Regulated function

Let X be a Banach space with norm || - ||_X. A function f : [0, T] → X is said to be a regulated function if one (and hence both) of the following two equivalent conditions holds true:{{harvnb|Dieudonné|1969|loc=§7.6}}

• for every t in the interval [0, T], both the left and right limits f(t−) and f(t+) exist in X (apart from, obviously, f(0−) and f(T+));
• there exists a sequence of step functions φ_n : [0, T] → X converging uniformly to f (i.e. with respect to the supremum norm || - ||_∞).

It requires a little work to show that these two conditions are equivalent. However, it is relatively easy to see that the second condition may be re-stated in the following equivalent ways:

• for every δ > 0, there is some step function φ_δ : [0, T] → X such that

::

• f lies in the closure of the space Step([0, T]; X) of all step functions from [0, T] into X (taking closure with respect to the supremum norm in the space B([0, T]; X) of all bounded functions from [0, T] into X).

Let Reg([0, T]; X) denote the set of all regulated functions f : [0, T] → X.

• Sums and scalar multiples of regulated functions are again regulated functions. In other words, Reg([0, T]; X) is a vector space over the same field K as the space X; typically, K will be the real or complex numbers. If X is equipped with an operation of multiplication, then products of regulated functions are again regulated functions. In other words, if X is a K-algebra, then so is Reg([0, T]; X).
• The supremum norm is a norm on Reg([0, T]; X), and Reg([0, T]; X) is a topological vector space with respect to the topology induced by the supremum norm.
• As noted above, Reg([0, T]; X) is the closure in B([0, T]; X) of Step([0, T]; X) with respect to the supremum norm.
• If X is a Banach space, then Reg([0, T]; X) is also a Banach space with respect to the supremum norm.
• Reg([0, T]; R) forms an infinite-dimensional real Banach algebra: finite linear combinations and products of regulated functions are again regulated functions.
• Since a continuous function defined on a compact space (such as [0, T]) is automatically uniformly continuous, every continuous function f : [0, T] → X is also regulated. In fact, with respect to the supremum norm, the space C⁰([0, T]; X) of continuous functions is a closed linear subspace of Reg([0, T]; X).
• If X is a Banach space, then the space BV([0, T]; X) of functions of bounded variation forms a dense linear subspace of Reg([0, T]; X):

::$\backslash mathrm\{Reg\}([0,\; T];\; X)\; =\; \backslash overline\{\backslash mathrm\{BV\}\; ([0,\; T];\; X)\}\; \backslash mbox\{\; w.r.t.\; \}\; \backslash |\; \backslash cdot\; \backslash |\_\{\backslash infty\}.$

• If X is a Banach space, then a function f : [0, T] → X is regulated if and only if it is of bounded φ-variation for some φ:

::$\backslash mathrm\{Reg\}([0,\; T];\; X)\; =\; \backslash bigcup\_\{\backslash varphi\}\; \backslash mathrm\{BV\}\_\{\backslash varphi\}\; ([0,\; T];\; X).$

• If X is a separable Hilbert space, then Reg([0, T]; X) satisfies a compactness theorem known as the Fraňková–Helly selection theorem.
• The set of discontinuities of a regulated function of bounded variation BV is countable for such functions have only jump-type of discontinuities. To see this it is sufficient to note that given $\backslash epsilon\; >\; 0$, the set of points at which the right and left limits differ by more than $\backslash epsilon$ is finite. In particular, the discontinuity set has measure zero, from which it follows that a regulated function has a well-defined Riemann integral.
• Remark: By the Baire Category theorem the set of points of discontinuity of such function $F\_\backslash sigma$ is either meager or else has nonempty interior. This is not always equivalent with countability.[https://math.stackexchange.com/q/84870 Stackexchange discussion]

• The integral, as defined on step functions in the obvious way, extends naturally to Reg([0, T]; X) by defining the integral of a regulated function to be the limit of the integrals of any sequence of step functions converging uniformly to it. This extension is well-defined and satisfies all of the usual properties of an integral. In particular, the regulated integral
• is a bounded linear function from Reg([0, T]; X) to X; hence, in the case X = R, the integral is an element of the space that is dual to Reg([0, T]; R);
• agrees with the Riemann integral.

{{Reflist}}

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• {{cite web |title=How to show that a set of discontinuous points of an increasing function is at most countable |date=November 23, 2011 |work=Stack Exchange |url=https://math.stackexchange.com/q/84870 }}
• {{cite web |title=Bounded variation functions have jump-type discontinuities |date=November 28, 2013 |work=Stack Exchange |url=https://math.stackexchange.com/q/584735 }}
• {{cite web |title=How discontinuous can a derivative be? |date=February 22, 2012 |work=Stack Exchange |url=https://math.stackexchange.com/q/112067 }}

Category:Real analysis

Category:Types of functions