User:Mgkrupa/Analysis of vector-valued curves

In mathematics, specifically in functional analysis and real/complex analysis, curves taking values in a topological vector space (TVS), such as $$\mathbb{R}^{n}$$ or a Banach space, play an important role in the more general theory of analysis of maps between two TVSs. This article describes the basic theory of analysis of curves valued in arbitrary TVSs; for analysis of curves valued in finite-dimensional spaces like $$\mathbb{R}^{n}$$, please see the articles vector calculus and multivariable calculus.

The analysis of vector-valued curves is significantly more straight forward than the analysis of TVS-valued functions having domains that are infinite-dimensional TVSs, where, for instance, there are multiple competing definitions of differentiability such as Fréchet differentiability and Gateaux differentiability. In particular, there is significant agreement about the definitions of integrals and derivatives of curves even if they take their values in infinite-dimensional TVSs. Consequently, the analysis of vector-valued curves is a useful tool for studying the more general case of analysis of functions between arbitrary TVSs.

Basic definitions
The pair (a, b) denotes an element of some product X &times; Y and use ]a, b[ := { r : a < r and r < b} (rather than (a, b)) to denote an open interval. The brackets [a, b] denote an interval of real numbers and X denotes a topological vector space. If a function f is constant on a set S then one can identify the singleton set f(S) with this constant value. The Lebesgue measure is denoted by 𝜆.

The continuous dual space of a TVS X is denoted by $$X^{\prime}$$. The dual $$X^{\prime}$$ is said to separate points on X if for every non-zero x in X, there exists some continuous linear functional L on X such that L(x) ≠ 0. Clearly, if $$X^{\prime}$$ separates points on X then X is necessarily Hausdorff and the weak topology on X induced by $$X^{\prime}$$ is a Hausdorff locally convex topology that is coarser than X's original topology.

Weak integration: the Pettis integral
If f is an X-valued map on an interval I then f is called Pettis integrable or weakly integrable if for all continuous linear functionals L on X, the scalar-valued map L ∘ f is measurable and Lebesgue integrable, and if there exists a unique y in X such that
 * $$\int_{I} \Lambda \circ f = \Lambda(y)$$ for all $$\Lambda \in X^{\prime}$$,

in which case we denote y by $$\int_{I} f$$. If the continuous dual space of X separates points on X then any such vector y is necessarily unique so that in this case one need not check for uniqueness; in addition, because of this one can henceforth assume that the continuous dual space of X separates points on X (which is a very mild requirement).

The defining condition of the value of the weak integral $$\int_{a}^{b} f$$ is:
 * $$\int_{I} \Lambda \circ f = \Lambda \left( \int_{I} f \right)$$ for all $$\Lambda \in X^{\prime}$$,

where it is clear that if X is Euclidean space then any X-valued Lebesgue integrable function is also weakly integrable. Relative to almost every other definition of integration of a vector-valued function, the Pettis integrability condition is relatively weak. In addition, no matter what definition of "integrable" one encounters (so long as the value of this integral is an element of X), if a function is integrable under that definition then it is extremely likely that it will also be weakly integrable; furthermore these two integrals are very likely to agree in their values. Because of this, once one has proven some equality for the Pettis integral (for instance, linearity of the integral, or change of variables formula) then one may almost immediately deduce that this property holds for this other definition of the integral. Due to the importance of linear functionals in modern functional analysis, it is rare to see a definition of integrability (where the integral is valued in X) that does not imply weak integrability (if such a definition even exists).

The vector space of all X-valued weakly integrable functions will be denoted by $$\operatorname{Pettis}(I; X)$$ where it is easy to see that the integral map $$\operatorname{Pettis}(I; X) \to X$$ defined by $$f \mapsto \int_I f$$ is a linear operator.

The Pettis integral can also be defined more generally for X valued maps whose domains are measure spaces.

Properties
If f is weakly integrable and p is a continuous seminorm on X then $$p \left( \int_I f \right) \leq \int_I p \circ f$$ and there exists a continuous linear functional L on X such that $$L\left( \int_{I} f \right) = p \left( \int_{I} f \right)$$ and $$\left| L (s) \right| \leq p \left( s \right)$$ for all s in I. In addition, if T : X &rarr; Y is a continuous linear operator into a TVS Y whose continuous dual space separates points on Y, then $$\int_{I} T \circ f$$ is weakly integrable and $$\int_{I} T \circ f = T \left( \int_I f \right)$$.

Conditions for existence
Suppose that f is an X-valued map from an interval I and that $$X^{\prime}$$ separates points on X.

If the domain of f is compact, f is continuous, and the closed convex hull of the image of f (denoted by $$\overline{\operatorname{co}}(f(I))$$) is compact (which is automatically true if X is a Fréchet spaces) then the weak integral of f exists and $$\int_I f \in \lambda(I) \overline{\operatorname{co}}(f(I))$$, where 𝜆 is Lebesgue measure.

If W and X are TVSs with X locally convex and Hausdorff, U is a subset of W, and $$f : U \times [a, b] \to X$$ is continuous, then the for all u in U, the weak integral $$\int_a^b f(u, t) \, \mathrm{d}t$$ exists and the map $$g : U \to X$$ defined by $$g(u) := \int_a^b f(u, t) \, \mathrm{d}t$$ is continuous.

Space of bounded maps
Given a seminorm p on X, let $$\overset{\infty}{p}$$ denote the map defined by sending an X-valued function f to $$\left\| p \circ f \right\|_\infty$$, where recall that for any real valued map g, $$\| g \|_\infty = \sup_{t \in \operatorname{Dom} g} \left| g(t) \right|$$. This map is a seminorm on the space of all X-valued functions such that $$p \circ f$$ is bounded. If X is a locally convex space then we let $$\operatorname{Bounded}([a, b]; X)$$ denote the vector space of all X-valued maps such that $$p \circ f$$ is bounded for every continuous seminorm p on X. If the topology of X is induced by the family of seminorms $$(p_\alpha)$$ then the seminorms $$\overset{\infty}{p}$$ induce a canonical locally convex topology on $$\operatorname{Bounded}([a, b]; X)$$ that we will henceforth associate with this space.

Integration of step functions
A partition of [a, b] is a finite sequence $$(a_i)_{i=0}^n$$ such that a0 = a, an = b, and ai &le; ai+1 for all i = 0, ..., n − 1. An X-valued step function on [a, b] with respect to a partition $$(a_i)_{i=0}^n$$ is a function f : [a, b] &rarr; X such that for all i = 0, ..., n − 1, f is constant on ]ai, ai+1[ (it may take on any value at the points a0, ..., an). Denote the space of all X-valued step functions on [a, b] by $$\operatorname{Step}([a, b]; X).$$

The integral of such a step function f, denoted by $$\int_a^b f(t) \, \mathrm{d}t$$, is the vector
 * $$\int_a^b f(t) \, \mathrm{d} t := \sum_{i=0}^{n-1} (a_{i+1} - a_i) f\left( \frac{a_{i} + a_{i+1}}{2} \right).$$

It may be shown that this value is actually independent of the choice of partition for f. Essentially all definitions of the integral of a step function agree with the definition just given. We use the notation $$\int_b^a f(t) \, \mathrm{d} t$$ to denote $$- \int_a^b f(t) \, \mathrm{d}t$$ and we may omit writing the symbols t and dt.

If a &le; s &le; t &le; b and f is defined on [a, b], then $$\int_{s}^{t} f$$ denotes $$\int_{s}^{t} f\big\vert_{[s, t]}$$. It may be shown that if b &le; c then $$\int_a^b f + \int_b^c f = \int_a^c f$$ for all step functions f defined on [a, c].

If p is a seminorm on X and every interval [ai, ai+1] is non-degenerate then
 * $$p\left( \int_a^b f \right) \leq (b - a) \sup_{i=0,\ldots,n-1} (p \circ f) \left( \frac{a_i + a_{i+1}}{2} \right).$$

If in addition at every ai, f is continuous from the left or from the right then $$p\left( \int_a^b f \right) \leq (b - a) \|p \circ f\|_\infty$$; in particular, if X is normed then $$\left\| \int_a^b f \right\| \leq (b - a) \left\| p \circ f \right\|_\infty$$;

Regulated functions
If X is locally convex then clearly $$\operatorname{Step}([a, b]; X) \subseteq \operatorname{Bounded}([a, b]; X)$$. The closure of $$\operatorname{Step}([a, b]; X)$$ in $$\operatorname{Bounded}([a, b]; X)$$ is called the space of regulated maps and is denoted by $$\operatorname{Regulated}([a, b]; X)$$.

Since the map $$\operatorname{Step}([a, b]; X) \mapsto X$$ defined by $$f \mapsto \int_a^b f$$ is a continuous linear operator, it has a unique continuous linear extension $$\operatorname{Regulated}([a, b]; X) \to X$$ that sends a function f to some value, which we will denote by $$\int_a^b f$$.

Integration of simple functions
A function f is called a simple measurable function if it takes on only a finite number of values and if each of its fibers is a measurable subset of its domain. We define the integral of a simple measurable function f by
 * $$\int_a^b f(t) \, \mathrm{d}t := \sum_{y \in \operatorname{Im} f} y \lambda(f^{-1}(y)).$$

Essentially all definitions of the integral of a simple measurable function agree with the definition just given.

More generally, if f : [a, b] &rarr; X is a function such that there exist non-empty measurable pairwise disjoint sets N, A0, ..., An whose union is [a, b] such that N has measure 0 and f takes on the constant value xi on each Ai, then we can define the integral of f by
 * $$\int_a^b f(t) \, \mathrm{d}t := \sum_i^n x_i \lambda(A_i).$$

Properties of integration
If f : [a, b] &rarr; X is a map, t0 is in the domain of f, and r > 0 is such that t + r &le; b, then
 * $$\int_{t_0}^{t_0 + r} f = \int_0^1 f(t_0 + r u) r \, \mathrm{d}u$$

where if one of these integrals exists then so too does the other one; in particular,
 * $$\int_a^b f = \int_0^1 f( a + (b - a)u) (b - a) \, \mathrm{d}u.$$

Differentiation
So that limits are unique, we will henceforth assume that all TVS topologies considered are Hausdorff.

Suppose that X is a Hausdorff TVS and f, which we'll also denote by $$f^{(0)}$$, is an X-valued map defined on a subset I of the real numbers. For any t0 in the domain of f, we say that f is 0-times differentiable at t0 if f is continuous at t0. We say that f is (once) differentiable at t0 if t0 is not an isolated point in the domain of f and if the limit
 * $$f^{\prime}(t_0) := \lim_{\stackrel{h \to 0}{t_0 + h \in \operatorname{Dom} f}} \frac{f(t_0 + h) - f(t_0)}{h}$$

exists in X, in which case we call this limit fs derivative at t0'. The derivative of f is the map, denoted by $$f^{\prime}$$ or $$f^{(1)}$$, whose domain consists of all t such that $$f^{\prime}(t)$$ exists and that sends a point t in its domain to $$f^{\prime}(t)$$. Having defined $$f^{(k)}$$, which we call the kth derivative of f, we define $$f^{(k+1)}$$ to be the derivative of $$f^{(k)}$$. We say that f is k-times differentiable at t0 if t0 belongs to the domain of $$f^{(k)}$$.

We say that f is $$C^{0}$$ or 0-times differentiable if it is continuous and we say that f is (once) differentiable if it is differentiable at every point of its domain (its domain, therefore, has no isolated points). For any k &isin; { 2, 3, ... }, we say that f is k-times differentiable if it is k-1 times differentiable and its (k − 1)th derivative is differentiable.

We say that f is $$C^1$$ or continuously differentiable if it is continuous, differentiable, and $$f^{\prime}$$ is continuous. For any positive integer k, we say that f is $$C^{k}$$ or k-times continuously differentiable if it is $$C^{k-1}$$ and its (k – 1)th derivative, $$f^{(k-1)}$$, is continuously differentiable. For any subset I of the real numbers, we denote the vector space of all X-valued Ck functions with domain I by Ck(I; X).

We call f a C0-embedding if it is a topological embedding. For any k &isin; { 1, 2, ... }, a Ck-embedding is a Ck map that is a topological embedding whose first derivative does not vanish at any point of its domain. We denote the set of all X-valued Ck embeddings with domain I by Embedk(I; X) and the set of all X-valued Ck embeddings from any non-empty subset of $$\mathbb{R}$$ by Embedk(X).

A curve or a C0-curve is a continuous X-valued map whose domain is a non-degenerate interval of real numbers and for any k &isin; { 0, 1, ... }, a Ck-curve is a curve that is k-times continuously differentiable. An arc or a C0-arc is a topological embedding whose domain is a non-degenerate closed and bounded interval, and for any k &isin; { 1, 2, ... }, a Ck-arc is a C0-arc that is also a Ck-embedding. If I is a non-degenerate compact interval then we denote the set of all X-valued Ck arcs with domain I by Arck(I; X) and we denote the set of all X-valued Ck arcs by Arck(X).

If X is a finite-dimensional Hausdorff TVS then each of the above definitions agrees with the usual definition of its counterpart as defined in standard finite-dimensional real analysis.

Definitions extended to other topologies
If we precede any of the above definitions with the word weak or weakly then we mean that definition applied to X when it has the weak topology induced by its continuous dual space. So for instance, f is weakly differentiable at t0 if f is differentiable at t0 when X is endowed with the weak topology induced by $$X^{\prime}$$, where one may show that this is true if and only if t0 is not an isolated point and there exists some v in X such that for all $$\lambda \in X^{\prime}$$,
 * $$\lambda(v) = \lim_{\stackrel{h \to 0}{t_0 + h \in \operatorname{Dom} f}} \frac{\lambda(f(t_0 + h)) - \lambda (f(t_0))}{h}$$

in which case v is called the weak derivative of f at t0. Note that even if f is not differentiable at a point, it may still be weakly differentiable at that point.

Similarly, if we precede any of the above definitions with the word Mackey (resp. bounded) then we mean that definition applied to X when it has the Mackey topology (resp. the topology of uniform convergence on bounded subsets of $X^{\prime}$) induced by its continuous dual space.

Fundamental theorem of calculus

 * First fundamental theorem of calculus: Let X be a TVS whose dual space separates points on X and let f : [a, b] &rarr; X be continuously differentiable. Then the weak integral $$\int_a^b f^{\prime}$$ exists and $$f(b) - f(a) = \int_a^b f^{\prime}$$.

It follows immediately that if f and g are continuously differentiable functions [a, b &rarr; X valued in a TVS whose continuous dual space separates points, then if f(a) = g(a) and $$f^{\prime} = g^{\prime}$$ then necessarily f = g. If in addition p is a continuous seminorm on X then we also have:
 * $$p(f(b) - f(a)) \leq (b - a) \left( \sup_{t \in \operatorname{Dom} f} p(f(t))\right).$$


 * Second fundamental theorem of Calculus: Let X be a locally convex Hausdorff TVS and let f : [a, b] &rarr; X be continuous. Then the function F : [a, b] &rarr; X defined by $$F(t) := \int_a^t f$$ is continuously differentiable and $$F^{\prime}(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left( \int_a^t f \right) = f(t).$$

Change of variables

 * Theorem: If X is a TVS whose continuous dual space separates points, f is a continuous X-valued map defined on a subset I of the real numbers, and $$\phi : [a, b] \to I$$ is continuously differentiable, then $$\int_{\phi(a)}^{\phi(b)} f = \int_a^b f(\phi(t)) \phi^{\prime}(t) \, \mathrm{d}t$$.

Ck curves and arcs and their relation to TVS topologies
If X is a set and if $$(g_\alpha)_{\alpha \in A}$$ is a collection of X-valued maps where the domain of each $$g_\alpha$$ is a topological space, then the final topology on X induced by $$(g_\alpha)_{\alpha \in A}$$ is the finest topology on X making all $$g_\alpha$$ continuous.

Relationship to the Gateaux derivative
Let X and Y be two Hausdorff TVSs, U an open subset of X, F : U &rarr; Y a map, and $$\mathrm{d}F : U \times X \to Y$$ be the Gateaux derivative of F. Recall that $$\mathrm{d}F(x,v)$$ is the directional derivative of F in the direction v. Indeed, $$\mathrm{d} F(x,v)$$ is by definition equal to the derivative at t = 0 of the Y-valued map defined by $$g (t) = g(x + tv)$$, where the domain of g is the set $$\left\{ t \in \mathbb{R} : x + t v \in U \right\}$$; that is,
 * $$\mathrm{d} F(x,v) = g^{\prime}(0) = \frac{\mathrm{d}}{\mathrm{d}t} \Bigg\vert_{t = 0}(F(x + t v)).$$

Through this observation, many basic properties of the Gateaux derivative of F may be deduced from the properties of Y-valued curves.