Functional Analysis/Banach spaces

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Functional Analysis
Chapter 2: Banach spaces
50% developed  as of Sep 1, 2007 (Sep 1, 2007)

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

(ii) implies that \|0\| = 0. This and (ii) then implies 0 = \|x-x\| \le \|x\| + \|-x\| = 2\|x\| 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) = \|x - y\| a normed space is a metric space. Note that (i) implies that:

\|x\| \le \|x - y\| + \|y\| and \|y\| \le \|x - y\| + \|x\|

and so: | \|x\| - \|y\| | \le \|x - y\|. (So, the map x \mapsto \|x\| 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: \mathcal{C}(K), the space of all continuous functions on a compact space K, can be identified with a Banach space by introducing the norm:

\|\cdot\| = \sup_K |\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 \|x\|_p < \infty, then

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

Proof. By Hölder's inequality,

\sup_{\|y\|_q=1} |\sum x_j \overline{y_j}| \le \|x\|_p

Conversely, if \|x\|_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 \|y\|_q = (\sum |x_j|^{(p-1)q})^{1/q} = 1.

since p = (p-1)q. More generally, if \|x\|_p \ne 0, then

{x \over \|x\|_p} = \sup_{\|y\|_q = 1} |\sum x_j\overline{y_j}| {1 \over \|x\|_p}.

Since the identity is obvious when x = 0, the proof is complete. \square

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}^\infty |(x_k)_n - (x_j)_n|^2 \to 0 as n, m \to \infty

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

\sum_{n=1}^\infty |(x_k)_n - (x_j)_n|^2 < \epsilon for k, j > N

Then, for any k > N,

\sum_{n=1}^\infty |(x_k)_n - y_n|^2 = \sup_{m \ge 1} \sum_{n=1}^m |(x_k)_n - y_n|^2 = \sup_{m \ge 1} \lim_{j \to \infty} \sum_{n=1}^m |(x_k)_n - (x_j)_n|^2 < \epsilon

Hence, x_k \to y with y = \sum_{n=1}^\infty y_n. y is in fact in l_p since \|y\|_2 \le \|y - x_n\|_2 + \|x_n\|_2 < \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 \mathcal{X} and \mathcal{Y}, denoted by \|f\|, by

\|f\| = \sup_{\|x\|_{\mathcal{X}} \le 1} \|f(x)\|_{\mathcal{Y}}

2 Theorem Let T be a linear operator from a normed space \mathcal{X} to a normed space \mathcal{Y}.

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

Proof: If \|T\| < \infty, then

\|T(x_n - x)\|_\mathcal{Y} \le \|T\| \|x_n - x\|_\mathcal{X} \to 0

as x_n \to x. Hence, T is continuous. Conversely, suppose \|T\| = \infty. Then we can find x_n \in \mathcal{X} with \|x_n\|_\mathcal{X} \le 1 and \|Tx_n\| \ge n. Then {x_n \over n} \to 0 while \left\| T \left({x_n \over n} \right) \right\| \not\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). \square

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)= XP' on the normed vector space of polynomials \mathbb{R}[X] with the suprenum norm \|P \|_{\infty}=\sup_{x\in [0,1]} |P(x)| ; since D(X^n)=nX^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 \Leftrightarrow \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 \mathbb{R} is not continuous. then by the previous theorem,

\forall c>0,\exists x(c) s.t.|Tx(c)|\geqslant c\|x(c)\| so in particular, one can define a sequence \{x_n\} such that |Tx_n|\geqslant n\|x_n\|>0. Then denote:
u_n:=\frac{x_n}{|x_n|}, one has defined a unit normed sequence (|u_n|=1) s.t. |Tu_n|\geqslant n. Furthermore, denote
v_n:=\dfrac{u_n}{|Tu_n|}. Since  \frac{|u_n|}{|Tu_n|}  \leqslant \dfrac{1}{n}, one can define a sequence that converges \{v_n\}\to 0 whilst |Tv_n|= 1.

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

\underbrace{a-v_n Ta}_{\in \ker T}\to a\notin \ker T and hence \ker T is not closed. \square

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 \Leftrightarrow 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 \mbox{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_1f_1(x)+\dots+a_rf_r(x)"
Proof: take a basis (a_1,\dots,a_r) of \mbox{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^* 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_if_i so

Tx=\sum_{i=1}^r f_i(x)a_i=(\sum_{i=2}^r\alpha_if_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_ia_1+a_i)_{i=2,\dots,r} spans \mbox{Im }T, so \dim \mbox{Im }T=r-1 which is a contradiction. Finally, one has a unique decomposition of a finite rank linear operator:
Tx=a_1f_1(x)+\dots a_r f_r(x) with f_i\in E^* \square

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 \mbox{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  \mbox{Im}T_i\subset \mbox{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. \square

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+\mathbb{K}a is closed when F is closed (where \mathbb{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,\|\cdot\|) so in (H,\|\cdot\|), 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 \mathbb{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 \leqslant 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+\bigoplus_{i=1}^p \mathbb{K}a_i + \mathbb{K}a_{p+1} and concludes easily.\square

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 \mbox{Im}T=\dim X holds, it comes that the null space is of finite dimension, so is closed (any \mathbb{K} vector subspace of finite dimension n is isomorphic to \mathbb{K}^n (where \mathbb{K} is a complete field), so the subspace is complete and closed). Then one applies the previous theorem.\square

2 Lemma (Riesz) A normed space \mathcal{X} is finite-dimensional if and only if its closed unit ball is compact.
Proof: Let T: \mathbf{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 \mathcal{X} | \|x\| \le 1 \} \subset T \{ y \in \mathbf{C}^n | \|y\| \le \|T^{-1}\| \}

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 \mathcal{X} is not finite dimensional, we can construct a sequence x_j such that:

1 = \|x_j\| \le \|x_j - \sum_{k=1}^{j-1} a_k x_k\| for any sequence of scalars a_k.

Thus, in particular, \|x_j - x_k\| \ge 1 for all j, k. (For the details of this argument, see : w:Riesz's_lemma for now) \square

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. \square

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

Tx = (f_1(x), f_2(x), ... f_n(x))

where f_1, ... 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.) \square

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: \mathcal{X} \to \mathcal{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 Tf_j is defined for every j; i.e., the sequence f_j is in the domain of T. Since

\| Tf_j - Tf_k \| \le \|T\| \|f_j - f_k\| \to 0,

T f_j is Cauchy. It follows that (f_j, Tf_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. \square

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 \mathcal{X}, \mathcal{Y} be Banach spaces. If T: \mathcal{X} \to \mathcal{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 \mathcal{X}; \|x\| < r \}. Since T is surjective, \cup_{n=1}^\infty T(B(n)) = T(\cup_{n=1}^\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. \square

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

\| I \cdot \|_1 = \|\cdot\|_1 \le (\|\cdot\|_1 + \|\cdot\|_2).

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

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

By the same argument we can show that \|\cdot\|_1 + \|\cdot\|_2 is dominated by \|\cdot\|_2 \square

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

|(x_1, ... 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 \operatorname{im}(T) then \|x\| \le K\|y\| 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 \mathcal{M} be a closed subspace of a normed space \mathcal{X}. The quotient space \mathcal{X} / \mathcal{M} is a normed space with norm:

\|\pi(x)\| = \inf \{ \|x + m\|; m \in \mathcal{M} \}

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

\|\pi(x + y)\| \le \|x+m\| + \|y+n\|

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

\|\pi(x + y)\| \le \|\pi(x)\| + \|\pi(y)\|

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

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

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

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

\sum_{n=1}^\infty \|\pi(x_n)\| < \infty

Then we can find a sequence y_n \in \mathcal{M} such that

\sum_{n=1}^\infty \|x_n + y_n\| < \infty

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

2 Lemma Let \mathcal{X} be a normed space. Then \mathcal{X} is complete (thus a Banach space) if and only if

\sum_{n=1}^\infty \|x_n\| < \infty implies \sum_{n=1}^\infty x_n converges.

Proof: (\Rightarrow) We have:

\| \sum_{n=k}^{k+m} x_n \| \le  \sum_{n=k}^{k+m} \|x_n\|.

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

\| x_{n_k} - x \| = \| \sum_{j=1}^n x_{k_{j+1}} - x_{k_j} - x \| \to 0 as n \to \infty,

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

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 \mathcal{X}, \mathcal{Y} be Banach spaces, and T: \mathcal{X} \to \mathcal{Y} a linear operator. The following are equivalent.

  • (i) T is continuous.
  • (ii) If x_j \to 0 and Tx_j is convergent, then Tx_j \to 0.
  • (iii) The graph of T is closed.

Proof: That (i) implies (ii) is clear. To show (iii), suppose (x_j, Tx_j) is convergent in X . Then x_j converges to some x_0 or x_j - x_0 \to 0, and Tx_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:

\|\cdot\| + \|T\cdot\| \le K \|\cdot\|

since by hypothesis the norm in the left-hand side is complete. Hence, if x_j \to x, then Tx_j \to Tx. \square

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

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


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

By hypothesis, the right-hand side goes to 0 as j, k \to \infty. Since T is closed, (u_j, Tu_j) converges to (u, Tu). \square

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

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

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

\|T_n(x) - T_m(x)\| \le \|T_n - T_m\|\|x\|

and \mathcal{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 \|T(x)\| \le \sup_n \|T_n\|\|x\|. Finally, \lim_{n \to \infty} \|T_n - T\| = \sup_{\|x\| \le 1} \|\lim_{n \to \infty} T_n(x) - T(x)\| and | \|T^n\| - \|T\| | \le \|T^n - T\| \to 0 as n \to \infty. (TODO: a proof for the converse.) \square

2 Theorem (uniform boundedness principle) Let \mathcal{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:

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

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

(a) \sup \{ \|f(x)\| : f \in \mathcal{F}, x \in G \} < \infty

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

(b) \sup \{ \|f\| : f \in \mathcal{F} \} < \infty

Proof: Let E_j = \cap_{f \in \mathcal{F}} \{ x \in X ; \|f(x)\| \le j \} be a sequence. By hypothesis, M \subset \bigcup_{j=1}^\infty E_j and each E_j is closed since \{ x \in X ; \|f(x)\| > 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 \mathcal{F} and any x \in X with \|x\| = 1,

\|f(x)\| = r^{-1}\|f(rx + y) - f(y)\| \le 2 r^{-1} N. \square

A family \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 \Gamma

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

2 Corollary Let \mathcal{X}, \mathcal{Y}, \mathcal{Z} be Banach spaces. Let T: \mathcal{X} \times \mathcal{Y} \to \mathcal{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 \mathcal{Y} is complete, then T is continuous.
Proof: For each y \in \mathcal{Y},

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

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

\|T(x, y)\|_\mathcal{Z} \le K\|y\|_\mathcal{Y} for every \|x\|_\mathcal{X} le 1 and every y \in \mathcal{Y}.

The theorem now follows since \mathcal{X} \times \mathcal{Y} is a metric space. \square

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

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

that is, \|z\| = \|w\|. \square

2. Corollary Let \mathcal{M} be a subspace of a normed linear space \mathcal{X}. Then x is in the closure of \mathcal{M} if and only if z(x) = 0 for any z \in \mathcal{X}^* that vanishes on \mathcal{M}.
Proof: By continuity z(\overline {\mathcal{M}}) \subset \overline{z(\mathcal{M})}. Thus, if x \in \overline{\mathcal{M}}, then z(x) \in \overline{z(\mathcal{M})} = \{0\}. Conversely, suppose x \not\in \overline{\mathcal{M}}. Then there is a \delta > 0 such that \|y - x\| \ge \delta for every y \in \mathcal{M}. Define a linear functional z(y + \lambda x) = \lambda for y \in \mathcal{M} and scalars \lambda. For any \lambda \ne 0, since -\lambda^{-1} y \in \mathcal{M},

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

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

Here is a classic application.

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

0 = \lim_{n \to \infty} z (T x_n) = z (y).

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

2 Theorem Let \mathcal{X} be a Banach space.

  • (i) Given E \subset \mathcal{X}, E is bounded if and only if \sup_E |f| < \infty for every f \in \mathcal{X}^*
  • (ii) Given x \in \mathcal{X}, if f(x) = 0 for every f \in \mathcal{X}^*, then x = 0.

Proof: (i) By continuity,

\sup \{ |f(x)|; x \in E \} \le \|f\| \sup_E \|\cdot\|.

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

|T_x f| \le \sup_E |f| for every x \in E.

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

|T_x f| \le K \|f\| for every x \in E, f \in \mathcal{X}^*

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

\|x\| = \sup \{ |f(x)|; f \in \mathcal{X}^*, \|f\|\le 1 \} \le K.

(ii) Suppose x \ne 0. Define f(s(x)) = s\|x\| 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 \mathcal{X}^*. \square

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

|f_j(x)| \le |f_j(x - y_j)| + |f_j(y_j)| \le (\sup_j \|f_j\|) \|x - y_j\| + |f_j(y_j)|

By denseness, we can take y_j so that \|y_j - x\| \to 0. \square

2 Theorem Let T be a continuous linear operator into a Banach space. If \| I - T \| < 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}^\infty \left( I - T \right)^k (x) for each x in the range of T.

Proof: For n \ge m, we have:

\| \sum_{k=m}^n \left( I - T \right)^k (x) \| \le \|x\| \sum_{k=m}^n \left \| I - T \right \| ^k.

Since the series is geometric by hypothesis, the right-hand side is finite. Let S_n = \sum_{k=0}^n \left( I - T \right)^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 \sup_{n \ge 1} \| S_n(x) \| < \infty, it follows from the principle of uniform boundedness that:

\sup_{n \ge 1} \| S_n \| \le \infty.

Thus, by the continuity of norms,

\| S(x) \| = \lim_{n \to \infty} \| S_n(x) \| \le (\sup_{n \ge 1} \| S_n \|) \|x\|.

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

\|TS (x) - x \| = \| \lim_{n \to \infty} -(I - T)^{n + 1} (x) \| \le \|x\| \lim_{n \to \infty} \| I - T \|^{n+1} = 0.

Hence, S is the inverse to T. \square

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

If \mathbf{F} is a scalar field and \mathcal{X} is a normed space, then L(\mathcal{X}, \mathbf{F}) is called a dual of \mathcal{X} and is denoted by \mathcal{X}^*. 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 \mathcal{X} be a reflexive Banach space and \mathcal{Y} be a Banach space. Then a linear operator T:\mathcal{X} \to \mathcal{Y} is a compact operator if and only if T sends weakly convergent sequence to norm convergent ones.
Proof: [1] Let x_n converges weakly to 0, and suppose Tx_n is not convergent. That is, there is an \epsilon > 0 such that T x_n \ge \epsilon 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 \mathcal{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 Tx_n is a Cauchy sequence in norm. Thus, T(E) is contained in a compact subset of \mathcal{Y}. \square

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. \square

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(\overline {\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 \epsilon > 0 be given. Then there is some compact operator T such that \| S - T \| < \epsilon / 2. Also, since T is a compact operator, we can cover T(\omega) by a finite number of open balls of radius \epsilon / 2 centered at z_1, z_2, ... z_n, respectively. It then follows: for x \in \omega, we can find some j so that \| Tx - z_j \| < \epsilon / 2 and so \| Sx - Tx \| \le \| Sx - z_j \| + \| z_j - Tx \| < \epsilon. This is to say, S(\omega) is totally bounded and since the completeness its closure is compact. \square

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

\|{}^t\!u(z)\| = \sup_{\|x\| \le 1} |(u \circ z) (x)| \le \|u\|\|z\|

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

\|{}^t\!u\| = \sup_{\|z\| \le 1}\|{}^t\!u(z)\| \ge \|{}^t\!u(z_0)\| = |z_0(u(x))| =  |u(x_0)| \ge (1-\epsilon)\|u\|.

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

{}^t\!u ( \{ z; z \in \mathcal{Y}^*, |z(y_1)| < \epsilon, ..., |z(y_n)| < \epsilon \} ) \subset V

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

2 Theorem Let T: \mathcal{X} \to \mathcal{Y} be a linear operator between normed spaces. Then T is compact if and only if its transpose T' 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^* 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'y_{n_k}(x) = y_{n_k}(Tx) is convergent for every x with \|x\|_\mathcal{X} \le 1, and so T' 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.)\square


  1. This proof and a few more related results appear in [1]