a mapping $T : \mathcal{D}_T \to Y$ is called a linear operator if for every $x,y \in \mathcal{D}_T$ and $\alpha, \beta \in K$ (the underlying field),

$\displaystyle T(\alpha x + \beta y) = \alpha Tx + \beta Ty$

If $T$ is a linear operator, then $T$ is injective if and only if $Tx = 0$ implies $x = 0$ .

Proof: Suppose $T$ is injective. Then $Tx = Ty \Rightarrow x = y$ , since $T$ is linear, it must map zero to zero, so $Tx = 0 \Rightarrow x = 0$ . Now suppose that $Tx = 0$ only for $x = 0$ . If $x,y \in \mathcal{D}_T$ and $Tx = Ty$ , then $Tx - Ty = 0$ and so $T(x - y) = 0$ (by linearity), and this means that $x - y = 0$ , so $x = y$ .

Now, we say that $T : \mathcal{D}_T \to Y$ is bounded if there exists a $c \in \mathbb{R}$ such that for every $x \in \mathcal{D}_T$ we have $\| Tx \| \leq c \|x\|$ . If such a $c$ exists (i.e., if $T$ is bounded), then we say that the least such $c$ is the operator norm of $T$ , and write $\|T\| = c$ . We can then find $c$ as follows.

$\displaystyle \|T\| = \sup_{x \in \mathcal{D}_T, \, x \neq 0} \frac{\|Tx\|}{\|x\|}$

(This follows from rewriting the above earlier inequality as $\frac{\|Tx\|}{\|x\|} \leq c$ .) Note that by letting $c = \|T\|$ , we have that $\|Tx\| \leq \|T\|\|x\|$ .

In a finite dimensional normed space, every linear operator is bounded. Suppose that $\dim X = n$ and that $\{ e_1, ..., e_n \}$ is a basis for $X$ . If we let $x \in X$ with $x = \sum_{i=1}^n \alpha_i e_i$ , then we have the following.

$\displaystyle \|Tx\| = \left\| T\left( \sum_{i=1}^n \alpha_i e_i \right) \right\|$

$\displaystyle \qquad = \left\|\sum_{i=1}^n \alpha_i Te_i \right\|$

$\displaystyle \qquad \leq \sum_{i=1}^n | \alpha_i | \|Te_i\|$

$\displaystyle \qquad \leq \frac{k}{c} \|x\|$

Where $k = \max \{ \|T e_1\|, ..., \|T e_n \| \}$ and $c$ is the value such that $\| \alpha_1 e_1 + ... + \alpha_n e_n \| \geq c (|\alpha_1| + ... + |\alpha_n|)$ .

Another important property is that a linear operator is bounded if and only if it is continuous.

Proof: Suppose $T$ is bounded and let $\epsilon > 0$ be given. Let $\delta = \frac{\epsilon}{\|T\|}$ . Let $x_0$ be a fixed point in $\mathcal{D}_T$ , and $x$ some other point in $\mathcal{D}_T$ such that $\| x - x_0 \| < \delta$ .

$\displaystyle \|Tx - Tx_0\| = \|T(x - x_0)\|$

$\displaystyle \qquad \leq \|T\| \|x - x_0\|$

$\displaystyle \qquad \leq \|T\| \delta$

$\displaystyle \qquad \leq \|T\| \frac{\epsilon}{\|T\|}$

$\displaystyle \qquad = \epsilon$

This shows that $T$ is continuous at $x_0$ . However, our $x_0$ was arbitrary, and $\delta$ didn’t depend on our choice of $x_0$ , so $T$ is uniformly continuous.

For the converse I’m going to copy the proof from Wikipedia, since I think it’s a bit clearer than Kreyszig’s proof.

Suppose that $T$ is continuous. Since $\mathcal{D}_T$ is a subspace of $X$ , it contains the zero vector. Since $T$ is continuous, it’s continuous at the zero vector. This means there exists a $\delta > 0$ such that for all $x \in \mathcal{D}_T$ satisfying $\|x\| \leq \delta$ we have that $\|Tx\| = \|T(x - 0)\| = \|Tx - T0\| < 1$ . Now let $y$ be any point in $\mathcal{D}_T$ .

$\displaystyle \|Ty\| = \left\| \frac{\|y\|}{\delta} T\left( \frac{\delta}{\|y\|} y\right )\right\|$

$\displaystyle \qquad = \frac{\|y\|}{\delta} \left\| T \left( \frac{\delta}{\|y\|} y \right) \right\|$

$\displaystyle \qquad \leq \frac{1}{\delta} \| y \|$

And so $T$ is bounded.

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Compact Closed Unit Ball Implies Finite Dimension

July 18, 2008

If $X$ is a normed space and the closed unit ball centered at zero is compact, then $X$ is finite dimensional.

Proof: Suppose $X$ is an infinite dimensional normed space and let $x_1$ be any point in $X$ with $\|x_1\| = 1$ and let $Y_1$ be the one-dimensional subspace of $X$ generated by $x_1$ . Recall that a finite dimensional subspace is always closed, and since $X$ is infinite dimensional, $Y_1$ is a proper subspace of $X$ . By Riesz’s Lemma, there exists an $x_2 \in X \setminus Y_1$ with $\|x_2\| = 1$ and $\|x_2 - y\| \geq \frac{1}{2}$ for all $y \in Y_1$ . Let $Y_2$ be the two-dimensional subspace generated by $x_1, x_2$ . There exists a $x_3 \in X \setminus Y_2$ such that $\|x_3\| = 1$ and $\|x_3 - y\| \geq \frac{1}{2}$ for all $y \in Y_2$ . Note that since $X$ is infinite dimensional, we can keep applying this procedure generating a sequence $(x_n)$ such that $\|x_n\| = 1$ but $\|x_m - x_n\| \geq \frac{1}{2}$ for all $m \neq n$ . This means that our sequence can not be Cauchy, and so it can not be convergent, and so the closed unit ball of radius one can not be compact. This means that, since all points in the sequence are at least distance $1/2$ from one another, no subsequence can be Cauchy, so no subsequence can be convergent. Hence, if the closed unit ball of radius one is compact, then the space is finite dimensional.

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Riesz’s Lemma

July 17, 2008

If $X$ is a normed space (of any dimension), $Z$ is a subspace of $X$ and $Y$ is a closed proper subspace of $Z$ , then for every $\theta \in [0, 1]$ there exists a $z \in Z$ such that $\|z\| = 1$ and $\|z - y\| \geq \theta$ for every $y \in Y$ .

Proof: Let $v \in Z \setminus Y$ and let $a = \inf_{y \in Y} \| v - y\|$ . As $Y$ is closed and $v \notin Y$ , we have $a > 0$ . Now let $\theta \in (0, 1)$ and note that there exists a $y_0 \in Y$ such that $a\leq \| v - y_0 \| \leq \frac{a}{\theta}$ (as $\theta < 1$ , we have $\frac{a}{\theta} > a$ ). Let

$\displaystyle z = \frac{v - y_0}{\| v - y_0 \|}$

Obviously $z \in Z$ and $\| z \| = 1$ . Let $y$ be any element of $Y$ . We have the following.

$\displaystyle \|z-y\| = \left\| \frac{1}{\|v - y_0\|} (v - y_0) - y\right\|$

$\displaystyle \qquad = \frac{1}{\|v - y_0\|} \| v - y_0 - (\|v - y_0\|) y\|$

$\displaystyle \qquad \geq \frac{a}{\| v - y_0 \|}$

$\displaystyle \qquad \geq \frac{a}{a/\theta}$

$\displaystyle \qquad = \theta$

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Question about LaTeX on WordPress

July 17, 2008

Hola. Does anyone know if there’s a way to do something like eqnarray inside of WordPress’ LaTeX? If not, what’s the preferred method of listing a long string of equalities?

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Mathematics Prelims

Analysis Prelim Passed!

The Banach Fixed Point Theorem

The Continuous Dual Space

The Monotone Convergence Theorem

Fatou’s Lemma, Part I

The Algebraic Dual Space

Linear Operators

Compact Closed Unit Ball Implies Finite Dimension

Riesz’s Lemma

Question about LaTeX on WordPress

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