ROUGH NOTES (!)
Integration of maps ${ f : [a, b] \to E }$; Mean value theorem
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\[{ \underline{\textbf{Integration of maps } f : [a, b] \to E} }\]The goal is to study 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). Link: Link. 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.
We can ask ourselves: How can we characterise the regulated maps ${ f : [a, b] \to E }$?
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
\[{ I(f) = \int _{a} ^{b} f .}\]We can ask ourselves: How does the integral behave with the usual operations on functions?
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
\[{ \boxed{ \int _a ^b f = \hat{F}(b) - \hat{F}(a) } }\]
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. ${ \blacksquare }$
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
\[{ \int _0 ^{\pi} A(t) \, dt = \int _0 ^{\pi} A _F (t) \, dt = \begin{pmatrix} 0 &\pi ^2 /2 \\ 2 &\pi ^3 /3 \end{pmatrix} . \, \, \blacksquare }\]Def: Let ${ \mathcal{P} = (\alpha _0, \ldots, \alpha _n) }$ be a partition of ${ [\alpha, \beta] . }$ Now
\[{ \Delta _{\mathcal{P}} = \max _{1 \leq j \leq n} (\alpha _j - \alpha _{j-1}) }\]is called the mesh of the partition, and any element ${ \xi _j \in [\alpha _{j-1}, \alpha _j] }$ is called a between point.
Thm [Regulated integral agrees with the Riemann integral]:
Let ${ E }$ be a complete normed space, and ${ f : [\alpha, \beta] \to E }$ a regulated map.
Then for every ${ \varepsilon > 0 }$ there is a ${ \delta > 0 }$ such that
for all partitions ${ \mathcal{P} = (\alpha _0, \ldots, \alpha _n) }$ of ${ [\alpha, \beta] }$ of mesh ${ \Delta _{\mathcal{P}} < \delta }$ and all between points ${ \xi _j \in [\alpha _{j-1}, \alpha _j] . }$
Pf: (i) Suppose first ${ f : [\alpha, \beta] \to E }$ is a step map. More specifically ${ f }$ is a step map wrt the partition ${ \hat{\mathcal{P}} = (\hat{\alpha} _0, \ldots, \hat{\alpha} _{\hat{n}}) }$ and ${ e _j = f \vert (\hat{\alpha} _{j-1}, \hat{\alpha} _j) . }$
Let ${ \varepsilon > 0 . }$ It suffices to find a ${ \delta > 0 }$ such that all Riemann sums of mesh less than ${ \delta }$ are within ${ \varepsilon }$ of ${ \int _{\alpha} ^{\beta} f \, dx . }$
Consider a partition ${ \mathcal{P} = (\alpha _0, \ldots, \alpha _n) }$ and between points ${ \xi _j \in [\alpha _{j-1}, \alpha _j]. }$
Now for ${ (\beta _0, \ldots, \beta _m) }$ ${ = \mathcal{P} \cup \hat{\mathcal{P}} }$ we have
where
\[{ {\begin{aligned} &\, e _k ^{’} : = f \vert ( \beta _{k-1}, \beta _k ) , \, \text{ and } \\ &\, e _k ^{’’} := f(\xi _j)\text{ where } j \text{ is such that } [\beta _{k-1}, \beta _k] \subseteq [\alpha _{j-1}, \alpha _j] . \end{aligned}} }\]Note that ${ e _k ^{’} = e _k ^{’’} }$ except perhaps at partition points ${ \lbrace \hat{\alpha} _0, \ldots, \hat{\alpha} _{\hat{n}} \rbrace . }$ Hence the above sum
\[{ \sum _{k=1} ^{m} (e _k ^{’} - e _k ^{’’}) (\beta _k - \beta _{k-1}) }\]has atmost ${ 2 \hat{n} }$ nonzero terms. For each such term we have
\[{ \vert (e _k ^{’} - e _k ^{’’} ) (\beta _k - \beta _{k-1}) \vert \leq 2 \lVert f \rVert _{\infty} \Delta _{\mathcal{P}} , }\]and hence
\[{ \left\vert \int _{\alpha} ^{\beta} f \, dx - \sum _{j=1} ^{n} f(\xi _j) (\alpha _j - \alpha _{j-1}) \right\vert \leq 4 \hat{n} \lVert f \rVert _{\infty} \Delta _{\mathcal{P}} . }\]Choosing ${ \delta := \varepsilon / (4 \hat{n} \lVert f \rVert _{\infty}), }$ we see
\[{ \Delta _{\mathcal{P}} < \delta \implies \left\vert \int _{\alpha} ^{\beta} f \, dx - \sum _{j=1} ^{n} f(\xi _j) (\alpha _j - \alpha _{j-1}) \right\vert < \varepsilon }\]as needed.
(ii) Now say ${ f : [\alpha, \beta] \to E }$ is a regulated map. Let ${ \varepsilon > 0 . }$
There is a step map ${ g : [\alpha, \beta] \to E }$ such that ${ \lVert f - g \rVert _{\infty} < \varepsilon . }$ From (i), for every partition ${ \mathcal{P} = (\alpha _0, \ldots, \alpha _n) }$ with mesh ${ \Delta _{\mathcal{P}} < \delta }$ and ${ \xi _j \in [\alpha _{j-1}, \alpha _j] }$ we have
Hence
\[{ {\begin{aligned} &\, \left\vert \int _{\alpha} ^{\beta} f \, dx - \sum _{i=1} ^{n} f(\xi _j) (\alpha _j - \alpha _{j-1}) \right\vert \\ \leq &\, \left\vert \int _{\alpha} ^{\beta} (f-g) \, dx \right\vert \\ + &\, \left\vert \int _{\alpha} ^{\beta} g \, dx - \sum _{i=1} ^{n} g(\xi _j) (\alpha _j - \alpha _{j-1}) \right\vert \\ + &\, \left\vert \sum _{i=1} ^{n} (g(\xi _j) - f(\xi _j)) (\alpha _j - \alpha _{j-1}) \right\vert \\ \leq &\, \varepsilon (\beta - \alpha) + \varepsilon + \varepsilon (\beta - \alpha) \end{aligned}} }\]as needed. ${ \blacksquare }$
Back to top.
\[{ \underline{\textbf{Mean value theorem}} }\]The goal is to study mean value theorem for maps ${ f : U (\subseteq E \text{ open}) \to F . }$
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.