Final - Jim Zenn

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1

Let $x_0 \in \overline{ E }, y_0 = f(x_0)$.

Since $x_0 \in E, \exists (x^{(n)})^\infty_{ n =1}, x_n \to x_0$.

Since $f$ is continuous, $x_n \to x_0 $ implies that $f(x_n) = f(x_0) $.

Since $f(x_n) \in f(E), \forall n, f(x_0) \in \overline{ f(E)} $

Definition: Inner Product Space

vector space with function $<, >: V \times V \to \C $

such that

$\inner{ v }{ w } = \overline{ \inner{ w }{ v }} $

$\inner{ v }{ v } \geq 0, \forall v \in V $

$\inner{ v }{ v } = 0 \iff v = 0$

$\inner{ cv + w }{ z } = c \inner{ v }{ z } + \inner{ w }{ z } $

Example

$C(\R / \Z , \C) = V $

vectors are functions

$\inner{ f }{ g } = \int_{ 0 }^{ 1 } f(x) \overline{ g(x)} \d x $

$\inner{ g }{ f } = \int_{ 0 }^{ 1 } g(x) \overline{ f(x)} \d x = \overline{ \int_{ 0 }^{ 1 } g(x) \overline{ f(x)} \d x } $

From those props, we have

(Cauchy-Scharz inequality) $\abs{ \inner{ f }{ g }} \leq \sqrt[ ]{ \inner{ f }{ f }} \sqrt[ ]{ \inner{ g }{ g }}$
For any inner vector space we define $2- $norm by

$$\norm{ v }_2 = \sqrt[ ]{ v, v } $$

Using Cauchy Schwaz,

$$\norm{ v+w }_2 \leq \norm{ v }_2 + \norm{ w }_2 $$

from other props, $\norm{ v }_2 \geq 0, \forall v \in V$

$$\norm{ v }_2 = 0 \implies v = 0_v $$

$$\norm{ aV }_2 = \abs{ a } \norm{ v }_2 $$

$$\norm{ av }_2 = \sqrt[ ]{ \inner{ av }{ av }} $$

$$\norm{ av }^2_2 = \inner{ av }{ av } = \br a \inner{ v }{ av } = a \overline{ \inner{ av }{ v }} = a \overline{ a } \overline{ \inner{ v }{ v }} = \abs{ a }^2 \norm{ v }_2$$

Theorem : Cauchy-Schwarz

$\abs{ \inner{ f }{ g }} \leq \norm{ f }_2 \norm{ g }_2$

$\inner{ v }{ v } \geq 0, v = f - tg, t \in \R$

$$\inner{ f-tg }{ f-tg } \geq 0 $$

$$\inner{ f }{ f-tg } - t \inner{ g }{ f-tg } $$

$$\inner{ f }{ f } - t \inner{ f }{ g } - t \inner{ g }{ f } + t^2 \inner{ g }{ g } $$

minimum occurs when

$t = \frac{ -b }{ 2a }, a = \norm{ g }_2^2 $

$b = - (\inner{ f }{ g } +? \inner{ g }{ f }) $

$b = -2 (\Re{(f, g)}) $

$$\norm{ f + g }_2 = \norm{ f }_2 + \norm{ g }_2 $$

$$\begin{align} \norm{ f+g }^2_2 &= \inner{ f }{ f } + \inner{ f }{ g } + \inner{ g }{ f } + \inner{ g }{ g } \br &= \inner{ f }{ f } + 2 \Re{ \inner{ f }{ g }} + \inner{ g }{ g } \br \leq \inner{ f }{ f } + 2 \abs{ \inner{ f }{ g }} + \inner{ g }{ g } \br \leq \inner{ f }{ f } + 2 \norm{ f }_2 \norm{ g }_2 + \inner{ g }{ g } \br = (\norm{ f }_2 + \norm{ g }_2)^2 \end{align}$$

know the definition of

$\inner{ f }{ g } V = C(\R / \Z , \C) $

Know how to compute

$$\inner{ e_n }{ e_m } = \begin{cases} \int_{ 0 }^{ 1 } \exp 2 \pi i (n - m)x\d x = \frac{ \exp (2 \pi i (n - m))}{ 2 \pi i (n - m)} \big |^1_0 = \frac{ e_n(1)^2 - e_k(0)}{ 2 \pi i (n - m)}\br \int_{ 0 }^{ 1 } 1 \d x \end{cases} $$

$e_n (0) = e_n (1) $ by periodicity

$\inner{ f }{ g } = 0 $

$$f+g ^2_2 = \norm{ f }^2_2 + \norm{ g }^2_2$$

Find function sequence such that

$f^{(n)}$ continuous, $f$ continuous.

$f^{(n)} \to f$, $L^2 $ but not uniformly
$f^{(n)} \to f $ L^2$ but not pointwise
pointwise but not $L^2 $

Idea:

$L^2 $ convergence measures area between functions uniform convergence measures max height diff

if $f $ is analytic find $c_n $

2

Example

$X$ connected. show $\forall, a, b,d(a, b) > 0, r \in [ 0, d(a, b) ], \exists x_r, d(a, x_r) = r $

Proof

  </span>
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<span class="proof__expand"><a>[expand]</a></span>

$s(a, r) = u $ open set $u \neq \emptyset $

$V = \set{ y \in X: d(a, y) > r } \neq \emptyset $

New question: Can you show $B(a,r) $ is open?

Proof

  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

12

$f: X \to X, k \in [0, 1)$

$$d(f(x), f(y)) \leq k d (x, y) $$

points get “closer”

Show $\exists$ unique point $x_0 $ such that $f(x_0) = x_0$

Two points such that $f(a) = a & f(b) = b$

$d(f(a), f(b)) \leq kd(a, b)$

$d(a, b) \leq k d(a, b)$

$d(a, b) = c$

Pick $x_0 \in X$

$x_n = f(x_{n - 1})$

$d(x_0, x_1) = d$

$d(x_1, x_2) = d(f(x_0), f(x_1)) \leq k d(x_0, d_1) = k d$

$d(x_2, x_3) = d(f(x_1, f(x_2))) \leq k d(x_1, x_2) \leq k^2 d$

$d(x_n, x_3) = k^n d $

Maybe this is Cauchy $n < m $

$d(x_n, x_m)$

$d(x_n, x_m) \leq k^n d$

$d(x_n, x_{n+2}) \leq d(x_n, x_{n+1}) + d(x_{n+1}, x_{n+2})$

$$\leq k^n d + k^{n +1} d $$

$$d(x_n, x_{n+3}) \leq k^n d + k^{n+1} d + k^{n+2} d $$

$$d(x_n, x_m) \leq k^n d + k^{n+1}d + … + k^{m - n - 1}d $$

$$\leq d (\sum_{ j = n }^{ \infty } k^j) $$

$\exists N$ such that if $n > N $

$\sum_{ j=n }^{ \infty }k^j < \epsilon / d $

$d(x_n, x_m) < \epsilon $ if $m> n> N$

$x_n $ is Cauchy, so it converges to a point called $x^* $

$f(x^*) = f(\limu{ n }{ \infty } x_n) \br $

$= \limu{ n }{ \infty } f(x_n) $\br

$= \limu{ n }{ \infty } x_{n+1} = x^* $

5

$C_c(\R^ -) =$ function such that $f : \R \to \R $

if $f(x) \neq 0 x \in$ compact set

given $f, \exists M $ such that if

$\abs{ x } \geq M, f(x) = 0 $

$C_0$

bell curve $e^{- \frac{1}{x^2}}$

$C_b(x) $

bounded continuous functions

If $f \in C_c(\R) $ f \in c_0(\R)

$g \in C_0 (\R) show $ show g is bounded.

$\limu{ x}{ \infty } g(x) = 0$ implies

$\exists M_1 $ such that $\abs{ g(x)} \leq 1 $ if $x > M_1 $

$\exists M_2 $ such that $\abs{ g(x)} \leq 1 $ if $x < M_2 $

$g([M_2, M_1]) is compact $

$\abs{ g(x)} \leq M_3 $ on $M_2, M_1 $

$\abs{ g(x)} \leq \max{ \set{ M_3, 1 }} $

$g $ is bounded

$C_c(\R) \leq C_0(\R) \leq C_b (\R) $

let $g(x) = e^{ - \frac{ 1 }{ x^2 }} \in C_0$

$h(x) = 2$