Let ${ F }$ be a normed space and ${ X }$ a set. For a map ${ X \overset{f}{\to} F },$ let ${ \lVert f \rVert _{\infty} }$ ${ := \sup _{ x \in X} \vert f (x) \vert }$ ${ \underline{ \in [0, \infty] } }.$
${ \lVert \ldots \rVert _{\infty}, }$ when restricted to the space ${ \mathcal{B}(X, F) }$ of bounded functions from ${ X }$ to ${ F, }$ becomes a norm.
Maps ${ X \overset{f _n}{\to} F }$ are uniformly Cauchy if for every ${ \epsilon \gt 0 }$ there is an ${ N }$ such that ${ \lVert f _n - f _m \rVert _{\infty} \lt \epsilon }$ whenever ${ m, n \geq N }.$
On restricting ${ f _n \in \mathcal{B}(X, F) }$ this just means ${ (f _n) }$ is Cauchy wrt ${ \lVert \ldots \rVert _{\infty} }$ norm.
Maps ${ X \overset{f _n}{\to} F }$ are uniformly convergent to ${ X \overset{f}{\to} F }$ if for every ${ \epsilon \gt 0 }$ there is an ${ N }$ such that ${ \lVert f _n - f \rVert _{\infty} \lt \epsilon }$ whenever ${ n \geq N }.$
If ${ F }$ is complete we have a completeness-like result.
Th: Let ${ F }$ be a Banach space (i.e a complete normed space) and ${ X }$ a set. Suppose maps ${ X \overset{f _n}{\to} F }$ are uniformly Cauchy.
Then ${ f _n }$s are uniformly convergent to some ${ X \overset{f}{\to} F }.$ Further :
- If the ${ f _n }$s are bounded, so is ${ f }.$
- If ${ X }$ is a metric space and ${ f _n }$s are continuous, so is ${ f }.$
Pf [Same proof]
i) For each ${ x \in X },$ seq ${ (f _n (x)) }$ is Cauchy (because ${ \vert f _n (x) - f _ m (x) \vert }$ ${ \leq \lVert f _n - f _m \rVert _{\infty} }$) and hence converges to a point, which we’ll call ${ f(x) }.$ We can show ${ f _n }$s are uniformly convergent to ${ f }.$
Let ${ \epsilon \gt 0 }.$ There is an ${ N }$ such that ${ \vert f _n (x) - f _m (x) \vert \lt \epsilon }$ whenever ${ m, n \geq N }$ and ${ x \in X }.$ So for every choice of ${ m \geq N, x \in X }$ we have ${ f _n (x) \in B[f _m (x), \epsilon] }$ ${ := \lbrace t \in F : \vert t - f _m (x) \vert \leq \epsilon \rbrace }$ for all ${ n \geq N }.$
Fix ${ m \geq N, x \in X }$ and take ${ n \to \infty.}$ Now ${ f (x) \in B[f _m (x), \epsilon] }$ for every choice of ${ m \geq N }$ and ${ x \in X }.$ That is, ${ \vert f(x) - f _m (x) \vert \leq \epsilon }$ for all ${ m \geq N }$ and ${ x \in X }.$
ii) Say ${ f _n }$s are also bounded. Pick an ${ N }$ such that ${ \lVert f - f _N \rVert _{\infty} \lt 1 }.$ Now ${ \lVert f \rVert _{\infty} }$ ${ \leq \lVert f - f _N \rVert _{\infty} }$ ${ + \lVert f _N \rVert _{\infty} }$ ${ \leq \lVert f _N \rVert _{\infty} + 1 }$ ${ \lt \infty },$ so ${ f }$ is bounded.
iii) Say ${ (X,d) }$ is a metric space and ${ f _n }$s are continuous. We can show ${ f }$ is continuous.
Let ${ p \in X }$ and ${ \epsilon \gt 0 }.$ Pick ${ N }$ such that ${ \lVert f - f _N \rVert _{\infty} \lt \epsilon }.$ Then pick ${ \delta \gt 0 }$ such that ${ \vert f _N (t) - f _N (p) \vert \lt \epsilon }$ whenever ${ t \in X, d(t,p) \lt \delta }.$
Now ${ \vert f (t) - f(p) \vert }$ ${ \leq \vert f (t) - f _N (t) \vert }$ ${ + \vert f _N (t) - f _N (p) \vert }$ ${ + \vert f _N (p) - f (p) \vert }$ ${ \lt 3\epsilon }$ whenever ${ t \in X, d(t,p) \lt \delta }.$
Cor: Consider a complete normed space ${ F }$ and a metric space ${ X }.$ Now ${ \mathcal{BC} ^{0} (X, F) }$ ${ \subseteq \mathcal{B}(X,F) }$ (i.e space of bounded continuous maps, and space of bounded maps) are both complete under sup norm.
Cor [Weierstrass M-test]: Consider a complete normed space ${ F },$ a set ${ X },$ and maps ${ X \overset{f _n}{\to} F }.$ Suppose there is a seq ${ (a _n) }$ such that ${ \lVert f _n \rVert _{\infty} \leq a _n }$ and ${ \sum _{1} ^{\infty} a _ n \lt \infty }.$
Then ${ \sum _{1} ^{\infty} f _n }$ converges uniformly (i.e the sums ${ S _n (x) := \sum _{1} ^n f _j (x) }$ are uniformly convergent).
Let ${ S _n (x) := \sum _{1} ^n f _j (x) }$ and ${ s _n := \sum _{1} ^{n} a _j }.$ For ${ m \gt n ,}$ ${ \vert S _m (x) - S _n (x) \vert }$ ${ \leq \sum _{n+1} ^{m} \vert f _j (x) \vert }$ ${ \leq \vert s _m - s _n \vert }.$ Also ${ (s _n) }$ is Cauchy. So ${ (S _n (x)) }$ is uniformly Cauchy, hence uniformly convergent.
Cor: Consider ${ \sum _{0} ^{\infty} a _n z ^n }$ with ${ a _n \in \mathbb{C} }.$ Say it converges absolutely at ${ z _0 }.$ Then it converges uniformly on ${ \vert z \vert \leq \vert z _0 \vert }$ (because ${ \vert a _n z ^n \vert }$ ${ \leq \vert a _n \vert \vert z _0 \vert ^n }$ on the set ${ \vert z \vert }$ ${ \leq \vert z _0 \vert },$ and ${ \sum _{0} ^{\infty} \vert a _n \vert \vert z _0 \vert ^n \lt \infty }$).
[Partial summation examples]
Consider ${ \sum _{1} ^{\infty} a _j z ^j }$ with ${ a _j \in \mathbb{C} }.$ Let ${ S _n (z) := \sum _{1} ^{n} a _j z ^j }.$ There are ${ 2 }$ similar ways to sum by parts (both of which will be used shortly):
(Removing ${ a _j }$s) Let ${ s _n := \sum _{1} ^{n} a _j }.$ Now ${ S _n (z) }$ ${ = s _1 z + \sum _{2} ^{n} (s _j - s _{j-1}) z ^j }$ ${ = \sum _{1} ^{n} s _j z ^j - \sum _{1} ^{n-1} s _j z ^{j+1} }$ ${ = {\color{green}{s _n z ^n + \sum _{1} ^{n-1} s _j (z ^j - z ^{j+1}) }} }.$ Especially for ${ m \gt n },$ we have ${ S _m (z) - S _n (z) }$ ${ = s _m z ^m - s _n z ^n }$ ${ + \sum _n ^{m-1} (\underline{s _j - s _n} + s _n) (z ^j - z ^{j+1}) }$ ${ = s _m z ^m - s _n z ^n }$ ${ + \sum _{n} ^{m-1} (s _j - s _n) (z ^j - z ^{j+1}) }$ ${ + s _n (z ^n - z ^m) }.$ That is, ${ S _m (z) - S _n (z) }$ ${ = {\color{purple}{(s _m - s _n) z ^m + \sum _{n} ^{m-1} (s _j - s _n) (z ^j - z ^{j+1})}} }$ for ${ m \gt n }.$
(Removing ${ z ^j }$s) Let ${ s _n (z) := \sum _{1} ^{n} z ^j }$ ${( = z (\frac{z ^n -1}{z-1}) }$ when ${ z \neq 1 ) }.$ Now ${ S _n (z) }$ ${ = a _1 s _1 (z) }$ ${ + \sum _{2} ^{n} a _j (s _j (z) - s _{j-1} (z)) }$ ${ = \sum _{1} ^{n} a _j s _j (z) }$ ${ - \sum _{1} ^{n-1} a _{j+1} s _j (z) }$ ${ = {\color{green}{ a _n s _n (z) + \sum _{1} ^{n-1} (a _j - a _{j+1}) s _j (z)}} }.$ Especially for ${ m \gt n ,}$ we have ${ S _m (z) - S _n (z) }$ ${ = a _m s _m (z) - a _n s _n (z) }$ ${ + \sum _{n} ^{m-1} (a _j - a _{j+1}) (\underline{s _j (z) - s _n (z)} + s _n (z)) }$ ${ = a _m s _m (z) - a _n s _n (z) }$ ${ + \sum _{n} ^{m-1} (a _j - a _{j+1}) (s _j (z) - s _n (z)) }$ ${ + s _n (z) (a _n - a _m) }.$ That is, ${ S _m (z) - S _n (z) }$ ${ = {\color{purple}{a _m (s _m (z) - s _n (z) ) + \sum _{n} ^{m-1} (a _j - a _{j+1}) (s _j (z) - s _n (z))}} }$ for ${ m \gt n }.$
Th [Abel]: Consider ${ \sum _{1} ^{\infty} a _j z ^j }$ with ${ a _j \in \mathbb{C} }.$ Say it converges at ${ z = 1 }$ (so ${ R \geq 1 }$).
Then it converges uniformly on ${ [0,1] }$ (so its continuous on ${ [0,1] },$ giving ${ \lim _{x \to 1 ^- } \sum _{1} ^{\infty} a _j x ^j }$ ${ = \sum _{1} ^{\infty} a _j }$).
Convergence on the full “disc” ${ \vert x \vert \leq 1 }$ isnt assured. For eg, ${ \sum _{1} ^{\infty} \frac{(-1) ^j}{j} x ^j }$ converges at ${ 1 }$ but not at ${ (-1) .}$
Pf: We know ${ s _n := \sum _{1} ^{n} a _j }$ converges. This suggests “removing ${ a _j }$s using ${ s _j }$s”. Let ${ S _n (z) := \sum _{1} ^{n} a _j z ^j }.$ Now ${ S _m (z) - S _n (z) }$ ${ = (s _m - s _n) z ^m }$ ${ + \sum _{n} ^{m-1} (s _j - s _n) (z ^j - z ^{j+1}) }$ for ${ m \gt n }.$
For ${ z \in [0,1] },$ ${ (z ^j - z ^{j+1}) }$ is always ${ \geq 0 }.$ So ${ \vert S _m (x) - S _n (x) \vert }$ ${ \leq \vert s _m - s _n \vert x ^m }$ ${ + \sum _{n} ^{m-1} \vert s _j - s _n \vert (x ^j - x ^{j+1}) }$ whenever ${ m \gt n, }$ ${ x \in [0,1] }.$ We can show ${ (S _n (x)) }$ converge uniformly on ${ [0,1] }.$
Let ${ \epsilon \gt 0 }.$ Pick ${ N }$ such that ${ \vert s _m - s _n \vert }$ ${ \lt \epsilon }$ whenever ${ m,n \geq N }.$ Now ${ \vert S _m (x) - S _n (x) \vert }$ ${ \leq \epsilon x ^m }$ ${ + \sum _{n} ^{m-1} \epsilon (x ^j - x ^{j+1}) }$ ${ = \epsilon x ^m + \epsilon (x ^n - x ^m) }$ ${ \leq \epsilon }$ whenever ${ m \gt n \geq N }$ and ${ x \in [0,1] }.$
Th: Suppose ${ (a _j) \subseteq \mathbb{R} _{\geq 0} }$ with ${ a _j \searrow 0 }.$ Then ${ \sum _{1} ^{\infty} a _j z ^j }$ converges uniformly on ${ D _{\delta} }$ ${ := \lbrace z : \vert z \vert \leq 1, \vert 1-z \vert \geq \delta \rbrace }$ for every ${ \delta \in (0,1) }.$
Convergence at ${ 1 }$ isnt assured. For eg, ${ \sum _{1} ^{\infty} \frac{1}{j} z ^j. }$
Pf: We know ${ a _j \searrow 0 . }$ This suggests “retaining ${ a _j }$s and removing ${ z ^j }$s”. Let ${ S _n (z) := \sum _{1} ^{n} a _j z ^j }$ and ${ s _n (z) := \sum _{1} ^{n} z ^j }.$ So ${ S _m (z) - S _n (z) }$ ${ = a _m (s _m (z) - s _n (z) ) }$ ${ + \sum _{n} ^{m-1} (a _j - a _{j+1}) (s _j (z) - s _n (z)) }$ for ${ m \gt n }.$
Fix ${ \delta \in (0,1) }.$ Note ${ \vert s _n (z) \vert }$ ${ = \vert z (\frac{z ^n -1}{z-1}) \vert }$ ${ \leq \frac{2}{\delta} }$ whenever ${ z \in D _{\delta} }.$ We can prove uniform convergence on ${ D _{\delta} }.$
Let ${ \epsilon \gt 0 }.$ Note ${ \vert S _m (z) - S _n (z) \vert }$ ${ \leq a _m (\frac{4}{\delta}) }$ ${ + \sum _{n} ^{m-1} (a _j - a _{j+1}) (\frac{4}{\delta}) }$ ${ = a _n (\frac{4}{\delta}) }$ whenever ${ m \gt n, }$ ${ z \in D _{\delta} }.$ Pick ${ N }$ such that ${ a _n (\frac{4}{\delta}) \lt \epsilon }$ whenever ${ n \geq N }.$ Now ${ \vert S _m (z) - S _n (z) \vert }$ ${ \lt \epsilon }$ whenever ${ m \gt n \geq N }$ and ${ z \in D _{\delta} }.$
Cor: For ${ \Delta \in (0, \pi) },$ ${ \sum _{1} ^{\infty} \frac{\cos(jx)}{j} , }$ ${ \sum _{1} ^{\infty} \frac{\sin(jx)}{j} }$ converge uniformly on ${ [\Delta, 2\pi - \Delta] }.$
Pick ${ \delta \in (0,1) }$ such that the arc ${ \lbrace e ^{ix} : x \in [\Delta, 2\pi - \Delta] \rbrace }$ is contained in ${ D _{\delta} }.$ We can now prove uniform convergence.
Let ${ \epsilon \gt 0 }.$ Pick ${ N }$ such that ${ \vert \sum _{n+1} ^{m} \frac{1}{j} z ^j \vert \lt \epsilon }$ whenever ${ m \gt n \geq N }$ and ${ z \in D _{\delta} }.$ Now ${ \vert \sum _{n+1} ^{m} \frac{1}{j} \cos(jx) \vert }$ ${ \leq \vert \sum _{n+1} ^{m} \frac{1}{j} (e ^{ix})^j \vert }$ ${ \lt \epsilon }$ whenever ${ m \gt n \geq N }$ and ${ x \in [\Delta, 2\pi - \Delta] }.$
[Applying ${ \frac{d}{dx} }$ or ${ \int }$ to ${ \lim _{n \to \infty} f _n }$]
Consider continuous maps ${ [a,b] \overset{f _n}{\to} \mathbb{R} }$ which are uniformly convergent. Now ${ f }$ ${:= \lim _{n \to \infty} f _n }$ is continuous, and ${ \vert \int _{a} ^{b} f _n - \int _{a} ^{b} f \vert }$ ${ \leq \lVert f _n - f \rVert _{\infty} (b-a) }$ ${ \to 0 }$ as ${ n \to \infty }.$ So ${ \lim _{n \to \infty} \int _{a} ^{b} f _n }$ ${ = \int _{a} ^{b}(\lim _{n \to \infty} f _n) }.$
Th: Consider ${ \mathcal{C} ^1 }$ maps ${ [a,b] \overset{f _n}{\to} \mathbb{R} }.$ Suppose derivatives ${( f _n ‘ ) }$ converge uniformly, and ${ (f _n (x _0) ) }$ converges at some ${ x _0 \in [a,b] }.$
Then even ${ (f _n) }$ converge uniformly, and the limit has a derivative ${ \frac{d}{dx} (\lim _{n \to \infty} f _n) }$ ${ = \lim _{n \to \infty} \frac{d}{dx} f _n }.$
Pf: We have ${ f _n (x) }$ ${ = f _n (x _0) + \int _{x _0} ^{x} f’ _n }$ whenever ${ x \in [a,b] }.$
For every choice of ${ x \in [a,b] },$ since ${ f _n (x _0) \to \lim f _n (x _0) }$ and ${ \int _{x _0} ^{x} f’ _n }$ ${ \to \int _{x _0} ^{x} (\lim f’ _n) },$ we get ${ f(x) := \lim _{n \to \infty} f _n (x) }$ exists. Fixing ${ x \in [a,b] }$ and taking ${ n \to \infty },$ ${ f(x) = f(x _0) }$ ${ + \int _{x _0} ^{x} (\lim f’ _n) }$ for every choice of ${ x \in [a,b] }.$
i) ${ f _n }$ converges uniformly to ${ f },$ because ${\lVert f _n - f \rVert _{\infty} }$ ${ = \lVert (f _n (x _0) - f(x _0)) + \int _{x _0} ^{x} (f’ _n - \lim f’ _n ) \rVert _{\infty} }$ ${ \leq \vert f _n (x _0) - f _n (x _0) \vert + (b-a) \lVert f’ _n - \lim f’ _n \rVert _{\infty} }$ ${ \to 0 }$ as ${ n \to \infty }.$
ii) As ${ f(x) = f(x _0) + \int _{x _0} ^{x} (\lim f’ _n) }$ for all ${ x \in [a,b] },$ FTC gives ${ f’(x) = (\lim f’ _n ) (x) }$ all ${ x \in [a,b] }.$ That is, ${ (\lim f _n)’ = \lim f’ _n .}$