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Week 10 Discussion Notes

Real Analytic Functions
Inner Products

Real Analytic Functions

Definition

Let $U \subseteq \R$ be an open set and $\func{f}{U}{\R}$ be a function. We say that $f$ is real analytic on $U$ if for every $a \in U$ , there exists $r > 0$ such that

f\p{x} = \sum_{n=0}^\infty c_{a,n}\p{x - a}^n \quad\text{for } x \in B\p{a, r}.

Remark.

One thing important to keep in mind is that the coefficients can be different depending on the center $a$ . For example, if $f = 0$ on $\p{0, 1}$ and $f = 1$ on $\p{2, 3}$ , then $f$ is real analytic, but

\begin{aligned} f\p{x} &= 0 + 0 + \cdots \quad\text{on } \p{0, 1} \\ f\p{x} &= 1 + 0 + \cdots \quad\text{on } \p{2, 3}. \end{aligned}

Example 1.

$\frac{1}{1 - x} = \sum_{n=0}^\infty x^n$ is real analytic on $\p{-1, 1}$ . (It's actually real analytic on $\p{-\infty, 1}$ , but of course, the power series expansion will be different from the above when $a < -1$ .)

Theorem

Let $f$ be a real analytic function on an interval $I \subseteq \R$ . Assume that the set $E = \set{x \st f\p{x} = 0}$ is non-empty and that there exists $a_0 \in E$ and a sequence $a_k \in E \setminus \set{a_0}$ such that $a_k \to a_0$ ( $a_0$ is called an accumulation point of $E$ ). Then $f = 0$ on $I$ .

Proof.

Claim: $f = 0$ in an open interval around $a_0$ .

Since $f$ is real analytic at $a_0$ , there exists $r > 0$ such that

f\p{x} = \sum_{n=0}^\infty c_n\p{x - a_0}^n \quad\text{for } x \in B\p{a_0, r}.

Suppose $f \neq 0$ on $B\p{a_0, r}$ . Then at least one of the coefficients $c_n$ must be non-zero or else the series expansion is just $0$ . Let $n_0$ be the first number such that $c_{n_0} \neq 0$ . Then

f\p{x} = \sum_{n=n_0}^\infty c_n\p{x - a_0}^n = \p{x - a_0}^{n_0} \underbrace{\sum_{n=n_0}^\infty c_n\p{x - a_0}^{n-n_0}}_{\eqqcolon\,g\p{x}}.

Notice that

\lim_{n\to\infty} \abs{c_n\p{x - a_0}^{n-n_0}}^{\frac{1}{n}} = \abs{x - a_0}\lim_{n\to\infty} \abs{c_n}^{\frac{1}{n}}

(the second limit exists since it's the reciprocal of the radius of convergence of $f$ at $a$ , which is at least $r$ ). Thus, $g$ is also real analytic at $a$ , and in particular, $g$ is continuous.

Since $a_k \in E$ ,

0 = f\p{a_k} = \p{x - a_0}^{n_0} g\p{a_k}.

Moreover, because $a_k \neq a_0$ for any $k$ , we can divide to get

g\p{a_k} = 0 \implies g\p{a_0} = \lim_{k\to\infty} g\p{a_k} = 0

by continuity of $g$ . But

g\p{a_0} = 0 \implies c_{n_0} = 0,

a contradiction. Thus, $c_n = 0$ for all $n$ , so $f$ is $0$ on $B\p{a_0, r}$ .

Claim: The set $A = \set{a \st f\p{x} = 0 \text{ in a neighborhood of } a}$ is non-empty, open, and closed.

From the first claim, we already have $a_0 \in A$ , so $A$ is non-empty.

To see that $A$ is open, let $a \in A$ , so by definition, there exists $r > 0$ such that $f = 0$ on $B\p{a, r}$ . Since open balls are open, for any $a' \in B\p{a, r}$ , there exists $r' > 0$ such that $B\p{a', r'} \subseteq B\p{a, r}$ . But this means that $f = 0$ on $B\p{a', r'}$ . Since $a'$ was arbitrary, this shows $B\p{a, r} \subseteq A$ , so $A$ is open.

Finally, to see that $A$ is closed, assume there exists $a \in \cl{A} \setminus A$ . By definition, there exists a sequence $b_k \in A \setminus \set{a}$ such that $b_k \to a$ . But notice that $a, b_k \in E$ by definition, so by the first claim, this means that $a \in A$ , a contradiction. Thus, $A = \cl{A}$ , so $A$ is closed.

Claim: $A = I$ , i.e., $f = 0$ on all of $I$ .

$A$ is open, and because $A$ is closed, $A^\comp$ is also open. Moreover, $A \cup A^\comp = I$ . Thus, because $I$ is connected and $A$ is non-empty, it follows that $A = I$ . (Otherwise, if $A^\comp \neq \emptyset$ , then $I$ is the union of two disjoint, non-empty, and open sets, which would imply that $I$ is disconnected.) $\square$

Inner Products

Definition

Let $V$ be a real or complex vector space. Then an inner product $\inner{\,\cdot\,, \,\cdot\,}$ is a function $V \times V \to \p{\R \text{ or } \C}$ which is:

(conjugate symmetric) $\inner{x, y} = \overline{\inner{y, x}}$
(linear in the first argument) $\inner{x + \lambda y, z} = \inner{x, y} + \lambda \inner{y, z}$
(positive definite) If $x \neq 0$ , then $\inner{x, x} \in \p{0, \infty}$

Example 2.

The dot product on $\R^n$ , i.e.,

x \cdot y = \sum_{i=1}^n x_iy_i

is an inner product on $\R^n$ . Recall that

x \cdot y = \norm{x} \norm{y} \cos{\theta},

so intuitively, inner products allow us to talk about angles in more general vector spaces.

Lemma

The function

\inner{f, g} = \int_0^1 f\p{x} \overline{g\p{x}} \,\diff{x}

defines an inner product on $C\p{\br{0, 1}}$ . Thus,

\norm{f}_2 \coloneqq \sqrt{\inner{f, f}} = \p{\int_0^1 \abs{f\p{x}}^2 \,\diff{x}}^{\frac{1}{2}}

defines a norm on $C\p{\br{0, 1}}$ called the $L^2$ -norm.

Remark.

$C\p{\br{0, 1}}$ is not complete with respect to the $L^2$ -norm.

Example 3.

Use Gram-Schmidt to find an orthonormal basis for $V = \span\set{1, x}$ , where the span is taken over $\R$ .

Solution.

We can take $e_1 = 1$ since $1$ is already normalized. Then

u_2 = x - \proj_{e_1}\p{x} = x - \inner{x, e_1}e_1.

Here,

\inner{x, e_1} = \int_0^1 x \cdot 1 \,\diff{x} = \frac{1}{2},

so $u_2 = x - \frac{1}{2}$ . Now we just need to normalize it:

\norm{u_2}_2 = \p{\int_0^1 \p{x - \frac{1}{2}}^2 \,\diff{x}}^{\frac{1}{2}} = \frac{1}{2\sqrt{3}} \implies e_2 = \frac{u_2}{\norm{u_2}_2} = 2x\sqrt{3} - \sqrt{3}.

Thus,

\boxed{\set{1, 2x\sqrt{3} - \sqrt{3}}}

is an orthonormal basis for $V$ .

Example 4.

Find the function in $V = \span\set{1, x}$ that minimizes the $L^2$ -distance to $x^2$ .

Solution.

Recall that the minimizer is the orthogonal projection of $x^2$ to $V$ . This is

\proj_V\p{x^2} = \inner{x^2, e_1}e_1 + \inner{x^2, e_2}e_2,

where $\set{e_1, e_2}$ is the orthonormal basis from the previous problem. Thus,

\begin{gathered} \inner{x^2, e_1} = \int_0^1 x^2 \cdot 1 \,\diff{x} = \frac{1}{3} \\ \inner{x^2, e_2} = \int_0^1 x^2\p{2x\sqrt{3} - \sqrt{3}} \,\diff{x} = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{3}, \end{gathered}

so the minimizer is

\boxed{\frac{1}{3} + \p{\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{3}}\p{2x\sqrt{3} - \sqrt{3}}}.

Week 10 Discussion Notes

Table of Contents

Real Analytic Functions

Definition

Remark.

Example 1.

Theorem

Proof.

Inner Products

Definition

Example 2.

Lemma

Remark.

Example 3.

Solution.

Example 4.

Solution.