Advanced Analysis, Notes 7: Banach spaces (dual spaces and duality, Lp spaces, the double dual, quotient spaces)

Today we continue our treatment of the dual space $latex X^*$ of a normed space (usually Banach) $latex X$. We start by considering a wide class of Banach spaces and their duals. 

1. $latex L^p$ spaces

Perhaps the best studied class of examples of infinite dimensional Banach spaces which are not Hilbert spaces are the $latex L^p$ spaces. In this section we collect some important facts about $latex L^p$ spaces which we shall use to illustrate some results or in applications to analysis.

In this section, we will assume some experience with measure theory.

Let $latex (X, mathcal{M}, mu)$ be a measure space. Recall that this means that $latex X$ is some set, $latex mathcal{M}$ is a $latex sigma$-algebra of subsets of $latex X$ (this just means that $latex mathcal{M}$ is a collection of subsets of $latex X$ that contains $latex X$ and is closed under taking complements and countable unions) and $latex mu : mathcal{M} rightarrow [0,infty]$ is a function that satisfies:

1. $latex mu(emptyset) = 0$, and
2. If $latex {E_i}$ is a countable sequence of disjoint sets, then $latex mu (cup_i E_i) = sum_i mu( E_i)$.

The function $latex mu$ is called a  measure, and elements of $latex mathcal{M}$ are called measurable sets. All our measure spaces will be assumed $latex sigma$ finite, meaning that there is a sequence $latex {X_n}$ of measurable sets such that $latex X = cup X_n$ and $latex mu(X_n)< infty$ for all $latex n$.

For every measurable function $latex f$ and every $latex pin [1, infty)$, we define

$latex |f|_p = (int |f|^p d mu )^{1/p}.$

We also define

$latex |f|_infty = inf{a : mu({|f|>a}) = 0} .$

For $latex p in [1, infty]$ we define $latex L^p = L^p(mu) = L^p(X, mathcal{M}, mu)$ to be the set of all measurable functions $latex f : X rightarrow mathbb{C}$ for which $latex |f|_p < infty$. Of course, we will identify  functions which are equal almost everywhere. The spaces $latex ell^p$ arise as $latex L^p$ spaces when $latex X = mathbb{N}$ and $latex mu$ is the counting measure.

Definition: A function $latex f : X rightarrow mathbb{C}$ is said to be simple if it as the form 

$latex f = sum_{k=1}^N a_k chi_{E_k}$

where $latex a_1, ldots, a_N in mathbb{C}$, $latex E_1, ldots, E_N in mathcal{M}$ all have finite measure, and $latex chi_{E_k}$ denotes the characteristic function of $latex E_k$. 

We shall denote by $latex mathcal{S}$ the vector space of all simple functions. clearly, $latex mathcal{S}$ is contained in $latex L^p$ for all $latex p$. It is a standard and important fact, that we shall use below, that if $latex p<infty$, then $latex mathcal{S}$ is dense in $latex L^p$

Theorem 1: For $latex 1 leq p leq infty$, $latex L^p$ is a Banach space. For $latex p<infty$, the space of simple functions of finite measure support are dense in $latex L^p$. The set of simple functions is dense in $latex L^infty$.

Proof: We prove the theorem for $latex 1leq p < infty$; $latex p = infty$ is left as an excercise. By Minkowski’s inequality $latex L^p$ is a normed space, and the issue we delve on is completeness.

We will use the fact that a normed space is complete if and only if every absolutely convergent sequence in the space is convergent (see the homework exercises). Thus, it suffices to show that if $latex f_k$ are elements of $latex L^p$ such that $latex sum |f_k|_p< infty$, then $latex sum f_k$ converges in the norm of $latex L^p$ to an element of this space.

Let $latex {f_k}$ be such a sequence. For every $latex N$, define

$latex F_N = sum_{k=1}^N |f_k| .$

The sequence $latex F_N$ converges everywhere to a to a function $latex F = sum |f_k|$ that has values in $latex [0,infty]$. For all $latex N$,

$latex |F_N|_p leq sum_{k=1}^N |f_k|_p leq sum_{k=1}^infty |f_k|_p < infty .$

By the monotone convergence theorem,

$latex |F|_p = lim_{Nrightarrow infty}|F_N|_p leq sum_{k=1}^infty |f_k|_p < infty$

thus $F in L^p$. In particular, the function $latex F = sum_{k=1}^infty |f_k|$ is finite (meaning that its value is not $latex infty$) almost everywhere. It follows that the sequence $latex sum f_k(x)$ converges absolutely for almost every $latex x in X$, hence it defines a measurable function. Since $latex |f| leq F$, we have that $latex f in L^p$.

Now we will show that $latex f = lim_{N rightarrow infty}sum_{k=1}^N f_k$ in the norm of $latex L^p$. For this we have to show that

(*)$latex int_X |f – sum_{k=1}^N f_k |^p d mu rightarrow 0$.

But $latex |f – sum_{k=1}^N f_k |^prightarrow 0$ pointwise almost everywhere. On the other hand,

$latex |f-sum_{k=1}^N f_k|^p = |sum_{k=N+1}^infty f_k|^p leq F^p in L^1$

for all $latex N$. Thus (*) follows from the dominated convergence theorem. That completes the proof.

Given $latex p in [1, infty]$, it is customary to denote $latex q = frac{p}{p-1}$, so that $latex 1/p + 1/q = 1$. $latex q$ is called the conjugate exponent of $latex p$.

Theorem 2 (Holder’s inequality): $latex |fg|_1 leq |f|_p|g|_q$.

For a proof, see this.

It is immediate from Holder’s inequality that every $latex g in L^q$ gives rise to a functional $latex phi_g in (L^p)^*$ by way of

$latex phi_g(f) = int fg d mu .$

By Holder’s inequality, $latex |phi_g| leq |g|_q$. A little more effort shows that $latex |phi_g| = |g|_q$ (this will be shown below). It is a deeper fact (to be shown further down below) that $latex g mapsto phi_g$ is actually an isometric isomorphism from $latex L^q$ onto $latex (L^p)^*$.

Theorem 3: Let $latex p in [1, infty)$. Every continuous linear functional on $latex L^p$ is of the form $latex phi_g$ for some $latex g in L^q$. 

We may summarize the above discussion as $latex (L^p)^* = L^q$ for all $latex p in [1, infty)$. The rest of this section is devoted to proving this fact. The proof will follow pretty much the proof from Folland’s book “Real Analysis”. We fix a $latex sigma$-finite measure space $latex (X, mu)$, and let $latex p,q in [1,infty]$ be related by $latex p^{-1}+q^{-1} = 1$.

Lemma: For all $latex g in L^q$, $latex |phi_g| = |g|_q$. 

Proof: We already know that $latex |phi_g| leq |g|_q$, so it remains to show the reverse. We may assume that $latex g neq 0$.

Assume first that $latex q< infty$. Define $latex = |g|_q^{1-q}|g|^{q-1}overline{sign(g)}$. Here $latex sign$ is the function that returns for every complex number $latex z = re^{-it}$ the phase $latex e^{-it}$. One checks readily that $latex f in L^p$ and that $latex |f|_p^p = 1$. Thus

$latex |phi_g| geq |int f g d mu | = |g|_q,$

so $latex |phi_g| geq |g|_q$.

Now we treat the case $latex q = infty$. Fix $latex epsilon > 0$. The set $latex {x: |g(x)| > |g|_infty – epsilon}$ has nonzero measure. Let $latex A$ be a subset of $latex {x: |g(x)| > |g|_infty – epsilon}$ such that $latex 0 < mu(A) < infty$. Let $latex f = mu(A)^{-1} overline{sign(g)}chi_A$. Then $latex f in L^1$, and $latex |f|_1 = 1$. Moreover,

$latex |phi_g| geq |int f g | geq |g|_infty – epsilon.$

Since this holds for all $latex epsilon > 0$, we find $latex |phi_g| geq |g|_infty$. The other inequality has already been established, thus $latex |phi| = |g|_infty$.

Lemma: Let $latex g : X rightarrow mathbb{C}$ be measurable and assume that

$latex M_q(g) := sup {int|fg| d mu : f in mathcal{S}, |f|_p=1 }$

satisfies $latex M_q(g) < infty$. Then $latex g in L^q$ and $latex M_q(g) = |g|_q$.

Proof: By Holder’s inequality. $latex M_q(g) leq |g|_q$, so we will only prove $latex |g|_q < infty$ (and this will also give us $latex g in L^q$). We treat the cases $latex q< infty$ and $latex q = infty$ separately.

The case $latex q<infty$:

There is a sequence of simple functions $latex g_n$ satisfying $latex |g_n| leq |g|$ and $latex g_n rightarrow g$ pointwise (such a sequence, which is easy to construct by hand, is constructed in any course on measure theory). Define $latex f_n = |g_n|_q^{1-q}|g_n|^{q-1}$. Then $latex f_n in mathcal{S}$, $latex |f_n|_p leq 1$. Then

$latex |g|_qleq liminf |g_n|_q = liminf int |f_n g_n| d mu leq M_q(g)$

where the first inequality follows from Fatou’s Lemma.

The case $latex q=infty$:

Let $latex epsilon > 0$, and put $latex A = {x: |g(x)| geq M_infty(g) + epsilon}$. We shall show that $latex mu(A) = 0$, from which it would follow that $latex |f|_infty leq M_infty(g)$.

Indeed, if $latex mu(A) > 0$, then (from $latex sigma$-finiteness) there is $latex B subseteq A$ such that $latex 0<mu(B) <infty$. Define $latex f = mu(B)^{-1}chi_B$. Then $latex |f|_1 = 1$, and $latex int|fg| d mu > M_infty(g) + epsilon$; this is a contradiction to the very definition of $latex M_infty(g)$. Thus $latex mu(A) = 0$, and the proof is complete.

Lemma: With the notation of the previous lemma,

$latex M_q(g) = sup { |int fg d mu | : f in mathcal{S}, |f|_p = 1}$.

Proof: It is clear that the right hand side is less than or equal to $latex M_q(g)$. For the reverse inequality, let $latex epsilon > 0$, and let $latex f in mathcal{S}$ have $latex p$-norm $latex 1$ such that

$latex int |f g| d mu > M_q(g) – epsilon. $

Let $latex h = |f| overline{sign(g)}$. Then $latex |int h g d mu| = int|hg| d mu > M_q(g) – epsilon$. Note, however, that $latex h$ is not necessarily simple. But there is a sequence of simple functions $latex h_n rightarrow h$ such that $latex |h_n| leq |h|$. By the dominated convergence theorem, $latex lim |int h_n g d mu| rightarrow |int hg d mu | > M_q(g) – epsilon$. This shows established the required inequality.

Proof of Theorem 3: We already know that the map $latex Phi: L^q rightarrow (L^p)^*$ given by $latex Phi(g) = phi_g$ is isometric (here as above $latex phi_g$ is the functional $latex phi_g(f) = int fg d mu$). It remains to show that this map is surjective. Recall that we are assuming that $latex p<infty$; it is your job to find where we use this assumption.

Assume first that $latex mu$ is a finite measure. Let $latex phi in (L^p)^*$. We seek a function $latex g in L^q$ such that $latex phi = phi_g$. To obtain this function, define a function $latex nu : mathcal{M} rightarrow mathbb{C}$ by

$latex nu(E) = phi(chi_E) $

for $latex E in mathcal{M}$. By finiteness of $latex mu$, the function $latex chi_E$ is in $latex L^p$ so $latex nu$ is well defined. In fact, it turns out that $latex nu$ is a complex valued measure. Ineed, if $latex E$ is the dsijoint union $latex cup E_n$, then $latex chi_E = sum chi_{E_n}$ pointwise and hence (by monotone convergence) in norm, whence

$latex nu(E) = phi(sum chi_{E_n}) =sum phi(chi_{E_n}) = sum nu(E_n).$

Now if $latex E$ has $latex mu$-measure zero, then $latex chi_E = 0$ in $latex L^p$, hence $latex nu(E) = phi(chi_E) = 0$. Thus $latex nu << mu$, so by the Radon-Nikodym Theorem, it follows that there is some $latex g in L^1$ such that $latex nu(E) = int_E g dmu$ for all $latex E in mathcal{M}$. From this we have that

(*)$latex phi(f) = int fg dmu$

for all $latex f in mathcal{S}$. The lemma above implies that $latex g in L^q$ and that $latex |g|_q leq |phi|$. Examining (*) again, we obtain by density of $latex mathcal{S}$ that $latex phi = phi_g$.

Finally, it remains to prove the theorem for a $latex sigma$-finite measure. As the heart of the proof is behind us, this is left as an exercise for the reader.

2. Duality and the double dual

Corollary 14 in the previous lecture is a generalization of the fact that in an inner product space, if a vector $latex x$ is orthogonal to all other vectors in the space then $latex x = 0$. This suggests the following notation. In some texts, when considering the action of $latex f in X^*$ on $latex x in X$, the notation $latex langle f,xrangle $ is used instead of $latex f(x)$. This invites one to think of the relationship between $latex X$ and $latex X^*$ as something more symmetrical. Sometimes, elements of the dual space $latex X^*$ are denoted as $latex x^*$. Let’s use this notation for a little while to make it familiar.

One can learn many things about a Banach space $latex X$ from its dual $latex X^*$. For example:

Exercise A: If $latex X^*$ is separable, then $latex X$ is separable, too.

Example: The example of $latex ell^infty = (ell^1)^*$ shows that the converse is false. In addition, the exercise shows that $latex ell^1 neq (ell^infty)^*$.

Let $latex S subseteq X$. We denote $latex S^perp : = {x^* in X^* : forall x in S . langle x^*, x rangle = 0 }$. If $latex T subseteq X^*$ then we denote $latex ^perp T = {x in X : forall x^* in T . langle x^*, x rangle = 0 }$. The spaces $latex S^perp$ and $latex ^perp T$ are called the annihilator of $latex S$ and the pre-annihilator of $latex T$, respectively. With this notation we can state Corollary 16 from the previous lecture as follows: A subspace $latex M$ contains $latex x$ in its closure if and only if $latex x in ^perp( M^perp)$. In other words, $latex overline{M} = ^perp (M^perp)$. 

Exercise B: True or false: $latex (^perp N )^perp = overline{N}$?

Every $latex x in X$ induces a function $latex hat{x} in X^{**} = (X^*)^*$ by way of

$latex hat{x}(x^*) = langle x^*, x rangle ,, , ,, x^* in X^*.$

For every normed space $latex Y$, we denote by $latex Y_1$ the (norm) closed unit ball of $latex Y$.

Proposition 4: The map $latex x mapsto hat{x}$ is an isometry from $latex X$ into $latex X^{**}$. 

Proof: Linearity is trivial. To see that the map is norm preserving,

$latex |hat{x}| =sup_{x^* in X^*_1} langle x^*,xrangle = |x|,$

where for the second equality we used Corollary 13 from the last lecture.

Definition 5: A Banach space $latex X$ is said to be reflexive if the map $latex x mapsto hat{x}$ is an isometry of $latex X$ onto $latex X^{**}$. 

Example: $latex L^p$ is reflexive for all $latex p in (1,infty)$. $latex L^1[0,1], ell^1, L^infty[0,1]$ and $latex ell^infty$ are examples of non-reflexive spaces (as Exercise A and the example following it show). It follows non of these spaces is isomorphic to an $latex L^p$ space for $latex 1<p<infty$.

3. Quotient spaces

Let $latex X$ be a normed space, and let $latex M subseteq X$ be a closed subspace. In linear algebra we learn how to form the quotient space $latex X/M$, which is defined to be the set of all cosets $latex x + M$, $latex x in X$, with the operations

$latex (x + M) + (y + M) = x+y+M .$

$latex c(x+M) = cx + M . $

It is known and also easy to show that these operations are well defined, and give $latex X/M$ the structure of a vector space. Our goal now is to make $latex X/M$ into a normed space, and to prove that when $latex X$ is complete, so is $latex M$.

To abbreviate, let us write $latex dot{x}$ for $latex x + M$. We define

$latex |dot{x}| = d(x,M) = inf {|x-m|: m in M}.$

Let us denote by $latex pi$ the quotient map $latex X rightarrow X/M$.

Theorem 6: With the above defined norm, the following hold:

1. $latex X/M$ is a normed space.
2. $latex pi$ is a contraction: $latex |pi(x) | leq |x|$ for all $latex x$.
3. For every $latex y in X/M$ such that $latex |y|<1$,  there is an $latex x in X$ with $latex |x|<1$ such that $latex pi(x) = y$.
4. $latex pi(U)$ is open if and only if $latex U$ is open.
5. If $latex F$ is a closed subspace containing $latex M$, then $latex pi(F)$ is closed.
6. If $latex X$ is complete, then so is $latex X/M$. 

Proof: 1. $latex |cdot{x}| = |c||dot{x}|$ is trivial. $latex |dot{x}| = 0 Leftrightarrow dot{x} = 0$ follows from the fact that $latex M$ is closed. For every $latex x,y in X$, let $latex m,n in M$ be such that $latex |x-m|$ and $latex |y-n|$ are very close to $latex |dot{x}|$ and $latex |dot{y}|$, respectively. Then

$latex |dot{x} + dot{y}| leq |x+y – m -n | leq |x-m| + |x-n| ,$

and the right hand side is very close to $latex |dot{x}| + |dot{y}|$, so $latex |dot{x} + dot{y}|$ can only be very slightly bigger than $latex |dot{x}| + |dot{y}|$. That proves the triangle inequality.

Exercise C: Complete the proof of the theorem.

Remark: Note that by 4 of the above theorem, the topology induced by the quotient norm is precisely the quotient topology which one defines in topology.

Theorem 7: Let $latex X$ be a normed space and $latex M subseteq X$ a closed subspace. Then

1. $latex M^*$ is isometrically isomorphic to $latex X^*/M^perp$. 
2. $latex (X/M)^*$ is isometrically isomorphic to $latex M^perp$. 

Proof: 1. Define $latex T : M^* rightarrow X/M^perp$ be

$latex T(m^*) = x^* + M^perp ,$

where $latex x^*$ is any extension of $latex m^*$ to $latex X$. This is well defined, because if $latex y^*$ is another extension of $latex m^*$ then $latex x^* – y^* in M^perp$. Well defined-ness implies linearity (think about it). $latex T$ is also surjective, sine $latex T(x^*big|_{M}) = x^* + M^perp$ for all $latex x^* in X^*$. It remains to prove that $latex T$ is isometric.

Let $latex m^* in M^*$ and let $latex x^*$ be an extension. Then $latex |T(m^*)| = inf {|x^*+n^*|:n^* in M^perp}$. But each functional $latex x^*+n^*$ extends $latex m^*$, so $latex |x^*+n^*| geq |m^*|$, whence $latex |T(m^*)| geq |m^*|$. On the other hand, as $latex n^*$ ranges over all $latex n^* in M^perp$, $latex x^* + n^*$ ranges over all extensions of $latex m^*$. By Hahn-Banach, there is some extension, say $latex y^* = x^* + n_1^*$, such that $latex |y^*| = |m^*|$. Thus the infimum is attained and $latex |T m^*| = |m^*|$.

2. Define a linear map $latex pi^*: (X/M)^* rightarrow X^*$ by

$latex pi^*(y^*) = y^* circ pi .$

It is obvious that $latex R(pi^*) subseteq M^perp$, since $latex pi$ vanishes on $latex M$. We need to show that $latex pi^*$ is a surjective isometry.

By 2 of Theorem 6, $latex |pi^*(y^*)|= |y^*circ pi| leq |y^*||pi| leq |y^*|$. On the other hand,

$latex |pi^*(y^*)| = sup {|langle y^* circ pi, xrangle|: |x|<1} = sup{|langle y^* , pi(x)rangle|: |x|<1} .$

Using 3 of Theorem 6, $latex pi$ maps the unit ball of $latex X$ onto the unit ball of $latex X/M$, so the right hand side is equal to

$latex sup{|langle y^* , yrangle|: |y|<1} = |y^*|.$

It remains to show that the range of $latex pi^*$ is equal to $latex M^perp$. Let $latex x^* in M^perp$. Then $latex N(x^*) supseteq M$. It follows that if $latex x_1 – x_2 in M$, then $latex langle x^*, x_1 rangle = langle x^*, x_2 rangle$, so we may define a functional $latex f$ on $latex X/M$ by

$latex f(dot{x}) = x^*(x).$

By definition $latex x^* = f circ pi$. The kernel of $latex f$ is equal to $latex pi(N(x^*))$, and is closed by Theorem 6. By Exercise F in Notes 6, $latex f$ is continuous. That completes the proof.

4. The adjoint of an operator

The idea of the proof in 2 of Theorem 7 — the construction of the map $latex pi^*$ — is one of general applicability.

Let $latex T in B(X,Y)$. Then we define $latex T^* : Y* rightarrow X^*$ by way of

$latex T^* y^* = y^* circ T .$

Theorem 8: Let $latex X$ and $latex Y$ be normed spaces, and let $latex T in B(X,Y)$. Then the above defined $latex T^*$ satisfies:

1. $latex T^* in B(Y^*,X^*)$, so $latex T^*$ is linear and bounded. 
2. $latex T^*$ is satisfies $latex langle Tx, y^* rangle = langle x, T^* y^* rangle$ for all $latex x in X, y^* in Y^*$. Moreover, $latex T^*$ is the unique function from $latex Y^*$ to $latex X^*$ with this property. 
3. $latex |T^*| = |T|$.

Exercise D: Prove Theorem 8.

Theorem 9: Let $latex X$ and $latex Y$ be normed spaces, and let $latex T in B(X,Y)$. Then $latex N(T^*) = R(T)^perp$ and $latex N(T) = ^perp R(T^*)$. 

Proof: Thanks to the notation, the proof is exactly the same as in the Hilbert space case (note, though, that both assertions require proof, and do not follow one from the other by conjugation).

Corollary 10: For $latex T in B(X,Y)$, $latex ^perp N(T^*) = overline{R(T)}$. 

Proof: This follows from Corollary 16 in the previous lecture.

Exercise E: Give an example of an operator for which $latex N(T)^perp neq overline{R(T^*)}$.