Functional Analysis/Banach spaces

←Chapter 1: Preliminaries

Functional Analysis Chapter 2: Banach spaces

Chapter 3: Hilbert spaces→

Let $𝒳$ be a linear space. A norm is a real-valued function $f$ on $𝒳$, with the notation $\Vert \cdot \Vert =f(\cdot )$, such that

• (i) $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ (w:triangular inequality)
• (ii) $\Vert \lambda x\Vert =|\lambda |\Vert x\Vert$ for any scalar $\lambda$
• (iii) $\Vert x\Vert =0$ implies $x=0$.

(ii) implies that $\Vert 0\Vert =0$. This and (i) then implies $0=\Vert x-x\Vert \le \Vert x\Vert +\Vert -x\Vert =2\Vert x\Vert$ for all $x$; that is, norms are always non-negative. A linear space with a norm is called a normed space. With the metric $d(x,y)=\Vert x-y\Vert$ a normed space is a metric space. Note that (i) implies that:

$\Vert x\Vert \le \Vert x-y\Vert +\Vert y\Vert$ and $\Vert y\Vert \le \Vert x-y\Vert +\Vert x\Vert$

and so: $|\Vert x\Vert -\Vert y\Vert |\le \Vert x-y\Vert$. (So, the map $x\mapsto \Vert x\Vert$ is continuous; in fact, 1-Lipschitz continuous.)

A complete normed space is called a Banach space. While there is seemingly no prototypical example of a Banach space, we still give one example of a Banach space: $𝒞(K)$, the space of all continuous functions on a compact space $K$, can be identified with a Banach space by introducing the norm:

$\Vert \cdot \Vert =\underset{K}{sup}|\cdot |$

It is a routine exercise to check that this is indeed a norm. The completeness holds since, from real analysis, we know that a uniform limit of a sequence of continuous functions is continuous. In concrete spaces like this one, one can directly show the completeness. More often than that, however, we will see that the completeness is a necessary condition for some results (especially, reflexivity), and thus the space has to be complete. The matter will be picked up in the later chapters.

Another example of Banach spaces, which is more historical, is an $l_p$ space; that is, the space of convergent series. (The geometric properties of $l_p$ spaces will be investigated in Chapter 4.) It is clear that $l_p$ is a linear space, since the sum of two p-convergent series is again p-convergent. That the $l_p$ norm is in fact a norm follows from

Lemma 2.1.

If $\Vert x{\Vert }_p<\mathrm{\infty }$, then

$\Vert x{\Vert }_p=\underset{\Vert y{\Vert }_q=1}{sup}|\sum x_j\overline{y_j}|$, where $1/p+1/q=1$

Proof. By Hölder's inequality,

$\underset{\Vert y{\Vert }_q=1}{sup}|\sum x_j\overline{y_j}|\le \Vert x{\Vert }_p$

Conversely, if $\Vert x{\Vert }_p=1$, then taking $y_j=\overline{x_j}|x_j{|}^{p-2}$ we have:

$\sum x_j\overline{y_j}=\sum |x_j{|}^p=1$, while $\Vert y{\Vert }_q=(\sum |x_j{|}^{(p-1)q}{)}^{1/q}=1$.

since $p=(p-1)q$. More generally, if $\Vert x{\Vert }_p\ne 0$, then

$\frac{x}{\Vert x{\Vert }_p}=\underset{\Vert y{\Vert }_q=1}{sup}|\sum x_j\overline{y_j}|\frac{1}{\Vert x{\Vert }_p}$.

Since the identity is obvious when $x=0$, the proof is complete. $◻$

Now, it remains to show that an $l_p$ space is complete. For that, let $x_k\in l_p$ be a Cauchy sequence. This means explicitly that

$\sum _{n=1}^{\mathrm{\infty }}|(x_k{)}_n-(x_j{)}_n{|}^2\to 0$ as $n,m\to \mathrm{\infty }$

For each $n$, by completeness, $\underset{k\to \mathrm{\infty }}{lim}(x_k{)}_n$ exists and we denote it by $y_n$. Let $ϵ>0$ be given. Since $x_k$ is Cauchy, there is $N$ such that

$\sum _{n=1}^{\mathrm{\infty }}|(x_k{)}_n-(x_j{)}_n{|}^2<ϵ$ for $k,j>N$

Then, for any $k>N$,

$\sum _{n=1}^{\mathrm{\infty }}|(x_k{)}_n-y_n{|}^2=\underset{m\ge 1}{sup}\sum _{n=1}^m|(x_k{)}_n-y_n{|}^2=\underset{m\ge 1}{sup}\underset{j\to \mathrm{\infty }}{lim}\sum _{n=1}^m|(x_k{)}_n-(x_j{)}_n{|}^2<ϵ$

Hence, $x_k\to y$ with $y=\sum _{n=1}^{\mathrm{\infty }}{y}_n$. $y$ is in fact in $l_p$ since $\Vert y{\Vert }_2\le \Vert y-x_n{\Vert }_2+\Vert x_n{\Vert }_2<\mathrm{\infty }$. (We stress the fact that the completeness of $l_p$ spaces come from the fact that the field of complex numbers is complete; in other words, $l_p$ spaces may fail to be complete if the base field is not complete.) $l_p$ is also separable; i.e., it has a countable dense subset. This follows from the fact that $l_p$ can be written as a union of subspaces with dimensions 1, 2, ..., which are separable. (TODO: need more details.)

We define the operator norm of a continuous linear operator $f$ between normed spaces $𝒳$ and $𝒴$, denoted by $\Vert f\Vert$, by

$\Vert f\Vert =\underset{\Vert x{\Vert }_{𝒳}\le 1}{sup}\Vert f(x){\Vert }_{𝒴}$

2 Theorem Let $T$ be a linear operator from a normed space $𝒳$ to a normed space $𝒴$.

• (i) $T$ is continuous if and only if there is a constant $C>0$ such that $\Vert Tx\Vert \le C\Vert x\Vert$ for all $x\in X$
• (ii) $\Vert f\Vert =inf\{$ any $C$ as in (i) $\}=\underset{\Vert x\Vert =1}{sup}\Vert f(x)\Vert$ if $𝒳$ has nonzero element. (Recall that the inf of the empty set is $\mathrm{\infty }$.)

Proof: If $\Vert T\Vert <\mathrm{\infty }$, then

$\Vert T(x_n-x){\Vert }_{𝒴}\le \Vert T\Vert \Vert x_n-x{\Vert }_{𝒳}\to 0$

as $x_n\to x$. Hence, $T$ is continuous. Conversely, suppose $\Vert T\Vert =\mathrm{\infty }$. Then we can find $x_n\in 𝒳$ with $\Vert x_n{\Vert }_{𝒳}\le 1$ and $\Vert T{x}_n\Vert \ge n$. Then $\frac{x_n}{n}\to 0$ while $\Vert T\left(\frac{x_n}{n}\right)\Vert \to ̸0$. Hence, $T$ is not continuous. The proof of (i) is complete. For (ii), see w:operator norm for now. (TODO: write an actual proof). $◻$

It is clear that an addition and a scalar multiplication are both continuous. (Use a sequence to check this.) Since the inverse of an addition is again addition, an addition is also an open mapping. Ditto to nonzero-scalar multiplications. In other words, translations and dilations of open (resp. closed) sets are again open (resp. closed).

Not all linear operators are continuous. Take the linear operator defined by $D(P)=X{P}^{\prime }$ on the normed vector space of polynomials $ℝ[X]$ with the suprenum norm $\Vert P{\Vert }_{\mathrm{\infty }}=\underset{x\in [0,1]}{sup}|P(x)|$ ; since $D(X^n)=n{X}^n$, the unit ball is not bounded and hence this linear operator is not continuous.

Notice that the kernel of this non continuous linear operator is closed: $\ker D=\{0\}$. However, when a linear operator is of finite rank, the closeness of the kernel is in fact synonymous to continuity. To see this, we start with the special case of linear forms.

2 Theorem A (non null) linear form is continuous iff it's kernel is closed.

$T$ continuous $\iff \ker T=\overline{\ker T}$

Proof: If the linear form $T$ on a normed vector space $X$ is continuous, then it's kernel is closed since it's the continuous inverse image of the closed set $\{0\}$.

Conversely, suppose a linear form $T:X\to ℝ$ is not continuous. then by the previous theorem,

$\mathrm{\forall }c>0,\mathrm{\exists }x(c)s.t.|Tx(c)|⩾c\Vert x(c)\Vert$ so in particular, one can define a sequence $\{x_n\}$ such that $|T{x}_n|⩾n\Vert x_n\Vert >0$. Then denote:

$u_n:=\frac{x_n}{|x_n|}$, one has defined a unit normed sequence ($|u_n|=1$) s.t. $|T{u}_n|⩾n$. Furthermore, denote

$v_n:=\frac{u_n}{|T{u}_n|}$. Since $\frac{|u_n|}{|T{u}_n|}⩽\frac{1}{n}$, one can define a sequence that converges $\{v_n\}\to 0$ whilst $|T{v}_n|=1$.

Now, since $\ker T\ne X$, then there exists $a$ such that $Ta\ne 0$. Then the sequence of general term converges

$\underset{\in \ker T}{\underbrace{a-v_{n}Ta}}\to a\notin \ker T$ and hence $\ker T$ is not closed. $◻$

Furthermore, if the linear form is continuous and the kernel is dense, then $\ker T=\overline{\ker T}=X\Rightarrow f=0$, hence a continuous & non null linear has a non dense kernel, and hence a linear form with a dense kernel is whether null or non continuous so a non null continuous linear form with a dense kernel is not continuous, and a linear form with a dense kernel is not continuous.

2 Corollary' " A non null linear form on a normed vector space is not continuous iff it's kernel is dense.

$\overline{\ker T}=X\iff T$ is not continuous"

More generally, we have: 2 Theorem " A non null linear operator of finite rank between normed vector spaces. then closeness of the null space is equivalent to continuity."
Proof: It remains to show that continuity implies closeness of the kernel. Suppose $T:X\to Y$ is not continuous. Denote $r:=\dim \text{Im }T$;

2 Lemma If $T:X\to Y$ is a linear operator between normed vector spaces, then $T$ is of finite rank $r$ iff there exists $r$ independent linear forms $(f_1,\dots ,f_r)$and $r$ independent vectors $(a_1,\dots ,a_r)$ such that $Tx=a_1{f}_1(x)+\dots +a_r{f}_r(x)$"
Proof: take a basis $(a_1,\dots ,a_r)$ of $\text{Im }T$, then from $Tx=\sum _{i=1}^r{a}_i{y}_i$, one can define $r$ mappings $f_i(x)=y_i$. Unicity and linearity of $T$ implies linearity of the $f_i$'s. Furthermore, the family $(f_1,\dots ,f_r)$ of linear forms of $X^{\ast }$ is linearly independent: suppose not, then there exist a non zero family $({\alpha }_1,\dots ,{\alpha }_r)$ such that e.g. $f_1=\sum _{i=2}^r{\alpha }_i{f}_i$ so

$Tx=\sum _{i=1}^r{f}_i(x)a_i=(\sum _{i=2}^r{\alpha }_i{f}_i(x))a_1+\sum _{i=2}^r{f}_i(x)a_i=({\alpha }_2{a}_1+a_2)f_2(x)+\dots +({\alpha }_n{a}_1+a_r)f_r(x)$ and the family $({\alpha }_i{a}_1+a_i{)}_{i=2,\dots ,r}$ spans $\text{Im }T$, so $\dim \text{Im }T=r-1$ which is a contradiction. Finally, one has a unique decomposition of a finite rank linear operator:

$Tx=a_1{f}_1(x)+\dots a_r{f}_r(x)$ with $f_i\in E^{\ast }$ $◻$

Take $x\in \ker T\Rightarrow x\in \bigcap _{i=1}^r\ker f_i\subset \ker f_i$. Then there exists a vector subspace $H_i$ such that $\ker f_i=\ker T\oplus H_i$. Denote $T_i:H_i\to \text{Im}T$ the restriction of $T$ to $H_i$. Since $ker{T}_i=\{x\in H_i:T_i(x)\}=0=\ker T\cap H_i=\{0\}$, the linear operator $T_i$ is injective so $\text{Im}{T}_i\subset \text{Im}T$ and $H_i$ is of finite dimension, and this for all $i=1,\dots ,r$.

By hypothesis $\ker T$ is closed. Since the sum of this closed subspace and a subspace of finite dimension ($H_i$) is closed (see lemma bellow), it follows that the kernel of each $r$ linear forms $\ker f_i$ is closed, so the $f_i$'s are all continuous by the first case and hence $T$ is continuous. $◻$

2 Lemma The sum of subspace of finite dimension with a closed subspace is closed.
Proof: by induction on the dimension.

Case $n=1$. Let's show that $H:=F+𝕂a$ is closed when $F$ is closed (where $𝕂$ is a complete field). Any $x\in H$ can be uniquely written as $x=y+\lambda a$ with $y\in F$. There exists a linear form $L$ s.t. $x=y+L(x)a$. Since $L$ is closed in $(X,\Vert \cdot \Vert )$ so in $(H,\Vert \cdot \Vert )$, then $f$ is continuous by the first case. Take a convergente sequence $x_n\to x\in E$ of $H$. He have $x_n=y_n+L(x_n)a$ with $y_n\in F$. Since the sequence $x_n$ is convergente, then it si Cauchy, so it's continuous image $L(x_n)$ is also Cauchy. Since $ℝ$ is complete, then $L(x_n)\to \lambda$. Finally, the sequence $y_n$ converges to $x-\lambda a$. Since $F$ is closed, then $x-\lambda a\in F$ and $x\in H$ so $H$ is closed.

Suppose the result holds for all subspaces of dimension $⩽p$. Let $G$ be a subspace of dimension $p+1$. Let $(a_1,\dots ,a_{p+1})$ be a basis of $G$. Then $H:=F+\underset{i=1}{\overset{p}{⨁}}𝕂a_i+𝕂a_{p+1}$ and concludes easily.$◻$

2 Corollary Any linear operator on a normed vector space of finite dimension onto a normed vector space is continuous.
Proof: Since $X$ is of finite dimension, then any linear operator is of finite rank. Then as $\dim \ker T+\dim \text{Im}T=\dim X$ holds, it comes that the null space is of finite dimension, so is closed (any $𝕂$ vector subspace of finite dimension $n$ is isomorphic to ${𝕂}^n$ (where $𝕂$ is a complete field), so the subspace is complete and closed). Then one applies the previous theorem.$◻$

2 Lemma (Riesz) A normed space $𝒳$ is finite-dimensional if and only if its closed unit ball is compact.
Proof: Let $T:{𝐂}^n\to X$ be a linear vector space isomorphism. Since $T$ has closed kernel, arguing as in the proof of the preceding theorem, we see that $T$ is continuous. By the same reasoning $T^{-1}$ is continuous. It follows:

$\{x\in 𝒳|\Vert x\Vert \le 1\}\subset T\{y\in {𝐂}^n|\Vert y\Vert \le \Vert T^{-1}\Vert \}$

In the above, the left-hand side is closed, and the right-hand is a continuous image of a closed ball, which is compact. Hence, the closed unit ball is a subset of a compact set and thus compact. Now, the converse. If $𝒳$ is not finite dimensional, we can construct a sequence $x_j$ such that:

$1=\Vert x_j\Vert \le \Vert x_j-\sum _{k=1}^{j-1}{a}_k{x}_k\Vert$ for any sequence of scalars $a_k$.

Thus, in particular, $\Vert x_j-x_k\Vert \ge 1$ for all $j,k$. (For the details of this argument, see : w:Riesz's lemma for now) $◻$

2 Corollary Every finite-dimensional normed space is a Banach space.
Proof: Let $x_n$ be a Cauchy sequence. Since it is bounded, it is contained in some closed ball, which is compact. $x_n$ thus has a convergent subsequence and so $x_n$ itself converges. $◻$

2 Theorem A normed space $𝒳$ is finite-dimensional if and only if every linear operator $T$ defined on $𝒳$ is continuous.
Proof: Identifying the range of $T$ with ${𝐂}^n$, we can write:

$Tx=(f_1(x),f_2(x),\mathrm{...}{f}_n(x))$

where $f_1,\mathrm{...}{f}_n$ are linear functionals. The dimensions of the kernels of $f_j$ are finite. Thus, $f_j$ all have complete and thus closed kernels. Hence, they are continuous and so $T$ is continuous. For the converse, we need Axiom of Choice. (TODO: complete the proof.) $◻$

The graph of any function $f$ on a set $E$ is the set $\{(x,f(x))|x\in E\}$. A continuous function between metric spaces has closed graph. In fact, suppose $(x_j,f(x_j))\to (x,y)$. By continuity, $f(x_j)\to f(x)$; in other words, $y=f(x)$ and so $(x,y)$ is in the graph of $f$. It follows (in the next theorem) that a continuous linear operator with closed graph has closed domain. (Note the continuity here is a key; we will shortly study a linear operator that has closed graph but has non-closed domain.)

2 Theorem Let $T:𝒳\to 𝒴$ be a continuous densely defined linear operator between Banach spaces. Then its domain is closed; i.e., $T$ is actually defined everywhere.
Proof: Suppose $f_j\to f$ and $T{f}_j$ is defined for every $j$; i.e., the sequence $f_j$ is in the domain of $T$. Since

$\Vert T{f}_j-T{f}_k\Vert \le \Vert T\Vert \Vert f_j-f_k\Vert \to 0$,

$T{f}_j$ is Cauchy. It follows that $(f_j,T{f}_j)$ is Cauchy and, by completeness, has limit $(g,Tg)$ since the graph of T is closed. Since $f=g$, $Tf$ is defined; i.e., $f$ is in the domain of $T$. $◻$

The theorem is frequently useful in application. Suppose we wish to prove some linear formula. We first show it holds for a function with compact support and of varying smoothness, which is usually easy to do because the function vanishes on the boundary, where much of complications reside. Because of th linear nature in the formula, the theorem then tells that the formula is true for the space where the above functions are dense.

We shall now turn our attention to the consequences of the fact that a complete metric space is a Baire space. They tend to be more significant than results obtained by directly appealing to the completeness. Note that not every normed space that is a Baire space is a Banach space.

2 Theorem (open mapping theorem) Let $𝒳,𝒴$ be Banach spaces. If $T:𝒳\to 𝒴$ is a continuous linear surjection, then it is an open mapping; i.e., it maps open sets to open sets.
Proof: Let $B(r)=\{x\in 𝒳;\Vert x\Vert <r\}$. Since $T$ is surjective, ${\cup }_{n=1}^{\mathrm{\infty }}T(B(n))=T({\cup }_{n=1}^{\mathrm{\infty }}B(n))=T(X)=Y$. Then by Baire's Theorem, some $B(k)$ contains an interior point; thus, it is a neighborhood of $0$. $◻$

2 Corollary If $(𝒳,\Vert \cdot {\Vert }_1)$ and $(𝒳,\Vert \cdot {\Vert }_2)$ are Banach spaces, then the norms $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent; i.e., each norm is dominated by the other.
Proof: Let $I:(𝒳,\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2)\to (𝒳,\Vert \cdot {\Vert }_1)$ be the identity map. Then we have:

$\Vert I\cdot {\Vert }_1=\Vert \cdot {\Vert }_1\le (\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2)$.

This is to say, $I$ is continuous. Since Cauchy sequences apparently converge in the norm $\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2$, the open mapping theorem says that the inverse of $I$ is also continuous, which means explicitly:

$\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2=\Vert I^{-1}\cdot {\Vert }_1+\Vert I^{-1}\cdot {\Vert }_2\le \Vert I^{-1}\Vert \Vert \cdot {\Vert }_1$.

By the same argument we can show that $\Vert \cdot {\Vert }_1+\Vert \cdot {\Vert }_2$ is dominated by $\Vert \cdot {\Vert }_2$ $◻$

2 Corollary Let $(𝒳,\Vert \cdot {\Vert }_{𝒳})$ be a Banach space with dimension $n$. Then the norm $\Vert \cdot {\Vert }_{𝒳}$ is equivalent to the standard Euclidean norm:

$|(x_1,\mathrm{...}{x}_n){|}^2=\sum _j|x_j{|}^2$

2 Corollary If $T$ is a continuous linear operator between Banach spaces with closed range, then there exists a $K>0$ such that if $y\in \mathrm{im}(T)$ then $\Vert x\Vert \le K\Vert y\Vert$ for some $x$ with $Tx=y$.
Proof: This is immediate once we have the notion of a quotient map, which we now define as follows.

Let $ℳ$ be a closed subspace of a normed space $𝒳$. The quotient space $𝒳/ℳ$ is a normed space with norm:

$\Vert \pi (x)\Vert =inf\{\Vert x+m\Vert ;m\in ℳ\}$

where $\pi :𝒳\to 𝒳/ℳ$ is a canonical projection. That $\Vert \cdot \Vert$ is a norm is obvious except for the triangular inequality. But since

$\Vert \pi (x+y)\Vert \le \Vert x+m\Vert +\Vert y+n\Vert$

for all $m,n\in ℳ$. Taking inf over $m,n$ separately we get:

$\Vert \pi (x+y)\Vert \le \Vert \pi (x)\Vert +\Vert \pi (y)\Vert$

Suppose, further, that $𝒳$ is also a commutative algebra and $ℳ$ is an ideal. Then $𝒳/ℳ$ becomes a quotient algebra. In fact, as above, we have:

$\Vert \pi (x)\pi (y)\Vert =\Vert \pi ((x+m)(y+n))\Vert \le \Vert x+m\Vert \Vert y+n\Vert$,

for all $m,n\in ℳ$ since $\pi$ is a homomorphism. Taking inf completes the proof.

So, the only nontrivial question is the completeness. It turns out that $𝒳/ℳ$ is a Banach space (or algebra) if $𝒳$ is Banach space (or algebra). In fact, suppose

$\sum _{n=1}^{\mathrm{\infty }}\Vert \pi (x_n)\Vert <\mathrm{\infty }$

Then we can find a sequence $y_n\in ℳ$ such that

$\sum _{n=1}^{\mathrm{\infty }}\Vert x_n+y_n\Vert <\mathrm{\infty }$

By completeness, $\sum _{n=1}^{\mathrm{\infty }}{x}_n+y_n$ converges, and since $\pi$ is continuous, $\sum _{n=1}^{\mathrm{\infty }}\pi (x_n)$ converges then. The completeness now follows from:

2 Lemma Let $𝒳$ be a normed space. Then $𝒳$ is complete (thus a Banach space) if and only if

$\sum _{n=1}^{\mathrm{\infty }}\Vert x_n\Vert <\mathrm{\infty }$ implies $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ converges.

Proof: ($\Rightarrow$) We have:

$\Vert \sum _{n=k}^{k+m}{x}_n\Vert \le \sum _{n=k}^{k+m}\Vert x_n\Vert$.

By hypothesis, the right-hand side goes to 0 as $n,m\to \mathrm{\infty }$. By completeness, $\sum _{n=1}^{\mathrm{\infty }}{x}_n$ converges. Conversely, suppose $x_j$ is a Cauchy sequence. Thus, for each $j=1,2,\mathrm{...}$, there exists an index $k_j$ such that $\Vert x_n-x_m\Vert <2^{-j}$ for any $n,m\ge k_j$. Let $x_{k_0}=0$. Then $\sum _{j=0}^{\mathrm{\infty }}\Vert x_{k_{j+1}}-x_{k_j}\Vert <2$. Hence, by assumption we can get the limit $x=\sum _{j=0}^{\mathrm{\infty }}{x}_{k_{j+1}}-x_{k_j}$, and since

$\Vert x_{n_k}-x\Vert =\Vert \sum _{j=1}^n{x}_{k_{j+1}}-x_{k_j}-x\Vert \to 0$ as $n\to \mathrm{\infty }$,

we conclude that $x_j$ has a subsequence converging to $x$; thus, it converges to $x$. $◻$

The next result is arguably the most important theorem in the theory of Banach spaces. (At least, it is used the most frequently in application.)

2 Theorem (closed graph theorem) Let $𝒳,𝒴$ be Banach spaces, and $T:𝒳\to 𝒴$ a linear operator. The following are equivalent.

• (i) $T$ is continuous.
• (ii) If $x_j\to 0$ and $T{x}_j$ is convergent, then $T{x}_j\to 0$.
• (iii) The graph of $T$ is closed.

Proof: That (i) implies (ii) is clear. To show (iii), suppose $(x_j,T{x}_j)$ is convergent in $X$. Then $x_j$ converges to some $x_0$ or $x_j-x_0\to 0$, and $T{x}_j-Tx$ is convergent. Thus, if (ii) holds, $T(x_j-x)\to 0$. Finally, to prove (iii) $\Rightarrow$ (i), we note that Corollary 2.something gives the inequality:

$\Vert \cdot \Vert +\Vert T\cdot \Vert \le K\Vert \cdot \Vert$

since by hypothesis the norm in the left-hand side is complete. Hence, if $x_j\to x$, then $T{x}_j\to Tx$. $◻$

Note that when the domain of a linear operator is not a Banach space (e.g., just dense in a Banach space), the condition (ii) is not sufficient for the graph of the operator to be closed. (It is not hard to find an example of this in other fields, but the reader might want to construct one himself as an exercise.)

Finally, note that an injective linear operator has closed graph if and only if its inverse is closed, since the map $(x_1,x_2)\mapsto x_2,x_1$ sends closed sets to closed sets.

2 Theorem Let $({𝒳}_j,\Vert \cdot {\Vert }_j)$ be Banach spaces. Let $T:{𝒳}_1\to {𝒳}_2$ be a closed densely defined operator and $S$ be a linear operator with $\mathrm{dom}(T)\subset \mathrm{dom}(S)$. If there are constants $a,b$ such that (i) $0\le a<1$ and $b>0$ and (ii) $\Vert Su\Vert \le a\Vert Tu\Vert +b\Vert u\Vert$ for every $u\in \mathrm{dom}(T)$, then $T+S$ is closed.
Proof: Suppose $\Vert u_j-u{\Vert }_1+\Vert (T+S)u_j-f{\Vert }_2\to 0$. Then

$\Vert T(u_j-u_k)\Vert \le \Vert (T+S)(u_j-u_k)\Vert +a\Vert T(u_j-u_k)\Vert +b\Vert u_j-u_k\Vert$

Thus,

$(1-a)\Vert T(u_j-u_k)\Vert \le \Vert (T+S)(u_j-u_k)\Vert +b\Vert u_j-u_k\Vert$

By hypothesis, the right-hand side goes to $0$ as $j,k\to \mathrm{\infty }$. Since $T$ is closed, $(u_j,T{u}_j)$ converges to $(u,Tu)$. $◻$

In particular, with $a=0$, the hypothesis of the theorem is fulfilled, if $S$ is continuous.

When $𝒳,𝒴$ are normed spaces, by $L(𝒳,𝒴)$ we denote the space of all continuous linear operators from $𝒳$ to $𝒴$.

2 Theorem If $𝒴$ is complete, then every Cauchy sequence $T_n$ in $L(𝒳,𝒴)$ converges to a limit $T$ and $\Vert T\Vert =\underset{n\to \mathrm{\infty }}{lim}\Vert T_n\Vert$. Conversely, if $L(𝒳,𝒴)$ is complete, then so is Y.
Proof: Let $T_n$ be a Cauchy sequence in operator norm. For each $x\in 𝒳$, since

$\Vert T_n(x)-T_m(x)\Vert \le \Vert T_n-T_m\Vert \Vert x\Vert$

and $𝒴$ is complete, there is a limit $y$ to which $T_n(x)$ converges. Define $T(x)=y$. $T$ is linear since the limit operations are linear. It is also continuous since $\Vert T(x)\Vert \le \underset{n}{sup}\Vert T_n\Vert \Vert x\Vert$. Finally, $\underset{n\to \mathrm{\infty }}{lim}\Vert T_n-T\Vert =\underset{\Vert x\Vert \le 1}{sup}\Vert \underset{n\to \mathrm{\infty }}{lim}{T}_n(x)-T(x)\Vert$ and $|\Vert T^n\Vert -\Vert T\Vert |\le \Vert T^n-T\Vert \to 0$ as $n\to \mathrm{\infty }$. (TODO: a proof for the converse.) $◻$

2 Theorem (uniform boundedness principle) Let $ℱ$ be a family of continuous functions $f:X\to Y$ where $Y$ is a normed linear space. Suppose that $M\subset X$ is non-meager and that:

<math>\sup \{ \|f(x)\| : f \in \mathcal{F} \} < \infty</math> for each $x\in M$

It then follows: there is some $G\subset X$ open and such that

(a) $sup\{\Vert f(x)\Vert :f\in ℱ,x\in G\}<\mathrm{\infty }$

If we assume in addition that each member of $ℱ$ is a linear operator and $X$ is a normed linear space, then

(b) $sup\{\Vert f\Vert :f\in ℱ\}<\mathrm{\infty }$

Proof: Let $E_j={\cap }_{f\in ℱ}\{x\in X;\Vert f(x)\Vert \le j\}$ be a sequence. By hypothesis, $M\subset \bigcup _{j=1}^{\mathrm{\infty }}{E}_j$ and each $E_j$ is closed since $\{x\in X;\Vert f(x)\Vert >j\}$ is open by continuity. It then follows that some $E_N$ has an interior point $y$; otherwise, $M$ fails to be non-meager. Hence, (a) holds. To show (b), making additional assumptions, we can find an open ball $B=B(2r,y)\subset E_N$. It then follows: for any $f\in ℱ$ and any $x\in X$ with $\Vert x\Vert =1$,

$\Vert f(x)\Vert =r^{-1}\Vert f(rx+y)-f(y)\Vert \le 2r^{-1}N$. $◻$

A family $\mathrm{\Gamma }$ of linear operators is said to be equicontinuous if given any neighborhood $W$ of $0$ we can find a neighborhood $V$ of $0$ such that:

$f(V)\subset W$ for every $f\in \mathrm{\Gamma }$

The conclusion of the theorem, therefore, means that the family satisfying the hypothesis of the theorem is equicontinuous.

2 Corollary Let $𝒳,𝒴,𝒵$ be Banach spaces. Let $T:𝒳\times 𝒴\to 𝒵$ be a bilinear or sesquilinear operator. If $T$ is separately continuous (i.e., the function is continuous when all but one variables are fixed) and $𝒴$ is complete, then $T$ is continuous.
Proof: For each $y\in 𝒴$,

$sup\{\Vert T(x,y){\Vert }_{𝒵};\Vert x{\Vert }_{𝒳}\le 1\}=\Vert T(\cdot ,y)\Vert$

where the right-hand side is finite by continuity. Hence, the application of the principle of uniform boundedness to the family $\{T(x,\cdot );\Vert x{\Vert }_{𝒳}\le 1\}$ shows the family is equicontinuous. That is, there is $K>0$ such that:

$\Vert T(x,y){\Vert }_{𝒵}\le K\Vert y{\Vert }_{𝒴}$ for every $\Vert x{\Vert }_{𝒳}le1$ and every $y\in 𝒴$.

The theorem now follows since $𝒳\times 𝒴$ is a metric space. $◻$

Since scalar multiplication is a continuous operation in normed spaces, the corollary says, in particular, that every linear operator on finite dimensional normed spaces is continuous. The next is one more example of the techniques discussed so far.

2. Theorem (Hahn-Banach) Let $(𝒳,\Vert \cdot \Vert )$ be normed space and $ℳ\subset 𝒳$ be a linear subspace. If $z$ is a linear functional continuous on $ℳ$, then there exists a continuous linear functional $w$ on $𝒳$ such that $z=w$ on $ℳ$ and $\Vert z\Vert =\Vert w\Vert$.
Proof: Apply the Hahn-Banach stated in Chapter 1 with $\Vert z\Vert \Vert \cdot \Vert$ as a sublinear functional dominating $z$. Then:

$\Vert z\Vert =sup\{\Vert w(x)\Vert ;x\in ℳ,\Vert x\Vert \le 1\}\le sup\{\Vert w(x)\Vert ;x\in 𝒳,\Vert x\Vert \le 1\}=\Vert w\Vert \le \Vert z\Vert$;

that is, $\Vert z\Vert =\Vert w\Vert$. $◻$

2. Corollary Let $ℳ$ be a subspace of a normed linear space $𝒳$. Then $x$ is in the closure of $ℳ$ if and only if $z(x)$ = 0 for any $z\in {𝒳}^{\ast }$ that vanishes on $ℳ$.
Proof: By continuity $z(\overline{ℳ})\subset \overline{z(ℳ)}$. Thus, if $x\in \overline{ℳ}$, then $z(x)\in \overline{z(ℳ)}=\{0\}$. Conversely, suppose $x\in ̸\overline{ℳ}$. Then there is a $\delta >0$ such that $\Vert y-x\Vert \ge \delta$ for every $y\in ℳ$. Define a linear functional $z(y+\lambda x)=\lambda$ for $y\in ℳ$ and scalars $\lambda$. For any $\lambda \ne 0$, since $-{\lambda }^{-1}y\in ℳ$,

$|z(y+\lambda x)|=|\lambda |{\delta }^{-1}\delta \le {\delta }^{-1}|\lambda ||{\lambda }^{-1}y+x\Vert =\Vert y+\lambda x\Vert$.

Since the inequality holds for $\lambda =0$ as well, $z$ is continuous. Hence, in view of the Hahn-Banach theorem, $z\in 𝒳$ while we still have $z=0$ on $ℳ$ and $z(x)\ne 0$. $◻$

Here is a classic application.

2 Theorem Let $𝒳,𝒴$ be Banach spaces, $T:𝒳\to 𝒴$ be a linear operator. If $x_n\to 0$ implies that $(z\circ T)x_n\to 0$ for every $z\in {𝒳}^{\ast }$, then $T$ is continuous.
Proof: Suppose $x_n\to 0$ and $T{x}_n\to y$. For every $z\in {𝒳}^{\ast }$, by hypothesis and the continuity of $z$,

$0=\underset{n\to \mathrm{\infty }}{lim}z(T{x}_n)=z(y)$.

Now, by the preceding corollary $y=0$ and the continuity follows from the closed graph theorem. $◻$

2 Theorem Let $𝒳$ be a Banach space.

• (i) Given $E\subset 𝒳$, $E$ is bounded if and only if $\underset{E}{sup}|f|<\mathrm{\infty }$ for every $f\in {𝒳}^{\ast }$
• (ii) Given $x\in 𝒳$, if $f(x)=0$ for every $f\in {𝒳}^{\ast }$, then $x=0$.

Proof: (i) By continuity,

$sup\{|f(x)|;x\in E\}\le \Vert f\Vert \underset{E}{sup}\Vert \cdot \Vert$.

This proves the direct part. For the converse, define $T_{x}f=f(x)$ for $x\in E,f\in {𝒳}^{\ast }$. By hypothesis

$|T_{x}f|\le \underset{E}{sup}|f|$ for every $x\in E$.

Thus, by the principle of uniform boundedness, there is $K>0$ such that:

$|T_{x}f|\le K\Vert f\Vert$ for every $x\in E,f\in {𝒳}^{\ast }$

Hence, in view of Theorem 2.something, for $x\in E$,

$\Vert x\Vert =sup\{|f(x)|;f\in {𝒳}^{\ast },\Vert f\Vert \le 1\}\le K$.

(ii) Suppose $x\ne 0$. Define $f(s(x))=s\Vert x\Vert$ for scalars $s$. Now, $f$ is continuous since its domain is finite-dimensional, and so by the Hahn-Banach theorem we could extend the domain of $f$ in such a way we have $f\in {𝒳}^{\ast }$. $◻$

2. Corollary Let $(𝒳,\Vert \cdot \Vert )$ be Banach, $f_j\in {𝒳}^{\ast }$ and $ℳ\subset 𝒳$ dense and linear. Then $f_j(x)\to 0$ for every $x\in 𝒳$ if and only if $\underset{j}{sup}\Vert f_j\Vert <\mathrm{\infty }$ and $f_j(y)\to 0$ for every $y\in ℳ$.
Proof: Since $f_j$ is Cauchy, it is bounded. This shows the direct part. To show the converse, let $x\in 𝒳$. If $y_j\in ℳ$, then

$|f_j(x)|\le |f_j(x-y_j)|+|f_j(y_j)|\le (\underset{j}{sup}\Vert f_j\Vert )\Vert x-y_j\Vert +|f_j(y_j)|$

By denseness, we can take $y_j$ so that $\Vert y_j-x\Vert \to 0$. $◻$

2 Theorem Let $T$ be a continuous linear operator into a Banach space. If $\Vert I-T\Vert <1$ where $I$ is the identity operator, then the inverse $T^{-1}$ exists, is continuous and can be written by:

$T^{-1}(x)=\sum _{k=0}^{\mathrm{\infty }}{(I-T)}^k(x)$ for each $x$ in the range of $T$.

Proof: For $n\ge m$, we have:

$\Vert \sum _{k=m}^n{(I-T)}^k(x)\Vert \le \Vert x\Vert \sum _{k=m}^n{\Vert I-T\Vert }^k$.

Since the series is geometric by hypothesis, the right-hand side is finite. Let $S_n=\sum _{k=0}^n{(I-T)}^k$. By the above, each time $x$ is fixed, $S_n(x)$ is a Cauchy sequence and the assumed completeness implies that the sequence converges to the limit, which we denote by $S(x)$. Since for each $x$ $\underset{n\ge 1}{sup}\Vert S_n(x)\Vert <\mathrm{\infty }$, it follows from the principle of uniform boundedness that:

$\underset{n\ge 1}{sup}\Vert S_n\Vert \le \mathrm{\infty }$.

Thus, by the continuity of norms,

$\Vert S(x)\Vert =\underset{n\to \mathrm{\infty }}{lim}\Vert S_n(x)\Vert \le (\underset{n\ge 1}{sup}\Vert S_n\Vert )\Vert x\Vert$.

This shows that $S$ is a continuous linear operator since the linearity is easily checked. Finally,

$\Vert TS(x)-x\Vert =\Vert \underset{n\to \mathrm{\infty }}{lim}-(I-T{)}^{n+1}(x)\Vert \le \Vert x\Vert \underset{n\to \mathrm{\infty }}{lim}\Vert I-T{\Vert }^{n+1}=0$.

Hence, $S$ is the inverse to $T$. $◻$

2 Corollary The space of invertible continuous linear operators $𝒳$ is an open subspace of $L(𝒳,𝒳)$.
Proof: If $T\in L(𝒳,𝒳)$ and $\Vert S-T\Vert <\frac{1}{\Vert T^{-1}\Vert }$, then $S$ is invertible. $◻$

If $𝐅$ is a scalar field and $𝒳$ is a normed space, then $L(𝒳,𝐅)$ is called a dual of $𝒳$ and is denoted by ${𝒳}^{\ast }$. In view of Theorem 2.something, it is a Banach space.

A linear operator $T$ is said to be a compact operator if the image of the open unit ball under $T$ is relatively compact. We recall that if a linear operator between normed spaces maps bounded sets to bounded sets, then it is continuous. Thus, every compact operator is continuous.

2 Theorem Let $𝒳$ be a reflexive Banach space and $𝒴$ be a Banach space. Then a linear operator $T:𝒳\to 𝒴$ is a compact operator if and only if $T$ sends weakly convergent sequence to norm convergent ones.
Proof: Let $x_n$ converges weakly to $0$, and suppose $T{x}_n$ is not convergent. That is, there is an $ϵ>0$ such that $T{x}_n\ge ϵ$ for infinitely many $n$. Denote this subsequence by $y_n$. By hypothesis we can then show (TODO: do this indeed) that it contains a subsequence $y_{n_k}$ such that $T{y}_{n_k}$ converges in norm, which is a contradiction. To show the converse, let $E$ be a bounded set. Then since $𝒳$ is reflexive every countable subset of $E$ contains a sequence $x_n$ that is Cauchy in the weak topology and so by the hypothesis $T{x}_n$ is a Cauchy sequence in norm. Thus, $T(E)$ is contained in a compact subset of $𝒴$. $◻$

2 Corollary

• (i) Every finite-rank linear operator $T$ (i.e., a linear operator with finite-dimensional range) is a compact operator.
• (ii) Every linear operator $T$ with the finite-dimensional domain is continuous.

Proof: (i) is clear, and (ii) follows from (i) since the range of a linear operator has dimension less than that of the domain. $◻$

2 Theorem The set of all compact operators into a Banach space forms a closed subspace of the set of all continuous linear operators in operator norm.
Proof: Let $T$ be a linear operator and $\omega$ be the open unit ball in the domain of $T$. If $T$ is compact, then $T(\bar{\omega })$ is bounded (try scalar multiplication); thus, $T$ is continuous. Since the sum of two compacts sets is again compact, the sum of two compact operators is again compact. For the similar reason, $\alpha T$ is compact for any scalar $\alpha$. We conclude that the set of all compact operators, which we denote by $E$, forms a subspace of continuous linear operators. To show the closedness, suppose $S$ is in the closure of $E$. Let $ϵ>0$ be given. Then there is some compact operator $T$ such that $\Vert S-T\Vert <ϵ/2$. Also, since $T$ is a compact operator, we can cover $T(\omega )$ by a finite number of open balls of radius $ϵ/2$ centered at $z_1,z_2,\mathrm{...}{z}_n$, respectively. It then follows: for $x\in \omega$, we can find some $j$ so that $\Vert Tx-z_j\Vert <ϵ/2$ and so $\Vert Sx-Tx\Vert \le \Vert Sx-z_j\Vert +\Vert z_j-Tx\Vert <ϵ$. This is to say, $S(\omega )$ is totally bounded and since the completeness its closure is compact. $◻$

2 Corollary If $T_n$ is a sequence of compact operators which converges in operator norm, then its limit is a compact operator.

2 Theorem (transpose) Let $𝒳,𝒴$ be Banach spaces, and $u:𝒳\to 𝒴$ be a continuous linear operator. Define ${}^{t}u:{𝒴}^{\ast }\to {𝒳}^{\ast }$ by the identity ${}^{t}u(z)(x)=u(z(x))$. Then ${}^{t}u$ is continuous both in operator norm and the weak-* topology, and $\Vert {}^{t}u\Vert =\Vert u\Vert$.
Proof: For any $z\in {𝒴}^{\ast }$

$\Vert {}^{t}u(z)\Vert =\underset{\Vert x\Vert \le 1}{sup}|(u\circ z)(x)|\le \Vert u\Vert \Vert z\Vert$

Thus, $\Vert {}^{t}u\Vert \le \Vert u\Vert$ and ${}^{t}u$ is continuous in operator norm. To show the opposite inequality, let $ϵ>0$ be given. Then there is $x_0\in 𝒳$ with $(1-ϵ)\Vert u\Vert \le |u(x_0)|$. Using the Hahn-Banach theorem we can also find $\Vert z_0\Vert =1$ and $z_0(u(x_0))=|u(x_0)|$. Hence,

$\Vert {}^{t}u\Vert =\underset{\Vert z\Vert \le 1}{sup}\Vert {}^{t}u(z)\Vert \ge \Vert {}^{t}u(z_0)\Vert =|z_0(u(x))|=|u(x_0)|\ge (1-ϵ)\Vert u\Vert$.

We conclude $\Vert {}^{t}u\Vert =\Vert u\Vert$. To show weak-* continuity let $V$ be a neighborhood of $0$ in ${𝒳}^{\ast }$; that is, $V=\{z;z\in {𝒳}^{\ast },|z(x_1)|<ϵ,\mathrm{...},|z(x_n)|<ϵ\}$ for some $ϵ>0,x_1,\mathrm{...},x_n\in 𝒳$. If we let $y_j=u(x_j)$, then

${}^{t}u(\{z;z\in {𝒴}^{\ast },|z(y_1)|<ϵ,\mathrm{...},|z(y_n)|<ϵ\})\subset V$

since $z(y_j)={}^{t}u(z)(x_j)$. This is to say, ${}^{t}u$ is weak-* continuous. $◻$

2 Theorem Let $T:𝒳\to 𝒴$ be a linear operator between normed spaces. Then $T$ is compact if and only if its transpose $T^{\prime }$ is compact.
Proof: Let $K$ be the closure of the image of the closed unit ball under $T$. If T is compact, then K is compact. Let $y_n\in Y^{\ast }$ be a bounded sequence. Then the restrictions of $y_n$ to K is a bounded equicontinuous sequence in $C(K)$; thus, it has a convergent subsequence $y_{n_k}$ by Ascoli's theorem. Thus, $T^{\prime }{y}_{n_k}(x)=y_{n_k}(Tx)$ is convergent for every x with $\Vert x{\Vert }_{𝒳}\le 1$, and so $T^{\prime }{y}_{n_k}$ is convergent. The converse follows from noting that every normed space can be embedded continuously into its second dual. (TODO: need more details.)$◻$