Demi-linear duality

Abstract

As is well known, there exist non-locally convex spaces with trivial dual and therefore the usual duality theory is invalid for this kind of spaces. In this article, for a topological vector space X, we study the family of continuous demi-linear functionals on X, which is called the demi-linear dual space of X. To be more precise, the spaces with non-trivial demi-linear dual (for which the usual dual may be trivial) are discussed and then many results on the usual duality theory are extended for the demi-linear duality. Especially, a version of Alaoglu-Bourbaki theorem for the demi-linear dual is established.

1 Introduction

Let $\mathbb{K}\in \left\{ℝ,ℂ\right\}$ and X be a locally convex space over $\mathbb{K}$ with the dual X'. There is a beautiful duality theory for the pair (X, X') (see [[1], Chapter 8]). However, it is possible that X' = {0} even for some Fréchet spaces such as L^p (0, 1) for 0 < p < 1. Then the usual duality theory would be useless and hence every reasonable extension of X' will be interesting.

Recently, ${ℒ}_{\gamma ,U}\left(X,Y\right)$, the family of demi-linear mappings between topological vector spaces X and Y is firstly introduced in [2]. ${ℒ}_{\gamma ,U}\left(X,Y\right)$ is a meaningful extension of the family of linear operators. The authors have established the equicontinuity theorem, the uniform boundedness principle and the Banach-Steinhaus closure theorem for the extension ${ℒ}_{\gamma ,U}\left(X,Y\right)$. Especially, for demi-linear functionals on the spaces of test functions, Ronglu Li et al have established a theory which is a natural generalization of the usual theory of distributions in their unpublished paper "Li, R, Chung, J, Kim, D: Demi-distributions, submitted".

Let X,Y be topological vector spaces over the scalar field $\mathbb{K}$ and $\mathcal{N}\left(X\right)$ the family of neighborhoods of 0 ∈ X. Let

$$C\left(0\right)=\left\{\gamma \in {\mathbb{K}}^{\mathbb{K}}:\underset{t\to 0}{lim}\gamma \left(t\right)=\gamma \left(0\right)=0,\mid \gamma \left(t\right)\mid \ge \mid t\mid \mathsf{\text{if}}\mid t\mid \le 1\right\}.$$

Definition 1.1 [2, Definition 2.1] A mapping f: X → Y is said to be demi-linear if f(0) = 0 and there exists γ ∈ C(0) and $U\in \mathcal{N}\left(X\right)$ such that every x ∈ X, u ∈ U and $t\in \left\{t\in \mathbb{K}:\mid t\mid \le 1\right\}$ yield $r,s\in \mathbb{K}$ for which |r - 1| ≤ | γ (t) |, |s| ≤ | γ (t)| and f(x + tu) = rf(x) + sf(u).

We denote by ${ℒ}_{\gamma ,U}\left(X,Y\right)$ the family of demi-linear mappings related to γ ∈ C(0) and $U\in \mathcal{N}\left(X\right)$, and by ${풦}_{\gamma ,U}\left(X,Y\right)$ the subfamily of ${ℒ}_{\gamma ,U}\left(X,Y\right)$ satisfying the following property: if x ∈ X, u ∈ U and |t| ≤ 1, then f(x + tu) = rf(x) + sf(u) for some s with |s| ≤ | γ (t)|. Let

$$X^{\left(\gamma ,U\right)}=\left\{f\in {ℒ}_{\gamma ,U}\left(X,\mathbb{K}\right):f\mathsf{\text{is}}\mathsf{\text{continuous}}\right\},$$

which is called the demi-linear dual space of X. Obviously, X' ⊂ X^{(γ, U)}.

In this article, first we discuss the spaces with non-trivial demi-linear dual, of which the usual dual may be trivial. Second we obtain a list of conclusions on the demi-linear dual pair (X, X^{(γ, U)}). Especially, the Alaoglu-Bourbaki theorem for the pair (X, X^{(γ, U)}) is established. We will see that many results in the usual duality theory of (X, X') can be extended to (X, X^{(γ, U)}).

Before we start, some existing conclusions about ${ℒ}_{\gamma ,U}\left(X,Y\right)$ are given as follows. In general, ${ℒ}_{\gamma ,U}\left(X,Y\right)$ is a large extension of L(X, Y). For instance, if ||·||: X → [0, +∞) is a norm, then $\parallel \cdot \parallel \in {ℒ}_{\gamma ,X}\left(X,ℝ\right)$ for every γ ∈ C(0). Moreover, we have the following

Proposition 1.2 ([2, Theorem 2.1]) Let X be a non-trivial normed space, C > 1, δ > 0 and U ={u ∈ X : ||u|| ≤ δ}, γ(t) = Ct for $t\in \mathbb{K}$. If Y is non-trivial, i.e.,Y≠{0}, then the family of nonlinear mappings in ${ℒ}_{\gamma ,U}\left(X,Y\right)$ is uncountable, and every non-zero linear operator T: X → Y produces uncountably many of nonlinear mappings in ${ℒ}_{\gamma ,U}\left(X,Y\right)$.

Definition 1.3 A family Г ⊂ Y^X is said to be equicontinuous at x ∈ X if for every $W\in \mathcal{N}\left(Y\right)$, there exists $V\in \mathcal{N}\left(X\right)$ such that f(x + V) ⊂ f(x) + W for all f ∈ Г, and Г is equicontinuous on X or, simply, equicontinuous if Г is equicontinuous at each x ∈ X.

As usual, Г ⊂ Y^X is said to be pointwise bounded on X if {f(x): f ∈ Г} is bounded at each x ∈ X, and f : X → Y is said to be bounded if f(B) is bounded for every bounded B ⊂ X.

The following results are substantial improvements of the equicontinuity theorem and the uniform boundedness principle in linear analysis.

Theorem 1.4 ([2, Theorem 3.1]) If X is of second category and $\mathrm{\Gamma }\subset {ℒ}_{\gamma ,U}\left(X,Y\right)$ is a pointwise bounded family of continuous demi-linear mappings, then Г is equicontinuous on X.

Theorem 1.5 ([2, Theorem 3.3]) If x is of second category and $\mathrm{\Gamma }\subset {ℒ}_{\gamma ,U}\left(X,Y\right)$ is a pointwise bounded family of continuous demi-linear mappings, then Г is uniformly bounded on each bounded subset of X, i.e.,{f(x): f ∈ Г, x ∈ B} is bounded for each bounded B ⊂ X.

If, in addition, X is metrizable, then the continuity of f ∈ Г can be replaced by boundedness of f ∈ Г.

2 Spaces with non-trivial demi-linear dual

Lemma 2.1 Let $f\in {ℒ}_{\gamma ,U}\left(X,\mathbb{K}\right)$. For each x∈ X, u ∈ U and |t| ≤ 1, we have

$$\mid f\left(tu\right)\mid \le \mid \gamma \left(t\right)\mid \mid f\left(u\right)\mid ;$$

(1)

$$\mid f\left(x+tu\right)-f\left(x\right)\mid \le \mid \gamma \left(t\right)\mid \left(\mid f\left(x\right)\mid +\mid f\left(u\right)\mid \right).$$

(2)

Proof. Since $f\in {ℒ}_{\gamma ,U}\left(X,\mathbb{K}\right)$, for each x ∈ X, u ∈ U and |t| ≤ 1, we have f(x + tu) = rf(x) + sf(u) where |r - 1| ≤ |γ(t)| and |s| ≤ |γ(t)|. Then

$$\mid f\left(x+tu\right)-f\left(x\right)\mid =\mid \left(r-1\right)f\left(x\right)+sf\left(u\right)\mid \le \mid r-1\mid \mid f\left(x\right)\mid +\mid s\mid \mid f\left(u\right)\mid \le \mid \gamma \left(t\right)\mid \left(\mid f\left(x\right)\mid +\mid f\left(u\right)\mid \right),$$

which implies (2). Then (1) holds by letting x = 0 in (2).

Theorem 2.2 Let X be a topological vector space and f : X → [0, +∞) a function satisfying

$$\left(*\right)f\left(0\right)=0,f\left(-x\right)=f\left(x\right)andf\left(x+y\right)\le f\left(x\right)+f\left(y\right)wheneverx,y\in X.$$

Then, for every γ ∈ C(0) and$U\in \mathcal{N}\left(X\right)$, the following (I), (II), and (III) are equivalent:

(I) $f\in {ℒ}_{\gamma ,U}\left(X,ℝ\right)$ ;

(II) f(tu) ≤ |γ(t)|f(u) whenever u ∈ U and |t| ≤ 1;

(III)$f\in {풦}_{\gamma ,U}\left(X,ℝ\right)$.

Proof. (I) ⇒ (II). By Lemma 2.1.

• (II) ⇒ (III). Let x ∈ X, u ∈ U and |t| ≤ 1. Then

  $$f\left(x\right)-\mid \gamma \left(t\right)\mid f\left(u\right)\le f\left(x\right)-f\left(tu\right)\le f\left(x+tu\right)\le f\left(x\right)+f\left(tu\right)\le f\left(x\right)+\mid \gamma \left(t\right)\mid f\left(u\right).$$

Define φ : [-|γ(t)|, |γ(t)|] → ℝ by φ(α) = f(x) + αf(u). Then φ is continuous and

$$\phi \left(-\mid \gamma \left(t\right)\mid \right)=f\left(x\right)-\mid \gamma \left(t\right)\mid f\left(u\right)\le f\left(x+tu\right)\le f\left(x\right)+\mid \gamma \left(t\right)\mid f\left(u\right)=\phi \left(\mid \gamma \left(t\right)\mid \right).$$

So there is s ∈[-|γ(t)|, |γ(t)|] such that f(x + tu) = γ(s) = f(x) + sf(u).

• (III) ⇒ (I). ${풦}_{\gamma ,U}\left(X,ℝ\right)\subset {ℒ}_{\gamma ,U}\left(X,ℝ\right)$.

In the following Theorem 2.2, we want to know whether a paranorm on a topological vector space X is in ${풦}_{\gamma ,U}\left(X,ℝ\right)$ for some γ and U. However, the following example shows that this is invalid.

Example 2.3 Let ω be the space of all sequences with the paranorm||·||:

$$\parallel x\parallel =\sum _{j=1}^{\infty }\frac{1}{2^j}\frac{\mid x_j\mid }{1+\mid x_j\mid },\forall x=\left(x_j\right)\in \omega .$$

Then, for every γ ∈ C(0) and U _ε = {u = (u _j ): ||u|| < ε}, we have$\parallel \cdot \parallel \notin {ℒ}_{\gamma ,U}\left(\omega ,ℝ\right)$. Otherwise, there exists γ ∈ C(0) and ε > 0 such that$\parallel \cdot \parallel \notin {ℒ}_{\gamma ,U}\left(\omega ,ℝ\right)$and hence

$$\parallel \frac{1}{n}u\parallel \le \mid \gamma \left(\frac{1}{n}\right)\mid \parallel u\parallel ,forallu\in U_{\epsilon }andn\in \mathbb{N}$$

by Theorem 2.2. Pick N ∈ ℕ with$\frac{1}{2^N}<\epsilon$. Let$u_n=\left(0,\cdots ,0,\stackrel{\left(N\right)}{n},0,\cdots \right)$, ∀n ∈ ℕ. Then$\parallel u_n\parallel =\frac{1}{2^N}\frac{n}{1+n}<\frac{1}{2^N}<\epsilon$implies u _n ∈ U _ε for each N ∈ ℕ. It follows from

$$\mid \gamma \left(\frac{1}{n}\right)\mid \ge \frac{\parallel \frac{1}{n}{u}_n\parallel }{\parallel u_n\parallel }=\left(\frac{1}{2^N}\frac{1}{1+1}\right)∕\left(\frac{1}{2^N}\frac{n}{1+n}\right)=\frac{1}{2}\frac{1+n}{n}>\frac{1}{2},\forall n\in \mathbb{N},$$

that$\gamma \left(\frac{1}{n}\right)\nrightarrow 0$as n → ∞, which contradicts γ ∈ C(0).

Note that the space ω in Example 2.3 has a Schauder basis. The following corollary shows that the set of nonlinear demi-linear continuous functionals on a Hausdorff topological vector space with a Schauder basis has an uncountable cardinality.

Corollary 2.4 Let X be a Hausdorff topological vector space with a Schauder basis. Then for every γ ∈ C(0) and $U\in \mathcal{N}\left(X\right)$, the demi-linear dual $X^{\left(\gamma ,U\right)}=\left\{f\in {\mathcal{L}}_{\gamma ,U}\left(X,ℝ\right):fiscontinuous\right\}$ is uncountable.

Proof. Let {b _k } be a Schauder basis of X. There is a family P of non-zero paranorms on X such that the vector topology on X is just σP, i.e., x _α → x in X if and only if ||x _α - x|| → 0 for each ||·|| ∈ P ([[1], p.55]).

Pick ||·|| ∈ P. Then $\parallel {\sum }_{k=1}^{\infty }{s}_k{b}_k\parallel \ne 0$ for some ${\sum }_{k=1}^{\infty }{s}_k{b}_k\in X$ and hence $\parallel s_{k_0}{b}_{k_0}\parallel \ne 0$ for some k₀ ∈ ℕ. For non-zero $c\in \mathbb{K}$, define f _c : X → [0, +∞) by

$$f_c\left(\sum _{k=1}^{\infty }{r}_k{b}_k\right)=\mid c{r}_{k_0}\mid \parallel s_{k_0}{b}_{k_0}\parallel .$$

Obviously, f _c is continuous and satisfies the condition (*) in Theorem 2.2. Let γ ∈ C(0), ${\sum }_{k=1}^{\infty }{r}_k{b}_k\in X$ and |t| ≤ 1. Then

$$f_c\left(t\sum _{k=1}^{\infty }{r}_k{b}_k\right)=\mid ct{r}_{k_0}\mid \parallel s_{k_0}{b}_{k_0}\parallel =\mid t\mid \mid c{r}_{k_0}\mid \parallel s_{k_0}{b}_{k_0}\parallel =\mid t\mid f_c\left(\sum _{k=1}^{\infty }{r}_k{b}_k\right)\le \mid \gamma \left(t\right)\mid f_c\left(\sum _{k=1}^{\infty }{r}_k{b}_k\right)$$

and hence $f_c\in {풦}_{\gamma ,U}\left(X,ℝ\right)\subset {ℒ}_{\gamma ,U}\left(X,ℝ\right)$ for all $U\in \mathcal{N}\left(X\right)$ by Theorem 2.2. Thus, $\left\{f_c:0\ne c\in \mathbb{K}\right\}\subset X^{\left(\gamma ,U\right)}$ for all γ ∈ C(0) and $U\in \mathcal{N}\left(X\right)$.

Example 2.5 As in Example 2.3, the space (ω, ||·||) is a Hausdorff topological vector space with the Schauder base $\left\{e_n=\left(0,\cdots ,0,\stackrel{\left(n\right)}{1},0,\cdots \right):n\in \mathbb{N}\right\}$ . Define f _c,n : ω → ℝ with f _c,n (u) = |cu _n | where u = (u _j ) ∈ ω. Then we have

$$\left\{f_{c,n}:0\ne c\in \mathbb{K},n\in \mathbb{N}\right\}\subset {\omega }^{\left(\gamma ,U\right)}=\left\{f\in {ℒ}_{\gamma ,U}\left(\omega ,ℝ\right):fiscontinuous\right\}$$

for every γ ∈ C(0) and $U\in \mathcal{N}\left(\omega \right)$by Corollary 2.4.

Recall that a p-seminorm ||·|| (0 < p ≤ 1) on a vector space E is characterized by ||x|| ≥ 0, ||tx|| = |t| ^p ||x|| and ||x + y|| ≤ ||x|| + ||y|| for all $t\in \mathbb{K}$ and x, y ∈ E. If, in addition, ||x|| = 0 implies x = 0, then, ||·|| is called a p-norm on E.

Definition 2.6 ([[3], p. 11][[4], Sec. 2]) A topological vector space X is semiconvex if and only if there is a family {p _α } of (continuous) k _α -seminorms ( 0 < k _α ≤ 1) such that the sets {x ∈ X : p _α (x) < 1} form a neighborhood basis at 0, that is,

$$\left\{\left\{x:p_{\alpha }\left(x\right)<\frac{1}{n}\right\}:p_{\alpha }\in P,n\in N\right\}$$

is a base of$\mathcal{N}\left(X\right)$, where P is the family of all continuous p-seminorms with 0 < p ≤ 1.

A topological vector space X is locally bounded if and only if its topology is given by a p-norm (0 < p ≤ 1) ([[5], §15, Sec. 10]).

Clearly, locally bounded spaces and locally convex spaces are both semiconvex.

Corollary 2.7 Let X be a semiconvex Hausdorff topological vector space and p₀a continuous k₀-seminorm ( 0 < k₀ ≤ 1) on X. Then for $U_0=\left\{x\in X:p_0\left(x\right)\le 1\right\}\in \mathcal{N}\left(X\right)$ and $\gamma \left(\cdot \right)=e\mid \cdot {\mid }^{{\mathbb{K}}_0}\in {\mathbb{K}}^{\mathbb{K}}$, the demi-linear dual

$$X^{\left(\gamma ,U_0\right)}=\left\{f\in {ℒ}_{\gamma ,U_0}\left(X,ℝ\right):fiscontinuous\right\}$$

is uncountable. Especially, $\left\{p_0\left(\cdot \right),\mathsf{\text{sin}}\left(p_0\left(\cdot \right)\right),e^{p_0\left(\cdot \right)}-\mathsf{\text{1}}\right\}\subset X^{\left(\gamma ,U_0\right)}$.

Proof. Let P be the family of all continuous k _α -seminorms with 0 < k _α ≤ 1. Obviously, the functionals in P satisfy the condition (*) in Theorem 2.2. Moreover, for each p _α ∈ P with k _α ≥ k₀, we have

$$c{p}_{\alpha }\left(tx\right)=c\mid t{\mid }^{k_{\alpha }}{p}_{\alpha }\left(x\right)\le c\mid t{\mid }^{k_0}{p}_{\alpha }\left(x\right)\le \mid \gamma \left(t\right)\mid c{p}_{\alpha }\left(x\right),\mathsf{\text{for}}\mathsf{\text{all}}x\in X,\mid t\mid \le 1\mathsf{\text{and}}c\in K,$$

and hence $\left\{c{p}_{\alpha }:c\in \mathbb{K},k_{\alpha }\ge k_0\right\}\subset X^{\left(\gamma ,U_0\right)}$ by Theorem 2.2.

Define f : X → ℝ by f(x) = sin(p₀(x)), ∀x ∈ X. For each x ∈ X, u ∈ U₀ and |t| ≤ 1, there exists $s\in \left[-\mid t{\mid }^{k_0},\mid t{\mid }^{k_0}\right]$ and θ ∈ [0,1] such that

$$sin\left(p_0\left(x+tu\right)\right)=sin\left(p_0\left(x\right)+s{p}_0\left(u\right)\right)=sin\left(p_0\left(x\right)\right)+cos\left(p_0\left(x\right)+\mathrm{\theta s}{p}_0\left(u\right)\right)s{p}_0\left(u\right),$$

i.e.,

$$f\left(x+tu\right)=f\left(x\right)+cos\left(p_0\left(x\right)+\mathrm{\theta s}{p}_0\left(u\right)\right)\frac{p_0\left(u\right)}{sin\left(p_0\left(u\right)\right)}sf\left(u\right),$$

where

$$\mid cos\left(p_0\left(x\right)+\mathrm{\theta s}{p}_0\left(u\right)\right)\frac{p_0\left(u\right)}{sin\left(p_0\left(u\right)\right)}s\mid \le \frac{\pi }{2}\mid t{\mid }^{k_0}\le e\mid t{\mid }^{k_0}=\mid \gamma \left(t\right)\mid ,$$

which implies that $f\left(\cdot \right)=sin\left(p_0\left(\cdot \right)\right)\in X^{\left(\gamma ,U_0\right)}$.

Define g : X → ℝ by $g\left(x\right)=e^{p_0\left(x\right)}-1$, ∀x ∈X. For each x ∈ X, u ∈ U₀ and |t| ≤ 1, there exists $s\in \left[-\mid t{\mid }^{k_0},\mid t{\mid }^{k_0}\right]$ such that

$$e^{p_0\left(x+tu\right)}-1=e^{p_0\left(x\right)+s{p}_0\left(u\right)}-1=e^{s{p}_0\left(u\right)}\left(e^{p_0\left(x\right)}-1\right)+\frac{e^{s{p}_0\left(u\right)}-1}{e^{p_0\left(x\right)}-1}\left(e^{p_0\left(x\right)}-1\right),$$

i.e.,

$$g\left(x+tu\right)=e^{s{p}_0\left(u\right)}g\left(x\right)+\frac{e^{s{p}_0\left(u\right)}-1}{e^{p_0\left(x\right)}-1}g\left(u\right).$$

Then, there exists θ,η ∈ [0,1] for which

$$\mid e^{s{p}_0\left(u\right)}-1\mid =\mid e^{\mathrm{\theta s}{p}_0\left(u\right)}s{p}_0\left(u\right)\mid \le e\mid s\mid \le e\mid t{\mid }^{k_0}=\mid \gamma \left(t\right)\mid$$

and

$$\mid \frac{e^{s{p}_0\left(u\right)}-1}{e^{p_0\left(x\right)}-1}\mid =\mid \frac{e^{\mathrm{\theta s}{p}_0\left(u\right)}s{p}_0\left(u\right)}{e^{\eta p_0\left(u\right)}{p}_0\left(u\right)}\mid \le e^{\mathrm{\theta s}{p}_0\left(u\right)}\mid s\mid \le e\mid s\mid \le e\mid t{\mid }^{k_0}=\mid \gamma \left(t\right)\mid .$$

Thus, $g\left(\cdot \right)=e^{p_0\left(\cdot \right)}-1\in X^{\left(\gamma ,U_0\right)}$.

Example 2.8 For 0 < p < 1, let L^p (0,1) be the space of equivalence classes of measurable functions on [0,1], with

$$\parallel f\parallel ={\int }_0^1{\mid f\left(t\right)\mid }^{p}dt<\infty .$$

Then (L^p (0,1), ||·||)' = {0} ([[1], p.25]). However, L^p (0,1) is locally bounded and hence semiconvex. By Corollary 2.7, if U₀ = {f : ||f|| ≤ 1} and γ(·) = e|·| ^p ∈ C(0), then the demi-linear dual${\left(L^p\left(0,1\right),\parallel \cdot \parallel \right)}^{\left(\gamma ,U_0\right)}$contains various non-zero functionals.

A conjecture is that each topological vector space has a nontrivial demi-linear dual space. However, this is invalid, even for separable Fréchet space.

Example 2.9 Let $ℳ\left(0,1\right)$ be the space of equivalence classes of measurable functions on[0,1], with

$$\parallel f\parallel ={\int }_0^1\frac{\mid f\left(t\right)\mid }{1+\mid f\left(t\right)\mid }\mathrm{dt}.$$

Then$ℳ\left(0,1\right)$is a separable Fréchet space with trivial dual. In fact, the demi-linear dual space of$ℳ\left(0,1\right)$is also trivial, that is,

$${\left(ℳ\left(0,1\right),\parallel \cdot \parallel \right)}^{\left(\gamma ,U\right)}=\left\{0\right\}foreach\gamma \in C\left(0\right)andU\in \mathcal{N}\left(ℳ\left(0,1\right)\right).$$

Let$u\in {\left(ℳ\left(0,1\right),\parallel \cdot \parallel \right)}^{\left(\gamma ,U\right)}$. Let N ∈ ℕ be such that$\parallel f\parallel \le \frac{1}{N}$implies f ∈ U and |u(f)| < 1. Given$f\in ℳ\left(0,1\right)$, write$f={\sum }_{k=1}^N{f}_k$where f _k = 0 off$\left[\frac{k-1}{N},\frac{k}{N}\right]$. Then$∥f_k∥\le \frac{1}{N}$so

$$\begin{array}{c}u\left(f\right)=u\left(\sum _{k=1}^N{f}_k\right)=u\left(\sum _{k=1}^{N-1}{f}_k+f_N\right)\hfill \\ =r_{N}u\left(\sum _{k=1}^{N-1}{f}_k\right)+s_{N}u\left(f_N\right)\hfill \\ =r_N{r}_{N-1}u\left(\sum _{k=1}^{N-2}{f}_k\right)+r_N{s}_{N-1}u\left(f_{N-1}\right)+s_{N}u\left(f_N\right)\hfill \\ =\cdot \cdot \cdot \hfill \\ =r_N\cdot \cdot \cdot r_3{r}_{2}u\left(f_1\right)+r_N\cdot \cdot \cdot r_3{s}_{2}u\left(f_2\right)+\cdot \cdot \cdot \hfill \\ +r_N{s}_{N-1}u\left(f_{N-1}\right)+s_{N}u\left(f_N\right),\hfill \end{array}$$

(3)

where |r _i - 1| ≤ |γ(1)| and |s _i | ≤ |γ(1)| for 2 ≤ I ≤ N. Then

$$\begin{array}{c}\mid u\left(f\right)\mid \le {\left(1+\mid \gamma \left(1\right)\mid \right)}^{N-1}\mid u\left(f_1\right)\mid +{\left(1+\mid \gamma \left(1\right)\mid \right)}^{N-2}\mid \gamma \left(1\right)\mid \mid u\left(f_2\right)\mid +\cdot \cdot \cdot \hfill \\ +\left(1+\mid \gamma \left(1\right)\mid \right)\mid \gamma \left(1\right)\mid \mid u\left(f_{N-1}\right)\mid +\mid \gamma \left(1\right)\mid \mid u\left(f_N\right)\mid \hfill \end{array}$$

(4)

$$\begin{array}{c}\le {\left(1+\mid \gamma \left(1\right)\mid \right)}^{N-1}+{\left(1+\mid \gamma \left(1\right)\mid \right)}^{N-2}\mid \gamma \left(1\right)\mid +\cdot \cdot \cdot \hfill \\ +\left(1+\mid \gamma \left(1\right)\mid \right)\mid \gamma \left(1\right)\mid +\mid \gamma \left(1\right)\mid \hfill \end{array}$$

(5)

$$=2{(1+\mid \gamma \left(1\right)\mid )}^{N-1}-1.$$

(6)

So${sup}_{f\in ℳ\left(0,1\right)}\mid u\left(f\right)\mid <+\infty$. Since$\parallel n{f}_k\parallel \le \frac{1}{N}$for each n ∈ ℕ and 1 ≤ k ≤ N, we have {nf _k : n ∈ ℕ, k ∈ ℕ} ⊂ U. Then by Lemma 2.1,

$$\mid u\left(f_k\right)\mid =\mid u\left(\frac{1}{n}\left(n{f}_k\right)\right)\mid \le \mid \gamma \left(\frac{1}{n}\right)\mid \mid u\left(n{f}_k\right)\mid \le \mid \gamma \left(\frac{1}{n}\right)\mid \underset{f\in ℳ\left(0,1\right)}{sup}\mid u\left(f\right)\mid$$

(7)

holds for all n ∈ ℕ and 1 ≤ k ≤ N. Letting n → ∞, (7) implies u(f _k ) = 0 for 1 ≤ k ≤ N. Hence, |u(f)| = 0 by (4). Thus, u = 0.

3 Conclusions on the demi-linear dual pair (X, X ^(γ,U))

Henceforth, X and Y are topological vector spaces over $\mathbb{K}$, $\mathcal{N}\left(X\right)$ is the family of neighborhoods of 0 ∈ X, and X^(γ,U)is the family of continuous demi-linear functionals in ${ℒ}_{\gamma ,U}\left(X,\mathbb{K}\right)$. Recall that for usual dual pair (X, X') and A ⊂ X, the polar of A, written as A^°, is given by

$$A^{\circ }=\left\{f\in X^{\prime }:\mid f\left(x\right)\mid \le 1,\forall x\in A\right\}.$$

In this article, for the demi-linear dual pair (X, X^(γ,U)) and A ⊂ X, we denote the polar of A by A^•, which is given by

$$A^{\bullet }=\left\{f\in X^{\left(\gamma ,U\right)}:\mid f\left(x\right)\mid \le 1,\forall x\in A\right\}.$$

Similarly, for S ⊂ X^(γ,U),

$$S^{\bullet }=\left\{x\in X:\mid f\left(x\right)\mid \le 1,\forall f\in S\right\}.$$

Lemma 3.1. Let $f\in {ℒ}_{\gamma ,U}\left(X,Y\right)$. For every u∈ U and n ∈ ℕ,

$$f\left(nu\right)=\alpha f\left(u\right),where\mid \alpha \mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{n-1}-1.$$

Proof. It is similar to the proof of (3)-(6) in Example 2.9.

Lemma 3.2. Let S ⊂ X^(γ,U). If S is equicontinuous at 0 ∈ X, then, $S^{\bullet }\in \mathcal{N}\left(X\right)$ and sup _f∈S,x∈B|f(x)| < +∞ for every bounded B ⊂ X.

Proof. Assume that S is equicontinuous at 0 ∈ X. There is $U\in \mathcal{N}\left(X\right)$ such that |f(x)| < 1 for all f ∈ S and x ∈ V. Then V ⊂ S^• and hence $S^{\bullet }\in \mathcal{N}\left(X\right)$.

Let B ⊂ X be bounded. Since $S^{\bullet }\cap U\in \mathcal{N}\left(X\right)$, we have $\frac{1}{m}B\subset S^{\bullet }\cap U$ for some m ∈ ℕ. Then for each f ∈ S and x ∈ B,

$$\mid f\left(x\right)\mid =\mid f\left(m\frac{x}{m}\right)\mid =\mid \alpha \mid \mid f\left(\frac{x}{m}\right)\mid \le \mid \alpha \mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{m-1}-1$$

by Lemma 3.1. Hence, sup _f∈S,x∈B|f(x)| ≤ 2(1 + |γ(1)|)^m-1- 1 < +∞.

Lemma 3.3. Let S ⊂ X^(γ,U). Then S is equicontinuous on X if and only if S is equicontinuous at 0 ∈ X.

Proof. Assume that S is equicontinuous at 0 ∈ X. There is $W\in \mathcal{N}\left(X\right)$ such that |f(ω)| < 1 for all f ∈ S and ω ∈ W.

Let x ∈ X and ε > 0. By Lemma 3.2, sup _{f ∈S}|f(x)| = M < +∞. Observing lim _{t →0} γ(t) = 0, pick δ ∈ (0, 1) such that $\mid \gamma \left(\frac{\delta }{2}\right)\mid <\frac{\epsilon }{2\left(M+1\right)}$. By Lemma 2.1, for f ∈ S and $u=\frac{\delta }{2}{u}_0\in \frac{\delta }{2}\left(W\cap U\right)$, we have

$$\mid f\left(x+u\right)-f\left(x\right)\mid =\mid f\left(x+\frac{\delta }{2}{u}_0\right)-f\left(x\right)\mid \le \mid \gamma \left(\frac{\delta }{2}\right)\mid \left(\mid f\left(x\right)\mid +\mid f\left(u_0\right)\mid \right)<\frac{\epsilon }{2\left(M+1\right)}\left(M+1\right)<\epsilon .$$

Thus, $f\left[x+\frac{\delta }{2}\left(W+U\right)\right]\subset f\left(x\right)+\left\{z\in \mathbb{K}:\mid z\mid <\epsilon \right\}$ for all f ∈ S, i.e., S is equicontinuous at x.

Theorem 3.4. Let S ⊂ X^(γ,U). Then S is equicontinuous on X if and only if $S^{\bullet }\in \mathcal{N}\left(X\right)$.

Proof. If S is equicontinuous, then $S^{\bullet }\in \mathcal{N}\left(X\right)$ by Lemma 3.2.

Assume that $S^{\bullet }\in \mathcal{N}\left(X\right)$ and ε > 0. Since lim _{t →0}γ(t) = γ(0) = 0, there is δ > 0 such that |γ(t)| < ε whenever |t| < δ. For f ∈ S and $x=\frac{\delta }{2}{x}_0\in \frac{\delta }{2}\left(S^{\bullet }\cap U\right)$, we have |f(x₀)| ≤ 1 and $\mid f\left(x\right)\mid =\mid f\left(\frac{\delta }{2}{x}_0\right)\mid \le \mid \gamma \left(\frac{\delta }{2}\right)\mid \mid f\left(x_0\right)\mid <\epsilon$ by Lemma 2.1. Thus, $f\left[\frac{\delta }{2}\left(S^{\bullet }\cap U\right)\right]\subset \left\{z\in \mathbb{K}:\mid z\mid <\epsilon \right\}$ for all f ∈ S, i.e., S is equicontinuous at 0 ∈ X. By Lemma 3.3, S is equicontinuous on X.

The following simple fact should be helpful for further discussions.

Example 3.5. Let (L^p (0, 1), ||·||) be as in Example 2.8, U = {f : ||f || ≤ 1} and γ(t) = e |t| ^p for $t\in \mathbb{K}$ . Then (L^p (0, 1), ||·||)^(γ,U)contains non-zero continuous functionals such as ||·||, sin ||·||, e^||·|| - 1, etc. Since (αf)(·) = αf(·) for $\alpha \in \mathbb{K}$ and f∈ (L^p (0, 1), ||·||)^(γ,U), it follows from e^||·|| - 1 ∈ (L^p (0, 1), ||·||)^(γ,U)that $\frac{1}{e}\left(e^{\parallel \cdot \parallel }-1\right)\in {\left(L^p\left(0,1\right),\parallel \cdot \parallel \right)}^{\left(\gamma ,U\right)}$. If u ∈ U, then ||u|| ≤ 1, |sin ||u||| ≤ ||u|| ≤ 1 and $\mid \frac{1}{e}\left(e^{\parallel u\parallel }-1\right)\mid \le \frac{e-1}{e}<1$. Thus, if V is a neighborhood of 0 ∈ L^p (0, 1) such that V ⊂ U, then V^•contains non-zero functionals such as ||·||, sin ||·||, $\frac{1}{e}\left(e^{\parallel \cdot \parallel }-1\right)$, etc.

Corollary 3.6. For every $U,V\in \mathcal{N}\left(X\right)$ and γ∈ C(0), V^• = {f ∈ X^(γ,U): |f(x)| ≤ 1, ∀x ∈ V} is equicontinuous on X.

Proof. Let x ∈ V. Then |f(x)| ≤ 1, ∀f ∈ V^•, i.e., x ∈ (V^•)^•. Thus, V ⊂ (V^•)^• and so ${\left(V^{\bullet }\right)}^{\bullet }\in \mathcal{N}\left(X\right)$. By Theorem 3.4, V^• is equicontinuous on X.

Corollary 3.7. If X is of second category and S ⊂ X^(γ,U)is pointwise bounded on X, then $S^{\bullet }\in \mathcal{N}\left(X\right)$.

Proof. By Theorem 1.4, S is equicontinuous on X. Then $S^{\bullet }\in \mathcal{N}\left(X\right)$ by Theorem 3.4.

Corollary 3.8. Let X be a semiconvex space and S ⊂ X^(γ,U). Then S is equicontinuous on x if and only if there exist finitely many continuous k _i -seminorm p _i 's ( 0 < k _i ≤ 1, 1 ≤ i ≤ n < +∞) on x such that

$$\underset{f\in S}{sup}\underset{p_i\left(x\right)\le 1,1\le i\le n}{sup}\mid f\left(x\right)\mid <+\infty .$$

(8)

In particular, for a p-seminormed space (X, ||·||) (||·|| is a p-seminorm for some p ∈ (0, 1], especially, a norm when p = 1) and S ⊂ X^(γ,U), S is equicontinuous on x if and only if

$$\underset{f\in S}{sup}\underset{\parallel x\parallel \le 1}{sup}\mid f\left(x\right)\mid <+\infty .$$

Proof. Assume that S is equicontinuous. Then $S^{\bullet }\in \mathcal{N}\left(X\right)$ by Theorem 3.4. According to Definition 2.6, there exist finitely many continuous k _i -seminorm p _i 's (0 < k _i ≤ 1, 1 ≤ i ≤ n < +∞) and ε > 0 such that

$$\left\{x\in X:p_i\left(x\right)<\epsilon ,1\le i\le n\right\}\subset S^{\bullet }\cap U.$$

Let f ∈ S and p _i (x) ≤ 1, 1 ≤ i ≤ n. Pick n₀ ∈ ℕ for which ${\left(\frac{1}{n_0}\right)}^{k_0}<\epsilon$, where k₀ = min_1≤i≤nk _i . Then

$$p_i\left(\frac{x}{n_0}\right)={\left(\frac{1}{n_0}\right)}^{k_i}{p}_i\left(x\right)\le {\left(\frac{1}{n_0}\right)}^{k_0}{p}_i\left(x\right)<\epsilon ,\mathsf{\text{for }}1\le i\le n,$$

which implies $\frac{x}{n_0}\in S^{\bullet }\cap U$ and hence $\mid f\left(\frac{x}{n_0}\right)\mid \le 1$. By Lemma 3.1,

$$\mid f\left(x\right)\mid =\mid f\left(n_0\frac{x}{n_0}\right)\mid =\mid \mathrm{\alpha f}\left(\frac{x}{n_0}\right)\mid \le \mid \alpha \mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{n_0-1}-1.$$

Thus, ${sup}_{f\in S}{sup}_{p_i\left(x\right)\le 1,1\le i\le n}\mid f\left(x\right)\mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{n_0-1}-1<+\infty$.

Conversely, suppose that p _i is a continuous k _i -seminorm with 0 < k _i ≤ 1 for 1 ≤ i ≤ n < +∞, and (8) holds. Let $A=\left\{\frac{1}{M+1}f:f\in S\right\}$. Then A ⊂ X^(γ,U)and

$$\underset{g\in A}{sup}\underset{p_i\left(x\right)\le 1,1\le i\le n}{sup}\mid g\left(x\right)\mid =\frac{1}{1+M}\underset{f\in S}{sup}\underset{p_i\left(x\right)\le 1,1\le i\le n}{sup}\mid f\left(x\right)\mid =\frac{M}{1+M}<1,$$

i.e., {x ∈ X : p _i (x) ≤ 1, 1 ≤ i ≤ n} ⊂ A^• and so $A^{\bullet }\in \mathcal{N}\left(X\right)$. By Theorem 3.4, A^• is equicontinuous on X and S = (1 + M)A is also equicontinuous on X.

Lemma 3.9. Let $C\left(X,\mathbb{K}\right)=\left\{f\in {\mathbb{K}}^X:fiscontinuous\right\}$. For $S\subset C\left(X,\mathbb{K}\right)$, the following (I) and (II) are equivalent.

(I) S is equicontinuous on X.

(II) If(x _α ) _{α ∈}I is a net in x such that x _α → x ∈ X, then lim _α f(x _α ) = f(x) uniformly for f ∈ S.

Proof. (I)⇒(II). Let ε > 0 and x_α → x in X. Since S is equicontinuous on X, there is $W\in \mathcal{N}\left(X\right)$ such that

$$\mid f\left(x+w\right)-f\left(x\right)\mid <\epsilon ,\mathsf{\text{for all }}f\in S\mathsf{\text{and }}w\in W.$$

Since x _α → x, there is an index α₀ such that x _α - x ∈ W for all α ≥ α₀. Then

$$\mid f\left(x_{\alpha }\right)-f\left(x\right)\mid =\mid f\left(x+x_{\alpha }-x\right)-f\left(x\right)\mid <\epsilon ,\mathsf{\text{for all }}f\in S\mathsf{\text{and }}\alpha >{\alpha }_0.$$

Thus, lim _α f(x _α ) = f(x) uniformly for f ∈ S.

(II)⇒(I). Suppose that (II) holds but there exists x ∈ X such that S is not equicontinuous at x.

Then there exists ε > 0 such that for every $V\in \mathcal{N}\left(X\right)$, we can choose f _v ∈ S and z _v ∈ V for which

$$\mid f_v\left(x+z_v\right)-f_v\left(x\right)\mid \ge \epsilon$$

(9)

Since $\left(\mathcal{N}\left(X\right),\supset \right)$ is a directed set, we have ${\left(x+z_v\right)}_{V\in \mathcal{N}\left(X\right)}$ is a net in X. For every $W\in \mathcal{N}\left(X\right)$,

$$x+z_v\in x+V\subset x+W\mathsf{\text{for all }}V\in \mathcal{N}\left(X\right)\mathsf{\text{with }}W\supset V,$$

that is, lim _v (x + z _v ) = x.

By (II), there exists $W_0\in \mathcal{N}\left(X\right)$ such that |f(x + z _v ) - f(x)| < ε for all f ∈ S and $V\in \mathcal{N}\left(X\right)$ with W₀ ⊃ V. Then |f _v (x + z _v ) - f _v (x)| < ε for all $V\in \mathcal{N}\left(X\right)$ with W₀ ⊃ V. This contradicts (9) established above. Therefore, (II) implies (I).

We also need the following generalization of the useful lemma on interchange of limit operations due to E. H. Moore, whose proof is similar to the proof of Moore lemma ([[6], p. 28]).

Lemma 3.10. Let D₁and D₂be directed sets, and suppose that D₁ × D₂is directed by the relation $\left(d_1,d_2\right)\le \left(d_1^{\prime },d_2^{\prime }\right)$, which is defined by $d_1\le d_1^{\prime }$ and $d_2\le d_2^{\prime }$. Let f: D₁ × D₂ → X be a net in the complete topological vector space X. Suppose that:

(a) for each d₂ ∈ D₂, the limit$g\left(d_2\right)={lim}_{D_1}f\left(d_1,d_2\right)$exists, and

(b) the limit$h\left(d_1\right)={lim}_{D_2}f\left(d_1,d_2\right)$exists uniformly on D₁.

Then, the three limits

$$\underset{D_2}{lim}g\left(d_2\right),\underset{D_1}{lim}h\left(d_1\right),\underset{D_1\times D_2}{lim}f\left(d_1,d_2\right)$$

all exist and are equal.

We now establish the Alaoglu-Bourbaki theorem ([[1], p. 130]) for the pair (X, X^(γ,U)), where X is an arbitrary non-trivial topological vector space.

Let ${\mathbb{K}}^X$ be the family of all scalar functions on X. With the pointwise operations (f + g)(x) = f(x) + g(x) and (t f)(x) = t f(x) for x ∈ X and $t\in \mathbb{K}$, we have $x:{\mathbb{K}}^X\to \mathbb{K}$ is a linear space and each x ∈ X defines a linear functional $x:{\mathbb{K}}^X\to \mathbb{K}$ by letting x( f) = f(x) for $f\in {\mathbb{K}}^X$. In fact, for $f,g\in {\mathbb{K}}^X$ and $\alpha ,\beta \in \mathbb{K}$,

$$x\left(\alpha f+\beta g\right)=\left(\alpha f+\beta g\right)\left(x\right)=\alpha x\left(f\right)+\beta x\left(g\right).$$

Then, each x ∈ X produces a vector topology ωx on ${\mathbb{K}}^X$ such that

$$f_{\alpha }\to f\mathsf{\text{in}}\left({\mathbb{K}}^X,\mathrm{\omega x}\right)\mathsf{\text{if and only if }}{f}_{\alpha }\left(x\right)\to f\left(x\right)\mathsf{\text{([1, p}}\mathsf{\text{.12, p}}\mathsf{\text{.38])}}.$$

The vector topology V {ωx : x ∈ X} is just the weak * topology in the pair $\left(X,{\mathbb{K}}^X\right)$, and f _α → f in $\left({\mathbb{K}}^X,weak*\right)$ if and only if f _α (x) → f(x) for each x ∈ X ( [[1], p. 12, p. 38]). Note that weak* is a Hausdorff locally convex topology on ${\mathbb{K}}^X$.

Definition 3.11. A subset A ⊂ X^(γ,U)is said to be weak * compact in the pair (X, X^(γ,U)) or, simply, weak * compact if A is compact in $\left({\mathbb{K}}^X,weak*\right)$, and A is said to be relatively weak* compact in the pair (X, X^γ,U ) or, simply, relatively weak* compact if in $\left({\mathbb{K}}^X,weak*\right)$ the closure $Ā$ is compact and $Ā\subset X^{\left(\gamma ,U\right)}$.

For A ⊂ X^(γ,U), ${Ā}^{weak*}$ stands for the closure of A in $\left({\mathbb{K}}^X,weak*\right)$.

Theorem 3.12. For every $V\in \mathcal{N}\left(X\right)$, V^• = {f ∈ X^(γ,U): |f(x)| ≤ 1, ∀x ∈ V} is weak* compact in the pair (X, X^(γ,U)), and every equicontinuous S ⊂ X^(γ,U)is relatively weak* compact in the pair (X, X^(γ,U)).

Proof. For each x ∈ X, let x(f) = f(x) for f ∈ ${\mathbb{K}}^X$, then $x:{\mathbb{K}}^X\to \mathbb{K}$ is a linear functional. Let $V\in \mathcal{N}\left(X\right)$. By Corollary 3.6, V^• is equicontinuous on X and, by Lemma 3.2, x(V^•) = {f(x): f ∈ V^•} is bounded in $\mathbb{K}$ for each x ∈ X, i.e., for each x ∈ X, x(V^•) is totally bounded in $\mathbb{K}$ and so V^• is totally bounded in $\left({\mathbb{K}}^X,\omega x\right)$ for each x ∈ X ( [[1], p. 84, Theorem 6]. But the weak* topology for ${\mathbb{K}}^X$ is just V {ωx : x ∈ X} and so V^• is totally bounded in $\left({\mathbb{K}}^X,weak*\right)$ ([[1], p. 85, Theorem 7].

Let (f _α )_α∈I⊂ V^• be a Cauchy net in $\left({\mathbb{K}}^X,weak*\right)$. Then lim _α f _α (x) = f(x) exists at each x ∈ X and so f _α → f in $\left({\mathbb{K}}^X,weak*\right)$. For x ∈ X, u ∈ U and $t\in \left\{z\in \mathbb{K}:\mid z\mid \le 1\right\}$,

$$f\left(x+tu\right)=\underset{\alpha }{lim}{f}_{\alpha }\left(x+tu\right)=\underset{\alpha }{lim}\left[r_{\alpha }{f}_{\alpha }\left(x\right)+s_{\alpha }{f}_{\alpha }\left(u\right)\right],\mathsf{\text{where}}\mid r_{\alpha }-1\mid \le \mid \gamma \left(t\right)\mid \mathsf{\text{and}}\mid s_{\alpha }\mid \le \mid \gamma \left(t\right)\mid ,\forall \alpha \in I.$$

By passing to a subnet if necessary, we assume that r _α → r and s _α → s in $\mathbb{K}$. Then |r - 1| = lim _α |r _α - 1| ≤ |γ(t)|, |s| = lim _α |s _α | ≤ |γ(t)|| and

$$f\left(x+tu\right)=\underset{\alpha }{lim}\left[r_{\alpha }{f}_{\alpha }\left(x\right)+s_{\alpha }{f}_{\alpha }\left(u\right)\right]=\mathrm{rf}\left(x\right)+sf\left(u\right).$$

This shows that $f\in {ℒ}_{\gamma ,U}\left(X,\mathbb{K}\right)$.

Let x _β → x in X. Since V^• is equicontinuous on X and f _α ∈ V^• for all α ∈ I, it follows from Lemma 3.9 that lim _β f _α (x _β ) = f _α (x) uniformly for α ∈ I. Then

$$\underset{\beta }{lim}f\left(x_{\beta }\right)=\underset{\beta }{lim}\underset{\alpha }{lim}{f}_{\alpha }\left(x_{\beta }\right)=\underset{\alpha }{lim}\underset{\beta }{lim}{f}_{\alpha }\left(x_{\beta }\right)=\underset{\alpha }{lim}{f}_{\alpha }\left(x\right)=f\left(x\right)$$

by Lemma 3.10, i.e., $f:X\to \mathbb{K}$ is continuous and hence f ∈ X^(γ,U). Moreover, |f(x)| = lim _α |f _α (x)| ≤ 1 for each x ∈ V, i.e., f ∈ V^•. Thus, V^• is complete in $\left({\mathbb{K}}^X,weak*\right)$. Since $\left({\mathbb{K}}^X,weak*\right)$ is a topological vector space and V^• is both totally bounded and complete in $\left({\mathbb{K}}^X,weak*\right)$, we have V^• is compact in $\left({\mathbb{K}}^X,weak*\right)$, i.e., V^• is weak* compact in the pair (X, X^(γ,U)) ( [[1], p. 88, Theorem 7]).

Assume that S ⊂ X^(γ,U)is equicontinuous on X. By Lemma 3.2,

$S^{\bullet }=\left\{x\in X:\mid f\left(x\right)\mid \le 1,\forall f\in S\right\}\in \mathcal{N}\left(X\right)$, it follows from what is established above that (S^•)^• = {f ∈ X^(γ,U): |f(x)| ≤ 1, ∀x ∈ S^•} is compact in the Hausdorff space $\left({\mathbb{K}}^X,weak*\right)$. Then S ⊂ (S^•)^• shows that ${\bar{S}}^{weak*}\subset {\left(S^{\bullet }\right)}^{\bullet }\subset X^{\left(\gamma ,U\right)}$ and S is relatively weak* compact in (X, X^(γ,U)).

Theorem 3.12 is a version of Alaoglu-Bourbaki theorem for the demi-linear dual pair (X, X^(γ,U)), by which we can establish an improved Banach-Alaoglu theorem ( [[1], p. 130] as follows.

Corollary 3.13 (Banach-Alaoglu). Let X be a seminormed space and M > 0. Then

$$S=\left\{f\in X^{\left(\gamma ,U\right)}:\underset{\parallel x\parallel \le 1}{sup}\mid f\left(x\right)\mid \le M\right\}$$

is weak* compact in the pair (X, X^(γ,U)).

Proof. Since sup_f∈Ssup_||x||≤1|f(x)| ≤ M < +∞, Corollary 3.8 shows that S is equicontinuous on X. By Theorem 3.12, ${\bar{S}}^{weak*}\subset X^{\left(\gamma ,U\right)}$ and ${\bar{S}}^{weak*}$ is compact in $\left({\mathbb{K}}^X,weak*\right)$.

Let (f _α )_α∈Ibe a net in S such that lim _α f _α (x) = f(x) at each x ∈ X. Then f ∈ X,^(γ,U)and

$$\underset{\parallel x\parallel \le 1}{sup}\mid f\left(x\right)\mid =\underset{\parallel x\parallel \le 1}{sup}\underset{\alpha }{lim}\mid f_{\alpha }\left(x\right)\mid \le M,$$

i.e., f ∈ S. Thus, ${\bar{S}}^{weak*}=S$.

Theorem 3.14. Let X be a separable space, K a weak* compact set in X^(γ,U), S an equicontinuous set in X^(γ,U), and $V\in \mathcal{N}\left(X\right)$, $V^{\bullet }=\left\{f\in X^{\left(\gamma ,U\right)}:\mid f\left(x\right)\mid \le 1,\forall x\in V\right\}$. Then(S, weak*) is metrizable, and both (K, weak*) and (V^•, weak*) are compact metric spaces.

Proof. Assume that ${\left\{x_n\right\}}_{n=1}^{\infty }$ is dense in X. Let

$$d\left(f,g\right)=\sum _{n=1}^{\infty }\frac{1}{2^n}\frac{\mid f\left(x_n\right)-g\left(x_n\right)\mid }{1+\mid f\left(x_n\right)-g\left(x_n\right)\mid },\forall f,g\in {\mathbb{K}}^X.$$

Then, d(·,·) is a pseudometric on ${\mathbb{K}}^X$. If f, g ∈ X^(γ,U)and d(f, g) = 0, then f(x _n ) = g(x _n ) for all n. Since both f and g are continuous on X and ${\left\{x_n\right\}}_{n=1}^{\infty }$ is dense in X, f(x) = g(x) for all x ∈ X, i.e., f = g. This shows that (X^(γ,U), d) is a metric space, and f _k → f in (X^(γ,U), d) if and only if lim _k f _k (x _n ) = f(x _n ) for each n ∈ ℕ. Hence, weak* is stronger than d(·, ·) and so the compact space (K, weak*) is homeomorphic to the (Hausdorff) metric space (K, d). Thus, (K, weak*) is a compact metric space.

By Theorem 3.12, in $\left({\mathbb{K}}^X,weak*\right)$ the closure ${\bar{S}}^{weak*}\subset X^{\left(\gamma ,U\right)}$, and both $\left({\bar{S}}^{weak*},weak*\right)$ and (V^•, weak*) are compact and so they are compact metric spaces.

The following special case of Theorem 3.14 is a well-known fact ([[1], p. 143]).

Corollary 3.15. Let X be a separable locally convex space with the dual X', K a weak* compact set in X', S an equicontinuous set in X', and $V\in \mathcal{N}\left(X\right)$, V^° = {f ∈ X⁰ : |f(x)| ≤ 1, ∀x ∈ V}. Then (S, weak*) is metrizable, and both (K, weak*) and (V^°, weak*) are compact metric spaces.

Corollary 3.16. Let X be a separable space and S an equicontinuous set in X^(γ,U). Every sequence {f _n } in S has a subsequence $\left\{f_{n_k}\right\}$ such that ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ exists at each X∈ X and the limit function f ∈ X^(γ,U), i.e., f is both continuous and demi-linear.

Proof. By Theorems 3.12 and 3.14, ${\bar{S}}^{weak*}\subset X^{\left(\gamma ,U\right)}$ and $\left({\bar{S}}^{weak*},weak*\right)$ is a compact metric space. Then $\left({\bar{S}}^{weak*},weak*\right)$ is sequentially compact.

Combining Theorem 1.4 and Corollary 3.16, we have the following

Corollary 3.17. Assume that X is of second category and separable, e.g., separable Fréchet spaces such as L^p (0, 1)(p > 0), C[0,1], c₀, c, l^p(p > 0), etc. If S ⊂ X^(γ,U)is pointwise bounded on X, then every sequence {f _n } in S has a subsequence $\left\{f_{n_k}\right\}$ such that ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ exists at each x∈ X, and f ∈ X^(γ,U).

For C ≥ 1 and δ > 0, letting γ(t) = Ct for t ∈ ℝ and U = (-δ,δ), we have γ ∈ C(0) and $U\in \mathcal{N}\left(ℝ\right)$. Then let ℝ^(C,δ)= ℝ^(γ,U). It is easy to see that every $f\in {ℒ}_{\gamma ,U}\left(ℝ,ℝ\right)$ is continuous and so ${ℝ}^{\left(C,\delta \right)}={ℒ}_{\gamma ,U}\left(ℝ,ℝ\right)$. Thus, ℝ^(C,δ)contains all linear functions and various nonlinear functions. It is noted that many functions in ℝ^(C,δ)have very complicated graphs.

For S ⊂ ℝ^(C,δ), there is an interesting fact: a local behavior in a small interval (-ε, ε) implies a nice behavior on (-∞, +∞).

Example 3.18. Let S ⊂ ℝ^(C,δ). If there exists M, ε > 0 such that |f(x)| ≤ M for every f ∈ S and x ∈ (-ε, ε), then every {f _n } ⊂ S has a subsequence $\left\{f_{n_{\mathbb{K}}}\right\}$ such that ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ exists at each x∈ ℝ, and f ∈ ℝ^(C,δ).

In fact, ℝ is separable and $\left(-\epsilon ,\epsilon \right)\in \mathcal{N}\left(ℝ\right)$. The assumption shows that

$$M^{-1}S\subset \left\{f\in {ℝ}^{\left(C,\delta \right)}:\mid f\left(x\right)\mid \le 1,\forall x\in \left(-\epsilon ,\epsilon \right)\right\}={\left(-\epsilon ,\epsilon \right)}^{\bullet }.$$

By Theorem 3.14, ((-ε, ε)^•, weak*) is a compact metric space and so it is sequentially compact. Similarly, we have

Example 3.19. Let p > 0 and S ⊂ (L^p (0, 1))^(γ,U). If there exists ε > 0 such that |f(x)| ≤ 1 whenever f ∈ S and X ∈ L^p (0, 1) with ||x|| < ε, then every {f _n } ⊂ S has a subsequence $\left\{f_{n_k}\right\}$ such that ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ exists for all x∈ L^p (0, 1), and f ∈ (L^p (0, 1))^(γ,U).

We shall show that the condition "sup _{f∈S,||x||<ε}| f(x)| ≤ 1" in Example 3.19 can be weakened as "sup_f∈S|f(x)| < +∞, ∀ ||x|| < ε" (see Corollary 3.20).

In general, combining Theorems 3.12 and 3.14, we have

Corollary 3.20. Let S ⊂ X^(γ,U). If there exists $V\in \mathcal{N}\left(X\right)$ such that sup _f∈S,x∈V|f(x)| < +∞, then

(a) S is equicontinuous on X,

(b) S is relatively weak * compact,

(c) every net(f _α ) in S has a subnet (f _ξ (α)) such that lim_ξ(α)f _ξ (α)(x) = f(x) exists for all x ∈ X, and f ∈ X^(γ,U).

If, in addition, x is separable, then

(d) every {f _n } ⊂ S has a subsequence $\left\{f_{n_k}\right\}$ such that ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ exists for all x∈ X, and f ∈ X^(γ,U).

In fact, for M = sup_f∈S,x∈V|f(x)|, we have $A=\left\{\frac{1}{M+1}f:f\in S\right\}\subset V^{\bullet }$ and (a)-(d) hold for A, i.e., S satisfies (a)-(d).

If X is of second category, then the condition "there exists $V\in \mathcal{N}\left(X\right)$ such that sup_f∈S,x∈V|f(x)| < +∞" in Corollary 3.20 can be weakened as "there exists $V\in \mathcal{N}\left(X\right)$ such that sup f∈S |f(x)| < +∞, ∀x ∈ V".

To see this, we first establish a simple fact.

Lemma 3.21. Let $\mathrm{\Gamma }\subset {ℒ}_{\gamma ,U}\left(X,Y\right)$. If there exists $V\in \mathcal{N}\left(X\right)$ such that{f(x): f ∈ Г} is bounded at each x ∈ V, then {f(x): f ∈ Г} is bounded at each x ∈ X.

Proof. Let x ∈ X. There exists n₀ ∈ ℕ such that $\frac{1}{n_0}x\in V\cap U$. By Lemma 3.1, for each f ∈ Г, we have

$$f\left(x\right)=f\left(n_0\frac{x}{n_0}\right)={\alpha }_{f}f\left(\frac{x}{n_0}\right),\mathsf{\text{where}}\mid {\alpha }_f\mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{n_0-1}-1.$$

Then

$$\left\{f\left(x\right):f\in \mathrm{\Gamma }\right\}\subset \left\{tf\left(\frac{x}{n_0}\right):f\in \mathrm{\Gamma },\mid t\mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{n_0-1}-1\right\}.$$

Since $\frac{x}{n_0}\in V$, $\left\{f\left(\frac{x}{n_0}\right):f\in \mathrm{\Gamma }\right\}$ is bounded and so $\left\{tf\left(\frac{x}{n_0}\right):f\in \mathrm{\Gamma },\mid t\mid \le {2\left(1+\mid \gamma \left(1\right)\mid \right)}^{n_0-1}-1\right\}$ is bounded.

Now we can improve Theorems 1.4 and 1.5 as follows.

Theorem 3.22. Assume that x is of second category and $\mathrm{\Gamma }\subset \left\{f\in {ℒ}_{\gamma ,U}\left(X,Y\right):fiscontinuous\right\}$. If there exists $V\in \mathcal{N}\left(X\right)$ such that Г is pointwise bounded on V, then Г is equicontinuous on X, and Г is uniformly bounded on each bounded subset of X.

Corollary 3.23. Assume that x is of second category and S ⊂ X^(γ,U). If there exists $V\in \mathcal{N}\left(X\right)$ such that sup_f∈S|f(x)| < +∞ at each x ∈ V, then (a)-(c) hold for S. If, in addition, X is separable, then (d) holds for S.

We now show that every equicontinuous S ⊂ X^(γ,U)has a nice behavior on any compact subset of X.

Theorem 3.24. Let X be a Hausdorff topological vector space. If S is an equicontinuous subset of X^(γ,U)and Ω is a compact subset of X, then every {f _n } ⊂ S has a subsequence $\left\{f_{n_k}\right\}$ such that ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ uniformly for x∈ Ω and f : $f:\mathrm{\Omega }\to \mathbb{K}$is continuous.

Proof. Let K = {f |_Ω: f ∈ S}. Then K ⊂ C(Ω) and K is equicontinuous at each x ∈ Ω. Suppose that sup _f∈K||f||_∞ = sup _f∈K,x∈Ω|f(x)| = +∞. Then there exist sequences {f _n } ⊂ S and {x _n } ⊂ Ω such that |f _n (x _n )| > n, ∀n ∈ ℕ. By Lemma 3.2, we may assume that x _n ≠ x _m for n ≠ m.

Since Ω is compact, ${\left\{x_n\right\}}_{n=1}^{\infty }$ has a cluster point x ∈ Ω.

Since S is equicontinuous at x, there exists $V\in \mathcal{N}\left(X\right)$ such that |f(y) - f(x)| < 1 for all f ∈ S and y ∈ x + V, i.e., |f(y)| < |f(x)| + 1 for all f ∈ S and y ∈ x + V. Observing that |f _n (x _n )| > n for all n ∈ ℕ and {f _n } ⊂ S, there exists n₀ ∈ ℕ such that x _n ∉ x + V for all n > n₀. Since (x + V) ∩ Ω contains some x _n with x _n ≠ x, it follows that

$$\varnothing \ne \left(\left\{x_n:x_n\ne x,x_n\in \left(x+V\right)\right\}\cap \mathrm{\Omega }\right)\subseteq \left\{x_n\ne x:n\le n_0\right\}=\left\{y_1,y_2,\cdots ,y_m\right\},$$

where m ≤ n₀. But X is Hausdorff, so Ω is also Hausdorff. Then there exists $V_0\in \mathcal{N}\left(X\right)$ such that V₀ ⊊ V and (x + V₀) ∩ (Ω ∩ {y₁, y₂, · · ·, y _m }) = ∅. Hence x _n ∈ (x + V₀) ∩ Ω implies that x _n = x.

This contradicts the fact that x is a cluster point of ${\left\{x_n\right\}}_{n=1}^{\infty }$. Hence, ${sup}_{f\in \mathbb{K}}\parallel f\mid {\mid }_{\infty }<+\infty$

By the Arzela-Ascoli theorem, K is relatively compact in the metric space (C(Ω), ||·||_∞). Hence, every {f _n } ⊂ S has a subsequence $\left\{f_{n_k}\right\}$ such that $\parallel {f_{n_K}\mid }_{\mathrm{\Omega }}-f{\parallel }_{\infty }\to 0$, where f ∈ C(Ω), i.e., ${lim}_k{f}_{n_k}\left(x\right)=f\left(x\right)$ uniformly for x ∈ Ω.

Corollary 3.25. Let X = ℝ ⁿ or ℂ ⁿ, ε > 0 and D _m = {x ∈ X : ||x|| ≤ mε}, ∀m ∈ ℕ. If S ⊂ X^(γ,U)is pointwise bounded on D₁, then every sequence {f _k } ⊂ S has a subsequence $\left\{f_{k_i}\right\}$ such that ${lim}_i{f}_{k_i}\left(x\right)=f\left(x\right)$ uniformly on each D _m , where f ∈ X^(γ,U).

Proof. Theorem 3.22 shows that S is equicontinuous on X and, by Theorem 3.24, {f _k } has a subsequence $\left\{f_{{\mathbb{K}}_i}\right\}$ such that ${lim}_i{f}_{k_i}\left(x\right)$ exists uniformly on D₁. Then ${\left\{f_{k_i}\right\}}_{i=2}^{\infty }$ has a subsequence $\left\{f_{k_{i_v}}\right\}$ such ${lim}_v{f}_{k_{i_v}}\left(x\right)$ exists uniformly on D₂. Proceeding inductively, the diagonal procedure yields a subsequence {g _i } of {f _k } such that lim _i g _i (x) exists uniformly on each D _m . Then lim _i g _i (x) = f(x) exists at each x ∈ X and $f\in {\bar{S}}^{weak*}$ in $\left({\mathbb{K}}^X,weak*\right)$. By Theorem 3.12, f ∈ X^(γ,U).