ROUGH NOTES (!)
Updated: 26/6/24
Integration of maps ${ f : [a, b] \to E }$; Mean value theorem
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\[{ \underline{\textbf{Integration of maps } f : [a, b] \to E} }\]Recall if a map is uniformly continuous and into a complete space, the domain of definition can be expanded to its closure while preserving continuity of the map (and uniquely so). The extended map also turns out to be uniformly continuous.
Thm [Extension of continuous linear maps]:
Consider normed vector spaces ${ E }$ and ${ F, }$ with ${ E }$ complete. Let ${ F _0 \subseteq F }$ be a subspace, and ${ \lambda : F _0 \to E }$ a continuous linear map.
Then ${ \lambda }$ has an extension ${ \overline{\lambda} : \overline{F _0} \to E }$ which is continuous linear. Also ${ \overline{\lambda} }$ is the unique continuous map ${ \overline{F _0} \to E }$ which extends ${ \lambda , }$ and ${ \left\lVert \overline{\lambda} \right\rVert = \lVert \lambda \rVert }.$
Pf: Firstly ${ \overline{F _0} }$ is also a vector subspace of ${ F :}$ Let ${ x, y \in \overline{F _0} }$ and ${ \alpha, \beta \in \mathbb{R} .}$ We are to show ${ \alpha x + \beta y \in \overline{F _0 } .}$ There are sequences ${ (x _n), (y _n) \subseteq F _0 }$ with ${ \lVert x _n - x \rVert \to 0, }$ ${ \lVert y _n - y \rVert \to 0 .}$ Now ${ \lVert (\alpha x _n + \beta y _n) - (\alpha x + \beta y) \rVert \to 0 }$ with sequence ${ (\alpha x _n + \beta y _n) \subseteq F _0 .}$ So ${ \alpha x + \beta y \in \overline{F _0 } .}$
Let ${ x \in \overline{F _0} .}$ (Equivalently, there is a sequence ${ (x _n) \subseteq F _0 }$ with ${ \lVert x _n - x \rVert \to 0 }$). If we show that
- For any sequence ${ (x _n) \subseteq F _0 }$ with ${ \lVert x _n - x \rVert \to 0 ,}$ the limit ${ \lim _{n \to \infty} \lambda (x _n) }$ exists
- For any two sequences ${ (x _n), (y _n) \subseteq F _0 }$ with ${ \lVert x _n - x \rVert \to 0, }$ ${ \lVert y _n - x \rVert \to 0 ,}$ the limits ${ \lim _{n \to \infty} \lambda(x _n) = \lim _{n \to \infty} \lambda(y _n) , }$
the natural extension map
\[{ \overline{\lambda} : \overline{F _0} \to E, }\] \[{ \overline{\lambda}(x) := \left(\begin{align*} &\lim _{n \to \infty} \lambda(x _n), \text{ where } (x _n) \text{ is any sequence } \\ &\text{with } (x _n) \subseteq F _0 \text{ and } \lVert x _n - x \rVert \to 0 \end{align*} \right) }\]is well defined.
These two properties are proved as in the previous result, so we have ${ \overline{\lambda} .}$ We are to show ${ \overline{\lambda} \in L(\overline{F _0}, E) .}$
Linearity of ${ \overline{\lambda} }$ is clear:
Let ${ x , y \in \overline{F _0} .}$ There are sequences ${ (x _n) , (y _n) \subseteq F _0 }$ with ${ \lVert x _n - x \rVert \to 0, }$ ${ \lVert y _n - y \rVert \to 0. }$ Now for any scalars ${ \alpha, \beta \in \mathbb{R}, }$
By the previous result, ${ \overline{\lambda} }$ is uniformly continuous. So ${ \overline{\lambda} \in L(\overline{F _0}, E) .}$
Uniqueness of the continuous extension ${ \overline{F _0} \to E }$ is clear. (Say ${ \lambda ^{’} : \overline{F _0} \to E }$ is another continuous extension of ${ \lambda .}$ Let ${ x \in \overline{F _0} .}$ There is a sequence ${ (x _n) \subseteq F _0 }$ with ${ \lVert x _n - x \rVert \to 0 .}$ Now ${ \lambda ^{’} (x) }$ ${ = \lim _{n \to \infty} \lambda ^{’} (x _n) }$ ${ = \lim _{n \to \infty} \lambda (x _n) }$ ${ = \overline{\lambda} (x) ,}$ as needed).
It is left to show that ${ \left\lVert \overline{\lambda} \right\rVert = \lVert \lambda \rVert . }$ Since
\[{ \left\lVert \overline{\lambda} \right\rVert = \sup _{v \in \overline{F _0}, v \neq 0} \frac{\left\lVert \overline{\lambda}(v) \right\rVert}{\lVert v \rVert} \geq \sup _{v \in F _0, v \neq 0} \frac{\left\lVert \overline{\lambda}(v) \right \rVert}{\lVert v \rVert} = \lVert \lambda \rVert }\]is clear, so we are to show ${ \left\lVert \overline{\lambda} \right\rVert \leq \lVert \lambda \rVert .}$
Let ${ v \in \overline{F _0 } .}$ There is a sequence ${ (x _n) \subseteq F _0 }$ with ${ \lVert x _n - v \rVert \to 0 .}$ As ${ \lVert \lambda (x _n) \rVert \leq \lVert \lambda \rVert \lVert x _n \rVert }$ and ${ \lim _{n \to \infty} \lambda(x _n) = \overline{\lambda}(v), }$ taking ${ n \to \infty }$ gives ${ \left\lVert \overline{\lambda}(v) \right\rVert }$ ${ \leq \lVert \lambda \rVert \lVert v \rVert .}$ So ${ \sup _{v \neq 0} \frac{\lVert \overline{\lambda} (v) \rVert}{\lVert v \rVert} \leq \lVert \lambda \rVert }$ that is ${ \left\lVert \overline{\lambda} \right\rVert \leq \lVert \lambda \rVert , }$ as needed. ${ \blacksquare }$
Consider reals ${ a < b , }$ and a complete normed space ${ E .}$ Recall the space of bounded functions ${ B([a, b], E) }$ with sup norm is complete.
We say ${ f : [a , b] \to E }$ is a step map if there is a partition ${ \mathcal{P} : a = x _0 < x _1 < \ldots < x _n = b }$ of ${ [a , b] }$ and elements ${ c _1, \ldots, c _n \in E }$ such that ${ f(t) = c _i }$ on each interval ${ (x _{i-1}, x _i) .}$
The set ${ \text{St}([a, b], E) }$ of step maps from ${ [a, b] }$ to ${ E }$ is a vector subspace of ${ B([a, b], E). }$ The closure ${ \overline{\text{St}}([a , b], E) }$ is written as
\[{ \text{Reg}([a, b], E) := \overline{\text{St}}([a , b], E) ,}\]and its elements are called regulated maps.
Thm [Characterising regulated maps]:
Consider reals ${ a < b }$ and a complete normed space ${ E .}$ Now the space of regulated maps
\[{ \text{Reg}([a, b], E) = \left\lbrace f \in B([a, b], E) \, : \, \begin{align*} &\lim _{t \to x ^{+}} f(t) \text{ exists } \forall x \in [a, b), \\ &\lim _{t \to x ^{-}} f(t) \text{ exists } \forall x \in (a,b] \end{align*} \right\rbrace . }\]Pf: [Ref: Coleman’s book]
${ \underline{\subseteq} }$: Let ${ f }$ be a regulated map. There is a sequence of step maps ${ (f _n) }$ converging to ${ f .}$
Let ${ x \in [a, b) .}$ We will show ${ \lim _{t \to x ^{+}} f(t) }$ exists. (A similar argument will show existence of left sided limits.)
We are to show
\[{ \text{To show: } \lim _{t \to x ^{+}} f(t) \text{ exists} }\]that is
\[{ \text{To show: } \quad \begin{align*} &\exists \ell \in E \text{ such that } \forall \text{ sequence } (t _n) \text{ in } (x, b] \\ &\text{with } \lim _{n \to \infty} t _n = x \text{ we have } \lim _{n \to \infty} f(t _n) = \ell \end{align*} }\]that is
\[{ \begin{align*} &\text{To show: } \\ &{\color{red}{1)}} \, \forall \text{ sequence } (t _n) \text{ in } (x, b] \text{ converging to } x \text{ we have } \\ &\text{existence of } \lim _{n \to \infty} f(t _n) \\ &{\color{blue}{2)}} \, \forall \text{ sequences } (s _n), (t _n) \text{ in } (x, b] \text{ converging to } x \text{ we have } \\ &\lim _{n \to \infty} f(s _n) = \lim _{n \to \infty} f(t _n) . \end{align*} }\]We will show a uniform-continuity-like condition
\[{ \text{Will show: } \quad \begin{align*} &\forall \varepsilon > 0 \, \exists \delta > 0 \text{ such that } \\ & p, q \in (x, x + \delta) \implies \lVert f(p) - f(q) \rVert < \varepsilon. \end{align*} }\]Note that this guarantees both ${ 1) }$ and ${ 2) }$ above.
For ${ {\color{red}{1)}} }$: Consider a sequence ${ (t _n) }$ in ${ (x, b] }$ with ${ \vert t _n - x \vert \to 0 .}$ We are to show sequence ${ (f(t _n)) }$ is Cauchy.
Let ${ \varepsilon > 0 .}$ Pick a ${ \delta > 0 }$ such that ${ p, q \in (x, x + \delta) \implies \lVert f(p) - f(q) \rVert < \varepsilon. }$ Pick an ${ N }$ such that ${ n \geq N \implies t _n \in (x, x + \delta) .}$ Now
as needed. ${ {\color{red}{\blacksquare}} }$
For ${ {\color{blue}{2)}} }$: Consider sequences ${ (s _n), (t _n) }$ in ${ (x, b] }$ with ${ \vert s _n - x \vert \to 0, }$ ${ \vert t _n - x \vert \to 0 .}$ By ${ 1) , }$ say ${ \lim _{n \to \infty} f(s _n) = \ell _s }$ and ${ \lim _{n \to \infty} f(t _n) = \ell _t .}$ We are to show ${ \ell _s = \ell _t .}$ Since
it suffices to show ${ \lVert f(s _n) - f(t _n) \rVert \to 0 .}$
Let ${ \varepsilon > 0 .}$ Pick a ${ \delta > 0 }$ such that ${ p, q \in (x, x + \delta) \implies \lVert f(p) - f(q) \rVert < \varepsilon. }$ Pick an ${ N }$ such that ${ n \geq N \implies s _n, t _n \in (x, x + \delta) .}$ Now
as needed. ${ {\color{blue}{\blacksquare}} }$
We will show the uniform-continuity-like condition used for proving ${ {\color{red}{1)}} }$ and ${ {\color{blue}{2)}} .}$
\[{ \text{To show: } \quad \begin{align*} &\forall \varepsilon > 0 \, \exists \delta > 0 \text{ such that } \\ & p, q \in (x, x + \delta) \implies \lVert f(p) - f(q) \rVert < \varepsilon. \end{align*} }\]Let ${ \varepsilon > 0 .}$ There is an ${ N }$ such that ${ \lVert f _N - f \rVert < \varepsilon .}$ As ${ f _N }$ is a step map, there is a ${ \delta > 0 }$ such that ${ f _N }$ is constant over ${ (x, x + \delta) .}$ Now for ${ p, q \in (x, x + \delta) , }$
\[{ \require{cancel} \begin{align*} &\lVert f(p) - f(q) \rVert \\ \leq \, &\underbrace{\lVert f(p) - f _N (p) \rVert} _{ < \varepsilon \text{ as } \lVert f _N - f \rVert < \varepsilon} + \underbrace{\cancel{\lVert f _N (p) - f _N (q) \rVert}} _{f _N \text{ constant over } (x , x + \delta) } + \underbrace{\lVert f _N (q) - f (q) \rVert} _{< \varepsilon \text{ as } \lVert f _N - f \rVert < \varepsilon} \\ < \, &2\varepsilon, \end{align*} }\]as needed.
${ \underline{\supseteq} }$: Say ${ f \in B([a, b], E) }$ satisfies the left and right limit criteria. Let ${ \varepsilon > 0 .}$ We are to show there is a step map ${ \phi }$ with ${ \lVert f - \phi \rVert < \varepsilon .}$
It suffices to show
\[{ \text{To show: } \quad \begin{align*} &\exists \text{ a partition } \mathcal{P} = (y _i) \text{ of } [a, b] \text{ such that } \\ &p, q \in (y _{i-1}, y _i) \implies \lVert f(p) - f(q) \rVert < \varepsilon, \end{align*} }\]because then for example the map ${ \phi }$ where
\[{ \phi (t) = \begin{cases} f\left( \frac{y _{i-1} + y _i}{2} \right) &\text{ for } t \in (y _{i-1}, y _i) \\ f(t) &\text{ for } t \in \mathcal{P} \end{cases} }\]works.
For each ${ x \in [a, b), }$ as ${ \lim _{t \to x ^{+} } f(t) }$ exists, there is a ${ \delta _x > 0 }$ such that
\[{ p, q \in (x, x + \delta _x) \implies \lVert f(p) - f(q) \rVert < \varepsilon. }\]For each ${ x \in (a, b], }$ as ${ \lim _{t \to x ^{-}} f(t) }$ exists, there is a ${ \Delta _x > 0 }$ such that
\[{ p, q \in (x - \Delta _x, x) \implies \lVert f(p) - f(q) \rVert < \varepsilon . }\]These give an open cover
\[{ [a, b] = [a, a + \delta _a) \cup (b - \Delta _b, b] \cup \bigcup _{x \in (a, b)} (x - \Delta _x, x + \delta _x). }\]The sets ${ [a, a + \delta _a) }$ and ${ (b - \Delta _b, b] }$ are open in the space ${ [a, b] .}$
As ${ [a, b] }$ is compact, the above open cover has a finite subcover. The finite subcover must have ${ [a, a + \delta _a) }$ and ${ (b - \Delta _b, b] }$: They are the only sets of the cover containing ${ a }$ and ${ b }$ respectively.
So there is a finite subcover
\[{ [a, b] = [a, a + \delta _a) \cup (b - \Delta _b, b] \cup \bigcup _{i=1} ^{n} (x _i - \Delta _{x _i}, x _i + \delta _{x _i}) }\]with ${ x _1, \ldots, x _n \in (a, b) .}$
Taking ${ \mathcal{P} }$ to be the partition produced by ordering the points
\[{ \lbrace a , a + \delta _a; b - \Delta _b, b; x _1 - \Delta _{x _1}, x _1, x _1 + \delta _{x _1} ; \ldots; x _n - \Delta _{x _n}, x _n, x _n + \delta _{x _n} \rbrace }\]works: Say ${ \mathcal{P} = (y _i). }$ Now each interval ${ (y _{i-1}, y _{i}) }$ is contained in a set of the form ${ (x - \Delta _x, x) }$ or ${ (x, x + \delta _x) }$ and hence satisfies
\[{ p, q \in (y _{i-1}, y _i) \implies \lVert f(p) - f(q) \rVert < \varepsilon, }\]as needed. ${ \blacksquare }$
A map ${ f : [a, b] \to E }$ is piecewise continuous if there is a partition ${ \mathcal{P} : a = x _0 < x _1 < \ldots < x _n = b }$ such that for each ${ i }$ the restriction ${ f : (x _{i-1}, x _i) \to E }$ admits a continuous extension ${ f _i : [x _{i-1}, x _i] \to E .}$
By the above result, piecewise continuous maps ${ [a, b] \to E }$ into a complete space are regulated.
We can now consider integrating step maps (This, along with linear extension theorem, will allow us to integrate regulated maps).
Def [Integral of a step map wrt a partition]:
Consider reals ${ a < b, }$ a complete normed space ${ E }$ and a step map ${ f : [a, b] \to E .}$
Say ${ \mathcal{P} : a = x _0 < \ldots < x _n = b }$ is a partition wrt which ${ f }$ is a step map, and ${ f(t) = c _i }$ on each interval ${ t \in (x _{i-1}, x _i) .}$ Then the integral of ${ f }$ wrt ${ \mathcal{P} }$ is defined as
Obs: Consider reals ${ a < b, }$ a complete normed space ${ E }$ and a step map ${ f : [a, b] \to E .}$
Say ${ f }$ is a step map wrt partition ${ \mathcal{P} : a = x _0 < \ldots < x _n = b ,}$ and ${ \hat{x} \in (a, b) .}$
Then ${ f }$ is a step map wrt the partition described by points ${ \mathcal{P} \cup \lbrace \hat{x} \rbrace, }$ and
Pf: If ${ \hat{x} }$ is some ${ x _j }$ then ${ \mathcal{P} \cup \lbrace \hat{x} \rbrace = \mathcal{P} }$ and we are done, so say ${ \hat{x} }$ lies in some ${ (x _{j-1}, x _j) .}$
Say ${ f(t) = c _i }$ on each interval ${ t \in (x _{i-1}, x _i ) .}$ We see ${ f }$ is a step map wrt partition ${ \mathcal{P} \cup \lbrace \hat{x} \rbrace, }$ and
as needed.
Obs: Consider reals ${ a < b, }$ a complete normed space ${ E }$ and a step map ${ f : [a, b] \to E .}$
If ${ f }$ is a step map wrt both partitions ${ \mathcal{P} }$ and ${ \mathcal{Q}, }$ then
Pf: Adding finitely many points to a partition ensures ${ f }$ is still a step map wrt new partition and integral of ${ f }$ wrt new partition is same as the old one.
So ${ f }$ is a step map wrt partition described by points ${ \mathcal{P} \cup \mathcal{Q}, }$ and
as needed.
Def [Integral of a step map]:
Consider reals ${ a < b, }$ a complete normed space ${ E }$ and a step map ${ f : [a, b] \to E .}$
By the above observation, we can define the integral of ${ f }$ as
where ${ \mathcal{P} }$ is any partition wrt which ${ f }$ is a step map.
Obs [Linearity of integral for step maps]:
Consider reals ${ a < b }$ and a complete normed space ${ E .}$ The integration operation on step maps
is continuous linear with ${ \lVert I(f) \rVert \leq (b-a) \lVert f \rVert. }$
Pf: Say ${ f , g : [a, b] \to E }$ are step maps wrt partitions ${ \mathcal{P} _f, \mathcal{P} _g }$ respectively, and ${ \alpha, \beta \in \mathbb{R} .}$
Now ${ \alpha f + \beta g , f, g }$ are step maps wrt the partition described by points ${ \mathcal{P} _f \cup \mathcal{P} _g .}$ Say the partition described by points ${ \mathcal{P} _f \cup \mathcal{P} _g }$ is ${ a = x _0 < \ldots < x _n = b , }$ and ${ f(t) = c _i, g(t) = d _i }$ on each interval ${ t \in (x _{i-1}, x _i) .}$ This gives
so ${ I }$ is linear. Further
\[{ \begin{align*} \lVert I(f) \rVert \leq &\sum _{i=1} ^{n} \lVert c _i \rVert (x _i - x _{i-1}) \\ \leq &\, \lVert f \rVert \sum _{i=1} ^{n} (x _i - x _{i-1}) \\ = &\, \lVert f \rVert (b-a) , \end{align*} }\]so especially ${ I }$ is continuous. ${ \blacksquare }$
Consider reals ${ a < b }$ and a complete normed space ${ E .}$ By linear extension theorem, we can extend the above map to a continuous linear map
\[{ I : \overline{\text{St}}([a, b], E) \longrightarrow E, }\] \[{ I(f) := \left( \begin{align*} &\lim _{n \to \infty} I(f _n), \text{ where } (f _n) \text{ is any sequence} \\ &\text{in } \text{St}([a, b], E) \text{ with } \lVert f _n - f \rVert \to 0 \end{align*} \right) . }\]We write ${ I(f) }$ as
\[{ I(f) = \int _{a} ^{b} f .}\]Obs: Consider reals ${ a < b ,}$ a complete normed space ${ E }$ and a step map ${ f : [a, b] \to E .}$
If ${ c \in [a, b], }$ then both ${ f \vert _{[a, c]} }$ and ${ f \vert _{[c, b]} }$ are step maps and
Pf: Say ${ f : [a, b] \to E }$ is a step map wrt partition ${ \mathcal{P} : a = x _0 < \ldots < x _n = b ,}$ and ${ c \in [a, b] .}$ If ${ c }$ is some ${ x _j }$ the result is clear, so say ${ c }$ lies in some ${ (x _{j-1}, x _j) .}$
Say ${ f(t) = c _i }$ on each interval ${ t \in (x _{i-1}, x _i) .}$ We see ${ f \vert _{[a, b]}, f \vert _{[a, c]} , f \vert _{[c, b]} }$ are step maps wrt partition ${ \mathcal{P} \cup \lbrace c \rbrace }$ restricted to ${ [a, b], [a, c], [c, b] }$ respectively. Also
\[{ \begin{align*} I(f \vert _{[a, b]}) = &\sum _{i=1} ^{n} c _i (x _i - x _{i-1}) \\ = &\sum _{i < j} c _i (x _i - x _{i-1}) + c _j (c - x _{j-1}) \\ &+ c _j (x _j - c) + \sum _{i > j} c _i (x _i - x _{i-1}) \\ = &I(f \vert _{[a, c]}) + I(f \vert _{[c, b]}) \end{align*} }\]as needed.
Obs [Monotonicity of integral for step maps]:
Consider reals ${ a < b , }$ and step maps ${ f, g : [a, b] \to \mathbb{R} .}$ Then
Pf: Say ${ f, g : [a, b] \to \mathbb{R} }$ are step maps wrt partitions ${ \mathcal{P} _f, \mathcal{P} _g }$ respectively. Now both ${ f, g }$ are step maps wrt the partition defined by points ${ \mathcal{P} _f \cup \mathcal{P} _g .}$ Say the partition described by points ${ \mathcal{P} _f \cup \mathcal{P} _g }$ is ${ a = x _0 < \ldots < x _n = b , }$ and ${ f(t) = c _i, g(t) = d _i }$ on each interval ${ t \in (x _{i-1}, x _i) .}$
Let ${ f \leq g . }$ So each ${ c _i \leq d _i .}$ Then
\[{ \begin{align*} I(f) = &\sum _{i=1} ^{n} c _i (x _i - x _{i-1}) \\ \leq &\sum _{i=1} ^{n} d _i (x _i - x _{i-1}) \\ = &\, I(g) , \end{align*} }\]as needed.
Def: Consider reals ${ \alpha < \beta , }$ a complete normed space ${ E }$ and a regulated map ${ f : [\alpha, \beta] \to E .}$
For ${ a \leq b }$ in ${ [\alpha, \beta], }$ note ${ f \vert _{[a, b]} }$ is regulated too. (If step maps ${ f _n }$ are uniformly convergent to ${ f ,}$ the step maps ${ f _n \vert _{[a, b]} }$ are uniformly convergent to ${ f \vert _{[a, b]} }$). We define
Thm: Consider reals ${ \alpha < \beta , }$ a complete normed space ${ E }$ and a regulated map ${ f : [\alpha, \beta] \to E .}$
Then for ${a, b, c \in [\alpha, \beta], }$
Pf: As ${ f }$ is regulated, there are step maps ${ f _n : [\alpha, \beta] \to E }$ uniformly converging to ${ f : [\alpha, \beta] \to E .}$
Consider first the case ${ a \leq c \leq b .}$ So step maps ${ f _n \vert _{[a, b]}, }$ ${ f _n \vert _{[a, c]}, }$ ${ f _n \vert _{[c, b]} }$ are uniformly convergent to ${ f \vert _{[a, b]} , }$ ${ f \vert _{[a, c]} , }$ ${ f \vert _{[c, b]} }$ respectively. We are to show ${ \int _{a} ^{b} f = \int _{a} ^{c} f + \int _{c} ^{b} f }$ that is ${ \lim I(f _n \vert _{[a, b]}) = \lim I(f _n \vert _{[a, c]}) + \lim I(f _n \vert _{[c, b]}), }$ which is true.
Consider the case ${ a \leq b \leq c .}$ We are to show ${ \int _{a} ^{b} f = \int _{a} ^{c} f + \int _{c} ^{b} f }$ that is ${ \lim I(f _n \vert _{[a, b]}) = \lim I(f _n \vert _{[a, c]}) + (-\lim I(f _n \vert _{[b, c]}) ), }$ which is true.
One can verify the truth of ${ \int _{a} ^{b} f = \int _{a} ^{c} f + \int _{c} ^{b} f }$ for the remaining ${ 3! - 2 = 4 }$ orderings of ${ a, b, c \in [\alpha, \beta] }$ similarly.
Thm [Monotonicity of integral]:
Consider reals ${ a < b , }$ and regulated maps ${ f, g : [a, b] \to \mathbb{R} .}$ Now
Pf: As ${ f, g }$ are regulated, so is ${ g - f .}$ The proof of characterisation of ${ \text{Reg}([a, b], E) }$ gives, for every ${ \varepsilon > 0 }$ a specific step map ${ \phi _{\varepsilon} }$ such that ${ \lVert (g - f) - \phi _{\varepsilon} \rVert < \varepsilon .}$
Let ${ g - f \geq 0 . }$ With this, the considered step maps ${ \phi _{\varepsilon} }$ are also ${ \geq 0 .}$
So the step maps ${ \phi _{\frac{1}{n}} }$ are ${ \geq 0 }$ and uniformly convergent to ${ g - f ,}$ giving
\[{ \int _{a} ^{b} (g-f) = \lim _{n \to \infty} I\left(\phi _{\frac{1}{n}}\right) \geq 0 }\]as needed.
Obs [Uniform limit and Integral]:
Consider reals ${ a < b, }$ a complete normed space ${ E ,}$ and a sequence of regulated maps ${ f _n : [a, b] \to E }$ with ${ \lVert f _n - f \rVert \to 0 }$ for some ${ f : [a, b] \to E .}$
Uniform limit of bounded maps is bounded, so ${ f }$ is in ${ B([a, b], E) .}$ As ${ \text{Reg}([a, b], E) }$ is closed in ${ B([a, b], E), }$ the limit ${ f }$ is in ${ \text{Reg}([a, b], E) . }$ As
is continuous linear,
\[{ \int _a ^b f _n \longrightarrow \int _{a} ^{b} f . }\]So uniform limit of regulated maps is regulated, and here integral of limit is limit of integrals.
Thm [Bound for integral]:
Consider reals ${ \alpha < \beta , }$ a complete normed space ${ E }$ and a regulated map ${ f : [\alpha, \beta] \to E . }$
Then for ${ x, c \in [\alpha, \beta], }$
For ${ c \in [\alpha, \beta] ,}$ the map ${ F: [\alpha, \beta] \to E , }$ ${ F(t) = \int _{c} ^{t} f }$ is uniformly continuous.
Pf: As ${ f }$ is regulated, there is a sequence of step maps ${ f _n : [\alpha, \beta] \to E }$ uniformly converging to ${ f : [\alpha, \beta] \to E .}$
i) It suffices to show the inequality for ${ c \leq x }$ case. Note step maps ${ f _n \vert _{[c, x]} }$ uniformly converge to ${ f \vert _{[c, x]} .}$ Each step map ${ f _n \vert _{[c, x]} }$ satisfies
so limit ${ n \to \infty }$ gives
\[{ \left\lVert \int _{c} ^{x} f \right\rVert \leq (x - c) \lVert f \rVert }\]as needed.
ii) The map ${ F: [\alpha, \beta] \to E , }$ ${ F(t) = \int _{c} ^{t} f }$ is uniformly continuous because for all ${ x, y \in [\alpha, \beta], }$
Thm: A map ${ f : [a, b] \to \mathbb{R} ^k , }$ ${ f = (f _1, \ldots, f _k) ^T }$ is regulated if and only if each ${ f _i }$ is regulated. In this case,
\[{ \int _a ^b f = \left( \int _a ^b f _1, \ldots, \int _a ^b f _k \right) .}\]Pf: ${ \underline{\Rightarrow} }$ Say ${ f : [a, b] \to \mathbb{R} ^k , }$ ${ f = (f _1, \ldots, f _k) ^T }$ is regulated. So there are step maps ${ s _n : [a, b] \to \mathbb{R} ^k, }$ ${ s _n = ((s _n) _1, \ldots, (s _n) _k) ^T }$ with ${ \lVert s _n - f \rVert \to 0 .}$
Now for each ${ i = 1, \ldots, k, }$
with ${ (s _n) _i }$ step maps. Especially each ${ f _i }$ is regulated.
Note if ${ s _n : [a, b] \to \mathbb{R} ^k }$ is step map wrt partition ${ \mathcal{P} _n ,}$ then ${ (s _n) _1, \ldots, (s _n) _k : [a, b] \to \mathbb{R} }$ are step maps wrt same partition, and
\[{ I _{\mathcal{P} _n} (s _n) = \left( I _{\mathcal{P} _n} ((s _n) _1), \ldots, I _{\mathcal{P} _n} ((s _n) _k ) \right) }\]that is
\[{ I (s _n) = \left( I ((s _n) _1), \ldots, I ((s _n) _k ) \right) . }\]Since ${ \lVert s _n - f \rVert \to 0 }$ and each ${ \lVert (s _n) _i - f _i \rVert \to 0, }$ taking ${ n \to \infty }$ gives
\[{ \int _a ^b f = \left( \int _a ^b f _1, \ldots, \int _a ^b f _k \right) .}\]${ \underline{\Leftarrow} }$: Say ${ f : [a, b] \to \mathbb{R} ^k , }$ ${ f = (f _1, \ldots, f _k) ^T }$ is such that each ${ f _i }$ is regulated. We are to show ${ f }$ is regulated.
For each ${ f _i }$ there are step maps ${ (s _i) _n : [a, b] \to \mathbb{R} }$ with ${ \lVert (s _i) _n - f _i \rVert \to 0 .}$
If ${ (s _1) _n, \ldots, (s _k) _n : [a, b] \to \mathbb{R} }$ are step maps wrt partitions ${ \mathcal{P} _{1, n}, \ldots, \mathcal{P} _{k,n} }$ then
is a step map wrt partition described by points ${ \mathcal{P} _n := \mathcal{P} _{1, n} \cup \ldots \cup \mathcal{P} _{k, n} .}$ Further,
\[{ \begin{align*} \lVert S _n - f \rVert = &\sup _{x \in [a, b]} \sqrt{\sum _{i=1} ^{k} [(s _i) _n (x) - f _i (x)] ^2 } \\ \leq &\sqrt{\sum _{i=1} ^{k} \lVert (s _i) _n (x) - f _i \rVert ^2 } \to 0 \end{align*} }\]so ${ f }$ is regulated, as needed.
Thm [Integral as antiderivative]:
Consider reals ${ a < b ,}$ a complete normed space ${ E }$ and a regulated map ${ f : [a, b] \to E .}$ Let
If ${ f }$ is continuous at ${ c \in [a, b] , }$ then
\[{ F ^{’} (c) = f(c) . }\]For ${ E = \mathbb{R} }$ it is intuitively clear: Area ${ F(c + h) \approx F(c) + f(c) h . }$
Pf: Say ${ f }$ is continuous at ${ c \in [a, b] .}$
For ${ h }$ such that ${ h \neq 0, }$ ${ c + h \in [a, b] }$ we have
Let ${ \varepsilon > 0 .}$ By continuity at ${ c,}$ there is a ${ \delta > 0 }$ such that
\[{ x \in [a, b], \, \vert x - c \vert < \delta \implies \lVert f(x) - f(c) \rVert < \varepsilon .}\]Recall segment notation ${ [[ \alpha, \beta ]] := \lbrace \alpha + t(\beta - \alpha) : t \in [0, 1] \rbrace .}$ Now
\[{ \begin{align*} \left\lVert \frac{F(c+h) - F(c)}{h} - f(c) \right\rVert = &\, \frac{1}{\vert h \vert} \left\lVert \int _{c} ^{c+h} (f - f(c)) \right\rVert \\ \leq &\sup _{x \in [[c, c+h]]} \lVert f(x) - f(c) \rVert \\ {\color{blue}{\leq}} &\, \varepsilon \end{align*} }\]for ${ h }$ such that ${ h \neq 0, }$ ${ c + h \in [a, b] }$ and ${ {\color{blue}{\vert h \vert < \delta}} . }$ So
\[{ \frac{F(c+h) - F(c)}{h} \longrightarrow f(c) }\]as ${ c + h \in [a, b], }$ ${ h \to 0 .}$ ${ \blacksquare }$
Thm [Integral as the only antiderivative, upto constants]:
Consider reals ${ a < b ,}$ a complete normed space ${ E }$ and a continuous map ${ f : [a, b] \to E .}$ Let
By the above observations, ${ F : [a, b] \to E }$ is a continuous antiderivative of ${ f , }$ that is
\[{ \left\lbrace \begin{align*} &F \text{ is continuous on } [a, b]; \\ &D F (x) = f(x) \text{ for all } x \in (a, b) \end{align*} \right\rbrace .}\]Say ${ \hat{F} : [a, b] \to E }$ is another continuous antiderivative of ${ f . }$ Then ${ F, \hat{F} : [a, b] \to E }$ differ by a constant.
So ${ \int _{a} ^{x} f = \hat{F}(x) + c ,}$ and setting ${ x = a }$ gives
\[{ \int _{a} ^{x} f = \hat{F}(x) - \hat{F}(a) .}\]Especially
\[{ \int _a ^b f = \hat{F}(b) - \hat{F}(a). }\]
Pf: We have ${ D (F - \hat{F}) (x) = 0 }$ for all ${ x \in (a, b) .}$ By the lemma below, ${ F - \hat{F} }$ is constant on ${ (a, b) .}$ But ${ F - \hat{F} }$ is continuous on ${ [a, b], }$ so it is constant on ${ [a, b] . }$
Lem: Consider reals ${ a < b , }$ a complete normed space ${ E }$ and a map ${ f : [a, b] \to E .}$
If ${ Df(t) = 0 }$ for all ${ t \in (a, b) , }$ then ${ f }$ is constant on ${ (a, b) .}$
Pf: Consider points ${ \alpha < \beta }$ in ${ (a, b) .}$ We will show ${ f(\alpha) = f(\beta) .}$
Let ${ \varepsilon > 0 .}$ For every ${ t \in [\alpha, \beta] }$ there is a ${ \delta _t > 0 }$ such that
\[{ \require{cancel} \vert h \vert < \delta _t \implies \lVert f(t + h) - f(t) - \cancel{Df(t) h} \rVert \leq \varepsilon \vert h \vert }\]that is
\[{ x \in (t - \delta _t, t + \delta _t) \implies \lVert f(x) - f(t) \rVert \leq \varepsilon \vert x - t \vert .}\]This is local information for every ${ t \in [\alpha, \beta], }$ but compactness lets us work with it.
Consider the open cover
\[{ [\alpha, \beta] \subseteq \bigcup _{t \in [\alpha, \beta]} (t - \delta _t, t + \delta _t). }\]By compactness of ${ [\alpha, \beta] , }$ it has a finite subcover
\[{ [\alpha, \beta] \subseteq \bigcup _{i=1} ^{n} (t _i - \delta _{t _i}, t _i + \delta _{t _i}) .}\]WLOG no element of the cover is contained in another element of the cover (because if so, the element contained can be removed). Order the centers ${ t _i }$ so that
\[{ \alpha \leq t _1 < \ldots < t _n \leq \beta .}\]Note if some ${ (t _i - \delta _i, t _i + \delta _i) }$ and ${ (t _{i+1} - \delta _{i+1}, t _{i+1} + \delta _{i+1}) }$ do not intersect, that is ${ t _i + \delta _i < t _{i+1} - \delta _{i+1} , }$ then the segment ${ [ t _i + \delta _i, t _{i+1} - \delta _{i+1} ] }$ doesnt intersect any of ${ (t _1 - \delta _1, t _1 + \delta _1), \ldots, (t _n - \delta _n, t _n + \delta _n) , }$ a contradiction.
So for each ${ i = 1, \ldots, n -1 }$ we can pick an ${ {\color{blue}{x _i}} \in (t _{i}, t _{i+1}) }$ in the intersection of ${ (t _i - \delta _i, t _i + \delta _i) }$ and ${ (t _{i+1} - \delta _{i+1}, t _{i+1} + \delta _{i+1}) . }$
This gives
\[{ \alpha \leq t _1 < {\color{blue}{x _1}} < t _2 < \ldots < t _{n-1} < {\color{blue}{x _{n-1}}} < t _n \leq \beta }\]with each ${ \lVert f({\color{blue}{x _i}}) - f(t _i) \rVert \leq \varepsilon ({\color{blue}{x _i}} - t _i) }$ and ${ \lVert f(t _{i+1}) - f({\color{blue}{x _i}}) \rVert \leq \varepsilon (t _{i+1} - {\color{blue}{x _i}}) .}$
Now each term
\[{ \begin{align*} \lVert f(t _{i+1}) - f(t _i) \rVert \leq &\, \lVert f(t _{i+1}) - f({\color{blue}{x _i}}) \rVert + \lVert f({\color{blue}{x _i}}) - f(t _i) \rVert \\ \leq &\, \varepsilon (t _{i+1} - {\color{blue}{x _i}}) + \varepsilon ({\color{blue}{x _i}} - t _i) \\ = &\, \varepsilon (t _{i+1} - t _i), \end{align*} }\]giving
\[{ \begin{align*} &\lVert f(\beta) - f(\alpha) \rVert \\ \leq &\, \lVert f(\beta) - f(t _n) \rVert + \sum _{i=1} ^{n-1} \lVert f(t _{i+1}) - f(t _i) \rVert + \lVert f(t _1) - f(\alpha) \rVert \\ \leq &\, \varepsilon(\beta - t _n) + \sum _{i=1} ^{n-1} \varepsilon (t _{i+1} - t _i) + \varepsilon (t _1 - \alpha) \\ = &\, \varepsilon (\beta - \alpha). \end{align*} }\]As ${ \varepsilon > 0 }$ was arbitrary, ${ f(\beta) = f(\alpha) }$ as needed. ${ \blacksquare }$
Thm [Integrating maps ${ [a, b] \to L(E, F) }$]:
Consider reals ${ a < b , }$ and normed spaces ${ E, F }$ with ${ F }$ complete. (So ${ L(E, F) }$ is complete). Consider a continuous map ${ \alpha : [a, b] \to L(E, F) .}$ Now
acts as
\[{ \left( \int _a ^b \alpha (t) \, dt \right) y = \int _a ^b \alpha(t) y \, dt }\]for all ${ y \in E . }$
Pf: Fix a ${ y \in E . }$ This gives a continuous linear map
Linearity is clear, and continuity is because
\[{ \sup _{\lambda \neq 0} \frac{\lVert T(\lambda) \rVert}{\lVert \lambda \rVert} = \sup _{\lambda \neq 0} \frac{\lVert \lambda(y) \rVert}{\lVert \lambda \rVert} \leq \lVert y \rVert . }\]We have continuous maps
\[{ [a, b] \overset{\alpha}{\longrightarrow} L(E, F) \overset{T}{\longrightarrow} F , }\] \[{ t \longmapsto \alpha(t) \longmapsto \alpha(t) y .}\]We are to show
\[{ \text{To show: } \int _a ^b (T \circ \alpha) (t) \, dt = T \circ \left( \int _a ^b \alpha(t) \, dt \right) . }\]There is a sequence of step maps ${ s _n : [a, b] \to L(E, F) }$ with ${ \lVert s _n - \alpha \rVert \to 0 .}$ Now ${ T \circ s _n : [a, b] \to F }$ are also step maps, and
\[{ \begin{align*} \lVert T\circ s _n - T \circ \alpha \rVert = &\, \sup _{t \in [a, b]} \lVert s _n (t) y - \alpha(t) y \rVert \\ \leq &\, \lVert s _n - \alpha \rVert \lVert y \rVert \to 0 . \end{align*} }\]It suffices to show
\[{ \text{Suffices to show: } \int _a ^b (T \circ s _n) (t) \, dt = T \circ \left( \int _a ^b s _n (t) \, dt \right) , }\]because then letting ${ n \to \infty }$ gives ${ \int (T \circ \alpha) = T \circ \int \alpha .}$
Say ${ s _n : [a, b] \to L(E, F) }$ is a step map wrt partition ${ \mathcal{P} _n : a = x _0 < \ldots < x _N = b ,}$ and ${ s _n (t) = c _i }$ on each interval ${ t \in (x _{i-1}, x _i) .}$
Now ${ T \circ s _n }$ is a step map wrt same partition, and
\[{ \begin{align*} \int _a ^b (T \circ s _n) (t) \, dt = &\sum _{i=1} ^{N} T(c _i) (x _i - x _{i-1}) \\ = &\, \sum _{i=1} ^{N} c _i y \, (x _i - x _{i-1}) \\ = &\, \left( \sum _{i=1} ^{N} c _i (x _i - x _{i-1}) \right) \, y \\ = &\, \left( \int _a ^b s _n (t) \, dt \right) \, y \\ = &\, T \circ \left( \int _a ^b s _n (t) \, dt \right) , \end{align*} }\]as needed. ${ \blacksquare }$
Thm [Integration by parts]:
Let ${ E, F, G }$ be complete normed spaces, and
a continuous bilinear map.
Let ${ I \subseteq \mathbb{R} }$ be an open interval, ${ a, b \in I , }$ and
be ${ C ^1 }$ maps. Then
\[{ \begin{align*} &\int _a ^b f(t) \bullet D g (t) \, dt \\ = &\, f(b) \bullet g(b) - f(a) \bullet g(a) - \int _a ^b Df(t) \bullet g(t) \, dt \end{align*} }\]Pf: By product rule,
\[{ D(f \bullet g)(t) = Df(t) \bullet g(t) + f(t) \bullet Dg(t) . }\]Integrating this gives the required identity.
Eg: Consider the map
\[{ A : [0, \pi] \longrightarrow (L(\mathbb{R} ^2, \mathbb{R} ^2), \lVert \ldots \rVert _{\text{op}}) , }\] \[{ t \longmapsto \begin{pmatrix} \cos(t) &t \\ \sin(t) &t ^2 \end{pmatrix} . }\]We can study ${ A ^{’} (t) }$ and ${ \int _0 ^\pi A(t) \, dt .}$
The map
\[{ A _F : [0, \pi] \longrightarrow (L(\mathbb{R} ^2, \mathbb{R} ^2), \lVert \ldots \rVert _{F}) \cong (\mathbb{R} ^4, \lVert \cdots \rVert _u) , }\] \[{ t \longmapsto \begin{pmatrix} \cos(t) &t \\ \sin(t) &t ^2 \end{pmatrix} }\]can be readily differentiated and integrated due to the results on component wise differentiation and integration of maps ${ [a, b] \to \mathbb{R} ^k :}$
\[{ \begin{align*} &A _F ^{’} (t) = \begin{pmatrix} -\sin(t) &1 \\ \cos(t) &2t \end{pmatrix} \text{ for } t \in (0, \pi) , \\ &\int _0 ^\pi A _F (t) \, dt = \begin{pmatrix} 0 &\pi ^2 /2 \\ 2 &\pi ^3 /3 \end{pmatrix} . \end{align*} }\]But ${ \lVert \cdots \rVert _{\text{op}} }$ and ${ \lVert \cdots \rVert _F }$ are equivalent norms on ${ L(\mathbb{R} ^2, \mathbb{R} ^2) .}$
Let ${ t \in (0, \pi) .}$ As
\[{ \begin{align*} &\frac{\lVert A(t+h) - A(t) - A _F ^{’}(t) h \rVert _{\text{op}} }{\vert h \vert} \\ \leq &\, K \frac{\lVert A(t+h) - A(t) - A _F ^{’}(t) h \rVert _{F} }{\vert h \vert} \to 0 \end{align*} }\]as ${ h \to 0 , }$ we see derivative
\[{ A ^{’} (t) = A _F ^{’} (t) = \begin{pmatrix} -\sin(t) &1 \\ \cos(t) &2t \end{pmatrix} \text{ for } t \in (0, \pi) .}\]There are step maps ${ s _n : [0, \pi] \to (L(\mathbb{R} ^2, \mathbb{R} ^2), \lVert \cdots \rVert _F) }$ with ${ \sup _{t \in [0, \pi]} \lVert s _n (t) - A (t) \rVert _F \to 0 }$ and
\[{ \left\lVert \int _0 ^\pi s _n (t) \, dt - \int _0 ^\pi A _F (t) \, dt \right\rVert _F \longrightarrow 0 . }\]Due to equivalence of norms, ${ \sup _{t \in [0, \pi]} \lVert s _n (t) - A (t) \rVert _{\text{op}} \to 0 }$ and
\[{ \left\lVert \int _0 ^\pi s _n (t) \, dt - \int _0 ^\pi A _F (t) \, dt \right\rVert _{\text{op}} \to 0 . }\]So integral
\[{ \int _0 ^{\pi} A(t) \, dt = \int _0 ^{\pi} A _F (t) \, dt = \begin{pmatrix} 0 &\pi ^2 /2 \\ 2 &\pi ^3 /3 \end{pmatrix} . }\]Back to top.
\[{ \underline{\textbf{Mean value theorem}} }\]Def [${ C ^1 }$ maps]: Let ${ E, F }$ be complete normed spaces, and ${ f : U (\subseteq E \text{ open}) \to F .}$
We say ${ f }$ is ${ C ^1 }$ on ${ U }$ if it is differentiable on ${ U }$ and the derivative
is continuous.
Obs [${ C ^1 }$ maps ${ \mathbb{R} ^n \to \mathbb{R} ^m }$ and partials]:
Consider ${ f : U (\subseteq \mathbb{R} ^n \text{ open}) \to \mathbb{R} ^m .}$
Now ${ f : U \to \mathbb{R} ^m }$ is ${ C ^1 }$ if and only if all partials ${ D _j f _i : U \to \mathbb{R} }$ exist and are continuous.
Pf: ${ \underline{\Rightarrow} }$ Say ${ f : U \to \mathbb{R} ^m }$ is ${ C ^1 . }$ So ${ Df : U \to L(\mathbb{R} ^n, \mathbb{R} ^m ) }$ exists as a continuous map.
By previous observation (on differentiating maps ${ \mathbb{R} ^n \to \mathbb{R} ^m }$), for every ${ x \in U }$ all partials ${ D _j f _i (x) }$ exist and ${ Df(x) }$ looks like
As ${ U \to (L(\mathbb{R} ^n, \mathbb{R} ^m), \lVert \cdots \rVert _{\text{op}}) ,}$ ${ x \mapsto Df(x) }$ is continuous, so is ${ U \to (L(\mathbb{R} ^n, \mathbb{R} ^m), \lVert \cdots \rVert _F ) \cong (\mathbb{R} ^{mn}, \lVert \cdots \rVert _u) ,}$ ${ x \mapsto Df(x) .}$
Hence each component ${ D _j f _i : U \to \mathbb{R} }$ is continuous.
${ \underline{\Leftarrow} }$ Say all partials ${ D _j f _i : U \to \mathbb{R} }$ exist and are continuous. By previous observation (on differentiating maps ${ \mathbb{R} ^n \to \mathbb{R} ^m }$), for every ${ x \in U , }$
\[{ \begin{pmatrix} D _1 f _1 (x) &\cdots &D _n f _1 (x) \\ \vdots &\ddots &\vdots \\ D _1 f _m (x) &\cdots &D _n f _m (x) \end{pmatrix} }\]works as the derivative of ${ f }$ at ${ x .}$ Continuity of components ${ D _j F _i : U \to \mathbb{R} }$ gives continuity of ${ U \to (L(\mathbb{R} ^n, \mathbb{R} ^m), \lVert \cdots \rVert _F ) , }$ ${ x \mapsto Df(x) .}$ Hence ${ U \to (L(\mathbb{R} ^n, \mathbb{R} ^m), \lVert \cdots \rVert _{\text{op}} ) , }$ ${ x \mapsto Df(x) }$ is continuous, that is ${ f }$ is a ${ C ^1 }$ map.
Thm [Mean value theorem]:
Let ${ E, F }$ be complete normed spaces, and ${ f : U (\subseteq E \text{ open}) \to F }$ a ${ C ^1 }$ map.
Consider a segment ${ [[x, y]] = \lbrace x + t(y-x) : t \in [0, 1] \rbrace }$ contained in ${ U .}$ Then
Especially
\[{ \begin{align*} \lVert f(y) - f(x) \rVert \leq &\sup _{t \in [0, 1]} \lVert Df(x + t(y-x)) \rVert \lVert y - x \rVert \\ = &\sup _{v \in [[x, y]]} \lVert Df(v) \rVert \lVert y - x \rVert . \end{align*} }\]As ${ t \mapsto \lVert Df(x + t(y-x)) \rVert }$ is continuous on the compact set ${ [0, 1] , }$ the term
\[{ \sup _{t \in [0, 1]} \lVert Df(x + t(y-x)) \rVert < \infty .}\]
Pf: Consider
\[{ g : [0, 1] \to F, \quad g(t) = f(x + t(y-x)). }\]As ${ g ^{’} (t) = Df(x+t(y-x)) \circ (y-x) }$ is continuous,
\[{ g(1) - g(0) = \int _0 ^1 g ^{’}(t) \, dt }\]that is
\[{ f(y) - f(x) = \int _0 ^1 Df(x+t(y-x)) (y-x) \, dt, }\]as needed.