$L^p$ spaces

Let $(X,\mathcal{S},\mu)$ be a complete measure space. Let $0 < p , \infty$ . Then,

$$L^p(X,\mathcal{S},\mu) = \{ f : X \rightarrow \mathbb{F}, mble : \int_X \vert f\vert^p d\mu < \infty \}.$$

We also want to define $L^\infty(X)$ .

Given $f : X \rightarrow \mathbb{F}$ ,

$$esssup \vert f\vert = \inf \{ t > 0 : m\{ x : \vert f(x)\vert > t \} = 0 \}.$$

Then,

$$L^\infty(X) = \{ f : X \rightarrow \mathbb{F}, mble : esssup\vert f\vert < \infty \}$$

For $1 \leq p < \infty$ , define

$$\Vert f \Vert _p = \left( \int_X \vert f\vert^p d\mu \right)^\frac1p$$

For $p = \infty$ , define

$$\Vert f\Vert _\infty = esssup \vert f\vert$$

Think of $\mathbb{R}^2$ as the set of functions on a two point space. Then, draw

$$\{ (x,y) \vert \Vert(x,y)\Vert _p = 1 \}, 1 \leq p \leq \infty.$$

Monday, 2-28-2005

Theorem 23   Minkowski's Inequality:
Let $(X,\mathcal{S},\mu)$ be a complete metric space. Let $1 \leq p \leq \infty$ . If $f,g \in L^p$ , then $f + g \in L^p$ and

$$\Vert f + g \Vert _p \leq \Vert f \Vert _p + \Vert g \Vert _p$$

Proof. If $f,g \in L^\infty$ , let $a > \Vert f \Vert _\infty$ and $b > \Vert g \Vert _\infty$ . Then,

$$\mu \{ x : \vert f(x)\vert > a \} = \mu\{ x : \vert g(x) \vert > b \} = 0.$$

By the triangle inequality,

$$\{ x : \vert f(x) + g(x)\vert > a + b \} \subset \{ x : \vert f(x) \vert > a \} \cup \{ x : \vert g(x) \vert > b \},$$

so

$$\mu\{ x : \vert f(x) + g(x)\vert > a + b \} = 0$$

Therefore $f + g \in L^\infty$ and $\Vert f + g \Vert _\infty < a + b$ . Taking the infinum over $a$ and $b$ , we establish Minkowski's Inequality for $p = \infty$ .

We must use the following:
Fact: For any real numbers $a,b$ and $p > 0$ , we have

$$\vert a + b \vert^p \leq 2^p (\vert a\vert^p + \vert b\vert^p)$$

Proof. For $s \in [0,1]$ , we know

$$\frac{1 + s}{2} \in [\frac12, 1]$$

So.

$$\left( \frac{1 + s}{2} \right)^p \in (0,1]$$

Thus,

$$\left( \frac{1 + s}{2} \right)^p \leq 1 + s^p.$$

Rearranging,

$$(1 + s)^p \leq 2^p( 1 + s^p )$$

WLOG, assume $\vert a\vert, \vert b\vert > 0$ . Then,

$$\vert a + b\vert^p = \vert a + \left(\frac{b}{a}\right)\vert^p = \vert a\vert^p \vert 1 + \frac{\vert b\vert}{\vert a\vert} \vert^p$$

$$\leq \vert a\vert^p 2^p\left(1 + \frac{\vert b\vert^p}{\vert a\vert^p}\right) = 2^p(\vert a\vert^p + \vert b\vert^p)$$

$\qedsymbol$

Recall that a function $\phi: \mathbb{R}\rightarrow \mathbb{R}$ is convex if

$$\phi(\lambda x + (1-\lambda)y) \leq \lambda \phi(x) + (1-\lambda)\phi(y)$$

whenever $x < y$ and $\lambda \in [0,]1$ .

If $\Vert f\Vert _p = 0$ or $\Vert g \Vert _p = 0$ , then there is nothing to do, so suppose

$$\alpha = \Vert f \Vert _p$$

$$\beta = \Vert g\Vert _p$$

are both nonzero. Put $f_0 = \frac{f}{\alpha}$ and $g_0 = \frac{g}{\beta}$ . Then $\Vert f_0\Vert _p = \Vert g_0\Vert _p = 1$ . Put

$$\lambda = \frac{\alpha}{\alpha + \beta}$$

and

$$(1-\lambda) = \frac{\beta}{\alpha + \beta}$$

Now,

$$\vert f(x) + g(x)\vert^p \leq (\vert f(x)\vert + \vert g(x)\vert)^p = (\alpha \vert f_0(x)\vert + \beta \vert g_0(x)\vert)^p$$

$$(\alpha + \beta)^p (\lambda \vert f_0(x)\vert + (1-\lambda)\vert ... ...alpha + \beta)^p \lambda \vert f_0(x)\vert^p + (1-\lambda)\vert g_0(x)\vert^p)$$

Hence,

$$\int_X \vert f(x) + g(x)\vert^p d\mu \leq (\alpha + \beta)^p \lef... ... \vert f_0(x)\vert^p d\mu + (1-\lambda)\int_X \vert g_0(x)\vert^p d\mu \right]$$

$$= (\alpha + \beta)^p = (\Vert f\Vert _p + \Vert g\Vert _p)^p$$

Taking $p^{th}$ roots, $f + g \in L^p$ and the inequality is established. $\qedsymbol$

Wednesday, 3-2-2005: Dr. Rammaha teaches

Definition 5.1   A normed space $(X, \Vert \cdot \Vert)$ is a linear space $X$ (closed under addition and multiplication by scalars over a field $\mathbb{F}$ ) and a function $\Vert \cdot \Vert : X \rightarrow \mathbb{R}$ with the properties:

1. $\Vert x \Vert \geq 0$ and $\Vert x\Vert = 0$ if and only if $x = 0$ .
2. $\Vert c x \Vert = \vert c\vert \Vert x \Vert$ , for all $c \in \mathbb{F}$ and all $x \in X$
3. Triangle Inequality:
   $$\Vert x + y \Vert \leq \Vert x \Vert + \Vert y \Vert, \forall x,y \in X$$

Let us agree that $(X,\mathcal{S},\mu)$ be any measure space. Then, for $1 \leq p < \infty$ ,

$$L^p(\mu) = L^p(X, \mathcal{S}, \mu) := \{ f : X \rightarrow \mathbb{F}: f$$    is measurable, $$\Vert f \Vert _p < \infty \}$$

where

$$\Vert f \Vert _p = \left( \int_X \vert f\vert^p d\mu \right)^{\frac1p}.$$

Because of the Minkowski inequality, it is now trivial to verify that $(L^p(\mu), \Vert \cdot \Vert _p)$ is a normed space for $1 \leq p \leq \infty$ .

Definition 5.2   Let $(X, \Vert \cdot \Vert)$ be a normed space. Let $x_n, x \in X$ for $n \in \mathbb{N}$ . Then, we say that $x_n \rightarrow x$ in $X$ if and only if

$$\Vert x_n - x \Vert _p \rightarrow 0$$

as $n \rightarrow \infty$ . We say $\{ x_n \}$ is Cauchy in $X$ if

$$\Vert x_n - x_m \Vert \rightarrow 0$$

as $n,m \rightarrow \infty$ . We say that $X$ is complete if all Cauchy sequences converge to a point in $X$ . In this case, $X$ is called a Banach space ($B$ -space).

Definition 5.3   Let $(X, \Vert \cdot \Vert)$ be a normed space and take a sequence $\{ u_n \}_1^\infty \subset X$ . We say the series

$$\sum_{n=1}^\infty u_n$$

converges absolutely if

$$\sum_{n=1}^\infty \Vert u_n\Vert$$

converges in $\mathbb{R}$ .

Theorem 24   A normed space $(X, \Vert \cdot \Vert)$ is complete if and only if every absolutely convergent series converges in $X$ .

Proof. Suppose $X$ is very complete - every Cauchy sequence converges. Let $\{ u_n \}_1^\infty \subset X$ such that

$$\sum_{n=1}^\infty \Vert u_n\Vert$$

is convergent - say to $M \in [0, \infty)$ . Let $\epsilon > 0$ be given. Then, there exists $N \in \mathbb{N}$ such that

$$\sum_{n=N}^\infty \Vert u_n \Vert < \epsilon$$

Set

$$\mathcal{S}_n = \sum_{k=1}^n u_k$$

Then, for all $n > m \geq N$ , we have

$$\Vert S_n - S_m \Vert = \Vert \sum_{k=m+1}^n u_k \Vert \leq \sum_{k=m+1}^n \Vert u_k \Vert$$

(where the last step is by the triangle inequality).

$$\leq \sum_{k=N}^\infty \Vert u_k \Vert < \epsilon$$

So, $S_n$ is Cauchy in $X$ , which is complete. Hence, $S_n$ converges to some point $u \in X$ . We write:

$$\lim_{n \rightarrow \infty} S_n =: \sum_{k=1}^\infty u_k =u \in X$$

The converse is a bit more tricky. Let $\{ u_n \}_{n=1}^\infty \subset X$ such that $\{ u_n \}$ is Cauchy. For every $k \in \mathbb{N}$ , there exists a strictly increasing sequence $\{n_k\}_{k = 1}^\infty \subset \mathbb{N}$ such that $\Vert u_n - u_m \Vert < 2^{-k}$ for all $n,m \geq n_k$ (we can do this by the definition of a Cauchy sequence).
So, we have a subsequence

$$\{ u_{n_k} \}_{k=1}^\infty \subset \{ u_n \}.$$

Set

$$g_1 = u_{n_1}$$

and

$$g_k = u_{n_k} - u_{n_{k-1}}$$

Set

$$S_j = \sum_{k=1}^j g_k = u_{n_j}$$

Also,

$$\sum_{k=1}^j \Vert g_k \Vert = \sum_{k=1}^j \Vert u_{n_k} - u_{n_{k-1}}\Vert \leq \sum_{k=1}^j 2^{-k}$$

Hence,

$$\sum_{k=1}^j \Vert g_k \Vert \leq \sum_{k=1}^\infty 2^{-k} = 1 < \infty.$$

Hence, by hypothesis, $S_k$ converges to some point, $u \in X$ . So,

$$\Vert u_{n_j} - u \Vert = \Vert S_j - u \Vert \rightarrow 0$$

Then, given any $\epsilon > 0$ , there exists a $N_1 > 0$ such that $\Vert u_{n_j} - u \Vert \leq \frac{\epsilon }{2}$ for $j \geq N_1$ and a $N_2 > 0$ such that $\Vert u_m - u_{n_j} \Vert \leq \frac{\epsilon }{2}$ for $m \geq N_2$ and $j \geq N_1$ . Hence, for any $m > N := \max\{ N_1, N_2 \}$ ,

$$\Vert u_m - u \Vert \leq \Vert u_m - u_{n_j}\Vert + \Vert u_{n_j} - u\Vert \leq \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon$$

Hence, the Cauchy sequence converges to $u \in X$ , and $X$ is complete. $\qedsymbol$

Theorem 25   Riesz-Fisher Theorem:
$(L^p(\mu), \Vert \cdot \Vert)$ is a Banach space for all $1 \leq p \leq \infty$ .

Proof. First, assume $1 \leq p < \infty$ . By the last theorem, we need to show every sequence $\{ f_n\}^\infty \subset L^p(\infty)$ such that $$\sum_{k=1}^\infty \Vert f_k \Vert$$ is absolutely convergent, then $\sum_{k=1}^\infty f_k$ is convergent in $L^p(\mu)$ . So, let $\{ f_k \} \subset L^p(\mu)$ such that

$$\sum_{k=1}^\infty \Vert f_k\Vert _p = B < \infty$$

Define

$$g_n(x) = \sum_{k=1}^n \vert f_k(x)\vert.$$

Then, $g_n$ is an increasing sequence on $X$ with values in $[0,\infty]$ . And so,

$$g(x) := \sum_{k=1}^\infty \vert f_k(x)\vert$$

is measurable and $g(x) \in [0,\infty]$ . Now, by the triangle (Minkowski) inequality,

$$\Vert g_n\Vert _p = \Vert \sum_{k=1}^n \vert g_k\vert \Vert _p \leq \sum_{k=1}^n \Vert g_k\Vert _p$$

I.e.,

$$\int_X \vert g_n\vert^p d\mu \leq B^p$$

So, by the Monotone Convergence Theorem, we have

$$\int_X \vert g(x)\vert^p d\mu = \lim_{n \rightarrow \infty} \int_X \vert g_n(x)\vert^p d\mu \leq B^p$$

Hence,

$$\Vert g \Vert _p \leq B.$$

So, $g \in L^p(\mu)$ . By an old theorem, $g < \infty$ almost everywhere and so the series

$$\sum_{k=1}^\infty f_k(x)$$

is absolutely convergent for almost all $x$ . Let $E$ be the set on which $f_k(x)$ converges absolutely. Then, define

$$F(x) = \begin{cases}\sum_{k=1}^\infty f_k(x), & x \in E \ 0, x \in E^c \end{cases}$$

Then,

$$\vert F(x)\vert \leq g(x)$$

for all $x \in X$ and $F$ is $\mu$ -measurable. Note

$$\int_X \vert F(x)\vert^p d\mu \leq \int_X g(x)^p d\mu \leq B^p < \infty$$

So, $F \in L^p(\mu)$ .

Finally, we note the following:

0    almost everywhere $$= \lim_{n \rightarrow \infty} \vert F - \sum_{k=1}^n f_k\vert^p$$

and, for all $n$ ,

$$\leq \left( \vert F\vert + \sum_{k=1}^n \vert f_k\vert\right)^p \leq (g + g)^p = 2^p g^p \in L^1$$

So, by the Dominated Convergence Theorem,

$$\int_X \vert F - \sum_{k=1}^\infty f_k \vert^p = 0$$

almost everywhere. Thus,

$$\lim_{n \rightarrow \infty} \Vert F - \sum_{k=1}^n f_k \Vert _p = 0.$$

I.e.,

$$\sum_{k=1}^\infty f_k = F.$$

in $L^p(\mu)$ . This proves the completeness of $L^p(\mu)$ for $p$ finite.

The case $p = \infty$ . Let $\{ f_n \}_1^\infty \subset L^\infty(\mu)$ be a Cauchy sequence. By the definition of $L^\infty(\mu)$ , for all $n \in \mathbb{N}$ , there exists a set $E_n$ such that $\vert f_n(x)\vert \leq \Vert f_n \Vert _\infty$ on $E_n^c$ and where $\mu(E_n) = 0$ . Similarly, $f_n - f_m \in L^\infty(\mu)$ and so for all $m,n \in \mathbb{N}$ , there exists a $F_{n,m}$ such that

$$\vert f_n(x) - f_m(x) \vert \leq \Vert f_n - f_m \Vert _\infty$$

on $F_{n,m}^c$ and where $\mu(F_{n,m}) = 0$ . Now, for all $n,m \in \mathbb{N}$ , set

$$A = \left( \bigcup_{n=1}^\infty E_n \right) \cup \left( \bigcup_{n,m=1}^\infty F_{n,m} \right)$$

Then, $\mu(A) = 0$ . Also, we still have

$$\vert f_n(x) \vert \leq \Vert f_n\Vert _\infty$$

and

$$\vert f_n(x) - f_m(x) \vert \leq \Vert f_n - f_m \Vert _\infty$$

on $A^c$ . Now, $\Vert f_n - f_m\Vert _\infty \rightarrow 0$ as $n,m \rightarrow \infty$ .
Therefore, $\{f_n(x)\}_{n=1}^\infty$ is uniformly Cauchy as $n,m \rightarrow \infty$ on $A^c$ . Thus, there exists a function $f$ such that $$f(x) = \lim_{n \rightarrow \infty} f_n(x)$$ for all $x \in A^c$ . Then, set

$$F(x) = \begin{cases}\displaystyle f(x) = \lim_{n \rightarrow \infty} f_n(x), & x \in A^c \ 0,& x \in A. \end{cases}$$

Now, there exists $N \in \mathbb{N}$ (using $f_n(x) \rightarrow f(x)$ uniformly on $A^c$ ) such that

$$\vert f(x) - f_N(x) \vert < 1, \forall x \in A^c.$$

So, for all $x \in A^c$ , we have

$$\vert F(x)\vert \leq \vert f(x) - f_N(x)\vert + \vert f_N(x)\vert < 1 + \Vert f_N \Vert _\infty < \infty.$$

So, $F \in L^\infty(\mu)$ . Now,

$$\Vert F - f_n \Vert _\infty = \sup_{x \in A^c} \vert F(x) - f_n(x) \vert = \sup_{x \in A^c} \vert f(x) - f_n(x) \vert \rightarrow 0$$

So, $f_n \rightarrow F$ in $L^\infty(\mu)$ . Hence, $L^\infty(\mu)$ is complete.
$\qedsymbol$

Monday, 3-7-2005:

Lemma 5.1   ``Young's Inequality":
If $a, b \geq 0$ and $0 \leq \lambda \leq 1$ , then

$$a^\lambda b^{1-\lambda} \leq \lambda a + (1 - \lambda) b$$ (2)

and equality holds if and only if $a = b$ .

History Young's original statement: Let $a: [0,\infty) \rightarrow [0,\infty)$ be continuous and strictly increasing. Let $\tilde a := a^{-1}$ . So, $\tilde a$ is continuous and strictly increasing on $[0,\infty)$ . Assume $$\lim_{x \rightarrow \infty} a(x) = +\infty$$ . For $x \geq 0$ , set

$$A(x) = \int_0^x a(t) dt$$    and $$\tilde A(x) = \int_0^x \tilde a (t) dt$$

Then, Young's Original Inequality says:

$$xy \leq A(x) + \tilde A(y), \forall x,y \geq 0$$

A special case of this inequality is with $a(t) = t^{p-1}$ where $1 < p < \infty$ . Here, $\tilde a(t) = t^\frac{1}{p-1}$ . So, the original inequality gives us that

$$xy \leq \frac1p x^p + \frac{p-1}{p} y^{\frac{p}{p-1}}$$

Definition 5.4   Let $1 \leq p, q \leq \infty$ . We say $p$ and $q$ are Holder conjugates if $\frac1p + \frac1q = 1$ . Here, if $1 < p < \infty$ , then $$q = \frac{p}{p-1}$$ is the Holder conjugate to $p$ .

So, we get that

$$x y \leq \frac{x^p}{p} + \frac{y^q}{q}, \forall x,y \geq 0$$    and $$p, q$$    are Holder conjugates$$$$

This is the statement we'd like to prove.

Proof. (Young's Inequality)
If $b = 0$ , then eq. 2 is trivial. Suppose $b > 0$ - then 2 is equivalent to

$$\left(\frac{a}{b}\right)^\lambda \leq \lambda \left(\frac{a}{b}\right) + 1 - \lambda$$

Setting $t = \frac{a}{b}$ . Then $t \geq 0$ and 2 is equivalent to

$$t^\lambda \leq \lambda t + 1 - \lambda, t \geq 0$$

and equality holds if and only if $t = 1$ .
Set

$$f(t) = t^\lambda - \lambda t, t \geq 0$$

Note,

$$f'(t) = \lambda t^{\lambda - 1} - \lambda = \lambda(t^{\lambda - 1} - 1)$$

$$= \begin{cases}< 0, & t > 1 \ > 0, & 0 < t < 1 \ 0, & t = 1 \end{cases}$$

i.e., $f$ has a global maximum at $t = 1$ and so

$$f(t) = t^\lambda - \lambda t \leq f(1) = 1 - \lambda$$

and equality holds if and only if $t = 1$ . $\qedsymbol$

Remark: Let $a, b \geq 0$ and $p > 1$ and $\frac{1}{q} = 1 - \frac1p$ (i.e., $p$ and $q$ are Holder conjugates.) Take $\tilde a = a^p$ and $\tilde b = b^q$ and $\lambda = \frac1p$ . Note that $\frac1q = 1 - \lambda$ . From eq. 2, we have

$$\tilde a^\lambda \tilde b^{1-\lambda} \leq \lambda \tilde a + (1 - \lambda) \tilde b$$

Rewriting, this is

$$a b \leq \frac{a^p}{p} + \frac{b^q}{q}$$

Example: Take $a,b,\epsilon > 0$ .

$$ab = a \epsilon ^{\frac1p} \epsilon ^{- \frac1p} b \leq \frac{1}{p} ( a \epsilon ^{\frac1p} )^p + \frac1q( \epsilon ^{-\frac1q} b)^q$$

$$= \frac{\epsilon a^p}{p} + \frac{1}{q} \epsilon ^{-\frac{p}{q}} b^q$$

Theorem 26   ``Holder's Inequality":
Let $1 < p < \infty$ and $\frac1p + \frac1q = 1$ . Let $f, g: X \rightarrow \mathbb{C}$ be $\mu$ -measurable functions. Then,

$$\Vert fg \Vert _1 = \int_X \vert fg\vert d\mu \leq \Vert f\Vert _... ...f\vert^p d\mu \right)^\frac1p \left( \int_X \vert g\vert^q d\mu \right)^\frac1q$$ (3)

In particular, if $f \in L^p(\mu)$ and $g \in L^q(\mu)$ , then $fg \in L^1(\mu)$ , and in this case, equality holds in 3 if and only if

$$\Vert g\Vert _q^q \vert f\vert^p = \Vert f\Vert _p^p \vert g\vert^q$$

$\mu$ -almost everywhere.

Proof. If $\Vert f\Vert _p = 0$ or $\Vert g\Vert _q=0$ , then $f = 0$ $\mu$ -almost everywhere or $g = 0$ $\mu$ -almost everywhere, and eq. 3 is trivial. Also, if $\Vert f\Vert _p = \infty$ or $\Vert g\Vert _q = \infty$ , then eq. 3 is again trivial. So, we assume that $0 < \Vert f\Vert _p, \Vert g\Vert _q < \infty$ .
So, eq. 3 is equivalent to

$$\frac{1}{\Vert f\Vert _p \Vert g\Vert _q} \int_X \vert fg\vert d\mu \leq 1$$ (4)

Let

$$a(x) = \frac{\vert f(x)\vert}{\Vert f\Vert _p}, b(x) = \frac{\vert g(x)\vert}{\Vert g\Vert _p}$$

Then, by Young's Inequality,

$$\frac{\vert fg\vert}{\Vert f\Vert _p \Vert g\Vert _q} = a b \leq ... ...t^p}{\Vert f\Vert _p^p} + \frac1q \frac{\vert g(x)\vert^q}{\Vert g\Vert _q^q}.$$

This holds $\mu$ -almost everywhere, with equality if and only if

$$\Vert g\Vert _q^q \vert f\vert^p = \Vert f\Vert _p^p \vert g\vert^q$$

By integrating, we have:

$$\frac{1}{\Vert f\Vert _p\Vert g\Vert _q} \int_X \vert fg\vert d\m... ... + \frac1q \frac{\Vert g\Vert _q^q}{\Vert g\Vert _q^q} = \frac1p + \frac1q = 1$$

So, we get

$$\Vert fg \Vert _1 \leq \Vert f\Vert _p \Vert g\Vert _q$$

$\qedsymbol$

Wednesday, 3-9-2005:

Proposition 5.1   For $1 \leq p < \infty$ , the set

$$\mathcal{S}:= \{ f = \sum_{j=1}^n a_j \chi_{E_j} : \mu(E_j) < \infty \}$$

is dense in $L^p$ .

Proof. As $\mu(E_j) < \infty$ , each element in the sum is in $L^p$ , so $\mathcal{S}\subset L^p$ . Let $f \in L^p$ . Find a sequence of simple functions $f_n$ such that $\vert f_n\vert \leq \vert f\vert$ and $f_n \rightarrow f$ pointwise. Writing these functions in standard form, $$f_n = \sum_{k=1}^r a_k \chi_{E_k}$$ , the condition that $\vert f_n\vert \leq f$ shows that $\mu(E_k) < \infty$ , so $f_n \in \mathcal{S}$ . Also,

$$\vert f_n(x) - f(x)\vert^p \leq 2^p (\vert f_n(x)\vert^p + \vert f(x)\vert^p) \leq 2^{p+1} \vert f(x)\vert^p$$

Hence, by the DCT, $\int \vert f_n - f\vert^p d\mu \rightarrow 0$ , and $\Vert f_n - f\Vert _p \rightarrow 0$ . $\qedsymbol$

Remark: A similar result is true for $p = \infty$ .

Definition 5.5   Let $X,Y$ be normed spaces over $\mathbb{F}$ and let $T: X \rightarrow Y$ be a linear map. We say that $T$ is a bounded linear map if

$$\sup \{ \Vert Tx\Vert _Y : \Vert x\Vert _X \leq 1 \} < \infty$$

If $T$ is bounded, we define

$$\Vert T\Vert := \sup \{ \Vert Tx\Vert _Y : \Vert x\Vert _X \leq 1 \}$$

Example: Let $X = \{ f : (0,1) \rightarrow \mathbb{R}:$    $f$ is bounded and has derivatives of all orders$\}$ . For $f \in X$ , let $\Vert f \Vert = \sup_{t \in (0,1)} \vert f(t)\vert$ . Then $X$ is a normed space. Define $D:X \rightarrow X$ by $Df = f'$ . Then $D$ isn't bounded. Reason: $f_n(x) = \sin nx$ . So, $\Vert f_n\Vert = 1$ but $\Vert D f_n\Vert = n$ .

Proposition 5.2   Let $X,Y$ be normed spaces and $T: X \rightarrow Y$ a linear map. The following are equivalent.

(a)

$T$ is continuous at 0.

(b)

$T$ is continuous.

(c)

$T$ is bounded.

Proof. $(b) \Rightarrow (a)$ is trivial. $(a) \Rightarrow (b)$ Suppose $x_n \in X$ and $x_n \rightarrow x \in X$ . Then, $x_n - x \rightarrow 0$ , so by hypotheses, $T(x_n - x) \rightarrow 0$ , i.e., $Tx_n \rightarrow Tx$ .
$(b) \Rightarrow (c)$ (Prove the contrapositive). Suppose $T$ isn't bounded. Then $\exists x_n \in X$ such that $\Vert x_n \Vert \leq 1$ , yet $\Vert T x_n \Vert > n$ . Then, $\Vert \frac{x_n}{n} \Vert \rightarrow 0$ so $\frac{x_n}{n} \rightarrow 0$ in $X$ . But, $\Vert T\frac{x_n}{n} \Vert = \frac{\Vert T x_n \Vert}{n} > 1$ , so $T(\frac{x_n}{n})$ does not converge to zero. Hence, $T$ isn't continuous at 0 so $T$ isn't continuous.
Finally, suppose that (c) holds. Let $x_n \rightarrow 0$ . Then,

$$\Vert Tx_n\Vert = \begin{cases}0, & x_n = 0 \ \Vert x_n\Vert \Vert T(\frac{x_n}{\Vert x_n\Vert}) \Vert, & x_n \neq 0 \end{cases}$$

Hence, $\Vert T x_n \Vert \leq \Vert T\Vert \Vert x_n\Vert \rightarrow 0$ . As $T x_n \rightarrow 0$ so $T$ is continuous at 0 . $\qedsymbol$

Remark:
If $T$ is a bounded linear transformation, then $\forall x \in X$ , $\Vert Tx\Vert \leq \Vert T\Vert \Vert x\Vert$ .

Definition 5.6   Linear Functional:
Let $X$ be a normed space. A linear functional is a linear map $f : X \rightarrow \mathbb{F}$ and is bounded if it is bounded as a linear transformation.

Example:
Let $1 < p < \infty$ and suppose $p,q$ are Holder conjugates. Let $X = L^p$ and fixed $g \in L^q$ . Define

$$\phi_g(f) = \int_X f g d\mu.$$

By Holder's inequality, $\phi_g$ is a linear functional on $L^p$ and

$$\vert \phi_g(f) \vert \leq \int_X \vert fg\vert d\mu \leq \Vert f\Vert _p \Vert g\Vert _q$$

Dividing by $\Vert f\Vert _p$ , we get that $\vert\phi_g(f)\vert \leq \Vert g\Vert _q$ . Taking the sup over $\Vert f\Vert _p \leq 1$ , we get that $\phi_g$ is bounded and $\Vert\phi_g\Vert \leq \Vert g\Vert _q$ .

Friday, 3-11-2005:
(Aside on notation:)

$$\ell^p := L^p(\mathbb{N},$$   counting measure$$)$$

That is,

$$= \{ (x_n)_{n=1}^\infty : \sum_{n=1}^\infty \vert x_n\vert^p < \}$$

Actually, when $1 < p \leq \infty$ , we get equality for any measure and equality also holds when $p=1$ and $\mu$ is semifinite.

Proof. For $1 < p < \infty$ :
We have

$$\Vert g\Vert _q^q = \int_X \vert g\vert^q d\mu = \in_X \vert g\vert^{q-1}\vert g\vert d\mu$$

Put $f_0 = \vert g\vert^{q-1}$ . Then,

$$\vert f_0\vert^p = \vert g\vert^{pq - p} = \vert g\vert^q$$

Therefore, $\vert f_0\vert \in L^p$ . Moreover,

$$\Vert f_0\Vert _p = \left( \int_X \vert f_0\vert^p \right)^\frac1... ...g\vert^q \right)^\frac1p = \Vert g\Vert _q^\frac{q}{p} = \Vert g\Vert _q^{q-1}$$

Put

$$f = \frac{\vert g\vert^{q-1}}{\Vert g\Vert _q^{-1}} \overline{sign(g)}$$

where $sign(g) :X \rightarrow \mathbb{F}$ satisfies

$$sign(g) \vert g\vert = g$$

Notice that $sign(g)$ is measurable. Then $\Vert f\Vert _p = 1$ and

$$\phi_g(f) = \int_X \frac{\vert g\vert^{q-1}}{\Vert g\Vert _q^{q-1}} \overline{sign(g)} g d\mu$$

$$= \int_X \frac{\vert g\vert^q}{\Vert g\Vert _q^{q-1}} d\mu = \frac{\Vert g\Vert _q^q}{\Vert g\Vert _q^{q-1}} = \Vert g\Vert _q.$$

Hence, $\Vert\phi_g\Vert = \Vert g\Vert _q$ . We leave the other cases for the interested reader. When $p = \infty$ , $q=1$ and take $f = \overline{sign(g)}$ . Then $f \in L^\infty$ and

$$\int_X fg = \int \vert g\vert = \Vert g\Vert _1.$$

$\qedsymbol$

Theorem 27   Reisz Representation Theorem Let $1 \leq p < \infty$ and let $q$ satisfy $\frac1p + \frac1q = 1$ . Let $\phi:L^p \rightarrow \mathbb{F}$ be a bounded linear functional. Then,

1. If $1 < p < \infty$ , then there exists a unique $g \in L^q$ such that for all $f \in L^p$ ,
   $$\displaystyle \phi(f) = \int_X fg d\mu.$$
   and $\Vert\phi\Vert = \Vert g\Vert _q$ .
2. If $X$ is $\sigma$ -finite and $g: L^1 \rightarrow \mathbb{F}$ is a bounded linear functional, then there exists a unique $g \in L^\infty$ such that
   $$\phi(f) = \int_X f g d\mu$$
   and $\Vert\phi\Vert = \Vert g\Vert _\infty$ .
   

Proof. First assume that $(X,\mathcal{S},\mu)$ is a finite measure. Define $\nu(E) = \phi(\chi_E)$ for $E \in \mathcal{S}$ . Then, $\Vert\chi_E\Vert _p = \mu(E)^\frac1p$ . Let $E = \bigcup_{j=1}^\infty E_j$ be a disjoint union and let $\alpha_i \in \mathbb{F}$ satisfy $\vert\alpha_i\vert = 1$ and $\alpha_i \nu(E_j) \geq 0$ . Let

$$f = \sum_{k=1}^\infty \alpha_j \chi_{E_j}$$

Note that the series converges in $L^p$ :

$$\Vert f - \sum_{j=1}^n \alpha_j \chi_{E_j} \Vert _p = \Vert\sum_{... ...^\infty \alpha_j \chi_{E_j} \Vert _p = \mu( \bigcup_{j=n+1}^\infty E_j)^\frac1p$$

Now, we are in a finite measure so the right hand side always exists. By the continuity theorem, we get that this term goes to zero. Hence, the partial sums converge in $L^p$ , so the whole series converge in $L^p$ .
Then

$$\phi(f) = \lim_{n \rightarrow \infty} \phi(\sum_{j=1}^n \alpha_j \chi_{E_j} )$$

$$= \lim_{n \rightarrow \infty} \sum_{j=1}^n \alpha_j \phi(\chi_{E_... ...j)\vert = \sum_1^\infty \vert\nu(E_j)\vert \leq \Vert\phi\Vert \Vert f\Vert _p$$

Hence, $$\sum_{j=1}^\infty \vert\nu(E_j)\vert$$ is absolutely convergent and $$\nu(E) = \sum_{j=1}^\infty \nu(E_j)$$ . Hence, $\nu$ is a complex measure.
If $\mu(E) = 0$ , then $\chi_E = 0 \in L^p$ ,

$$\nu(E) = \phi(\chi_E) = \phi(0) = 0$$

Hence, $\nu \ll \mu$ . By the Radon-Nikodym theorem, there exists a function $g \in L^1(\mu)$ such that

$$\phi(\chi_E) = \nu(E) = \int_X g \chi_E d\mu$$

By linearity, if $f = \sum_{j=1}^n c_j \chi_{E_j}$ is a simple function, $\phi(f) = \int_E g f d\mu$ . As simple functions are dense in $L^p$ and $\phi$ is continuous, we see this holds for all $f \in L^p$ .
We next show that $g \in L^q$ . Write $g = sign(g)\vert g\vert$ , where

$$sign(g)(x) = \begin{cases}\frac{g(x)}{\vert g(x)\vert}, & g(x) \neq 0 \ 1, & g(x) = 0 \end{cases}$$

Let $\psi_n$ be a sequence of simple functions with $0 \leq \psi_n \leq \vert g\vert^q$ and $$\lim_{n \rightarrow \infty} \psi_n(x) = \vert g(x)\vert^q$$ . Because the measure is finite, we conclude that each $\psi_n$ is integrable.
Then, $\psi_n^\frac1p \overline{sign(g)} \in L^p$ and when $p \in (1,\infty)$ ,

$$\psi_n = \psi_n^\frac1p \psi_n^\frac1q \leq \psi_n^\frac1p\vert g\vert = \psi_n^\frac1p \overline{sign(n)} g,$$

so

$$\int \psi_n d\mu \leq \int \psi_n^\frac1p \overline{sign(g)} g d\mu$$

$$= \phi(\psi_n^\frac1p \overline{sign(g)}) \leq \Vert \phi \Vert \Vert \psi_n^\frac1p \overline{sign(g)}\Vert _p$$

$$= \Vert \phi \Vert \left( \int_X \psi_n d\mu \right)^\frac1p$$

Therefore,

$$\left( \int_X \psi_n d\mu \right)^\frac1q \leq \Vert \phi\Vert$$

By the Monotone Convergence Theorem, we can take the limit to get that $g \in L^q$ and $\Vert g\Vert _q \leq \Vert\phi\Vert$ . By our previous work, $\Vert \phi \Vert \leq \Vert g \Vert _q$ , so $\Vert\phi\Vert = \Vert g\Vert _q$ .
We now establish uniqueness:
Let $g_1 \in L^q$ and

$$\int_X fg d\mu = \int_X fg_1 d\mu$$

for all $f \in L^p$ . Then,

$$\int_X f(g - g_1) d\mu = 0$$

for all $f \in L^p$ . Take $f = \overline{sign(g-g_1)} \in L^p$ ($\mu$ is finite), so

$$\int \vert g - g_1\vert = 0$$

Hence, $g = g_1$ almost everywhere.
We now finish the proof for (b). When $p=1$ , argue as before to obtain $g \in L^1$ such that

$$\phi(f) = \int_X f g d\mu$$

Let $0 < M < esssup\vert g\vert$ (if $esssup\vert g\vert = 0$ , then we really have nothing to do). Let

$$E := \{ x \in X : \vert g(x)\vert > M \}$$

Note that $\mu(E) > 0$ . Put

$$f = \frac{\chi_E}{\mu(E)} \overline{sign(g)}$$

Then,

$$\Vert f\Vert _1 = 1$$

and

$$\phi(f) = \int_X \frac{\chi_E}{\mu(E)} \overline{sign(g)} g d\mu = \int_X \frac{\chi_E}{\mu(E)} \vert g\vert d\mu$$

$$\geq \frac{M}{\mu(E)} \mu(E) = M$$

Therefore,

$$M < \phi(f) \leq \Vert\phi\Vert \Vert f\Vert _1 = \Vert\phi\Vert$$

Taking the sup over possible values of $M$ , we get that

$$esssup\vert g\vert \leq \Vert\phi\Vert,$$

i.e., $g \in L^\infty$ . Uniqueness follows similarly. Hence, this completes the theorem for the case of a finite measure.

Wednesday, 3-23-2005:
We now proceed from the finite case to the $\sigma$ -finite case.
Write $X = \displaystyle \cup_{n=1}^\infty X_n$ where $\mu(X_n) < \infty$ and $X_n \subset X_{n+1}$ . Consider the linear functional $\phi_n$ on $L^p(X, \mu)$ defined by

$$\phi_n(f) = \phi(f \chi_{X_n})$$

Note that $\phi_n$ is bounded since

$$\vert\phi_n(f)\vert = \vert\phi(f \chi_{X_n}) \leq \Vert\phi\Vert \Vert f\chi_{X_n}\Vert _p \leq \Vert \phi \Vert \Vert f \Vert _p$$

This also shows that $\Vert\phi_n\Vert \leq \Vert\phi\Vert$ .
By our previous work, there exists a unique $g_n \in L^q$ such that

$$\phi_n(f) = \int_X f g_n d\mu$$

and $\Vert g_n\Vert _q = \Vert\phi_n\Vert \leq \Vert\phi\Vert$ . But,

$$g_{n+1}\vert _{X_n} = g_n$$

by the uniqueness assumption. Define $g: X \rightarrow \mathbb{F}$ by

$$g(x) = g_n(x)$$

if $x \in X_n$ . We have $\vert g \chi_{X_n}\vert = \vert g_n \chi_{X_n}\vert$ and thus $\vert g \chi_{X_n}\vert \rightarrow \vert g\vert$ pointwise. Further,

$$\int_X \vert g\vert^q d\mu = \lim \int_X \vert g \chi_{X_n} \vert^q d\mu$$

$$= \lim \Vert g_n\Vert _q^q \leq \phi^q$$

Therefore, $g \in L^q$ and $\vert\vert g\Vert _q \leq \Vert\phi\Vert$ . Finally, for $f \in L^p$ , put $f_n = f \chi_{X_n}$ . Then $f_n \rightarrow f$ pointwise and

$$\vert f_ng\vert \leq \vert fg\vert \in L^1$$

by Holder's Inequality. By the DCT:

$$\int_X fg d\mu = \lim_{n \rightarrow \infty} \int f_n g d\mu = \l... ...\infty} \int f g \chi_{X_n} d\mu = \lim_{n \rightarrow \infty} \int f g_n d\mu$$

$$= \lim_{n \rightarrow \infty} \phi_n9f) = \lim_{n \rightarrow \infty} \phi(f \chi_{X_n}) = \phi(f),$$

since $\Vert f \chi_{X_n} - f \Vert _p \rightarrow 0$ and by the continuity of $\phi$ . Uniqueness of $g$ follows from the uniqueness of $g_n$ .
This shows the theorem holds for $1 < p < \infty$ and $\mu$ $\sigma$ -finite.

Now if $p=1$ and $\mu$ is $\sigma$ -finite, argue as above to obtain unique $g_n \in L^\infty$ and construct $g$ as well. Then, $\Vert g_n\Vert _\infty \leq \Vert\phi\Vert$ for all $n$ , and hence $\Vert g\Vert _\infty = \sup_n \Vert g_n\Vert _\infty \leq \Vert\phi\Vert$ .

Now, let $\mu$ be an arbitrary measure.
For each set $E \subset X$ with $E$ of $\sigma$ -finite measure, let $g_E \in L^q$ be such that

$$\phi(f \chi_E) = \int_X f g_E \chi_E d\mu.$$

Notice that if $E$ and $F$ are both of $\sigma$ -finite measure and $E \subset F$ , then $g_F \vert _E = g_E$ and

$$\Vert g_E\Vert _q \leq \Vert\phi\Vert$$

Let

$$M := \sup \{ \Vert g_E\Vert _q : E$$    has $\sigma$ -finite measure$$\}$$

So $M \leq \Vert\phi\Vert$ . Let $\{ E_N \}$ be a sequence of $\sigma$ -finite measure such that $\Vert g_{E_n}\Vert _q \rightarrow M$ . Set

$$F = \displaystyle \cup_{n=1}^\infty E_n.$$

Then $F$ is of $\sigma$ -finite measure. Since $F \supset E_n$ for all $n$ , and $\Vert g_F\Vert _q \geq \Vert g_{E_n}\Vert _q$ for all $n$ , we have $\Vert g_F\Vert _q = M$ .
If $A \supset F$ and $A$ has $\sigma$ -finite measure, then $M \geq \Vert g_A\Vert _q \geq \Vert g_F\Vert _q = M$ , so $\Vert g_A\Vert _q = M$ too.
Also, $g_A$ extends $g_F$ and so

$$M^q = \int_A \vert g_A\vert^q d\mu = \int_{A \setminus F} \vert g_A\vert^q + \int_F \vert g_A\vert^q$$

Therefore, $g_A = 0$ almost everywhere on $A \setminus F$ and $g_A = g_F$ almost everywhere.
Finally, for $f \in L^p$ ,

$$N := \{ x \in X : \vert f(x)\vert > 0 \}$$

is of $\sigma$ -finite measure. So, take $A = N \cup F$ . Thus,

$$\phi(f) = \int_X f g_A d\mu = \int_X f g_F d\mu$$

Thus, if $g = g_F$ , we get

$$\phi(f) = \int_X f g d\mu$$

$\qedsymbol$