# 3.3. real analysisÂ¶

Some notes on real analysis, following the textbook Understanding analysis

## 3.3.1. ch 1 - the real numbersÂ¶

there is no rational number whose square is 2 (proof by contradiction)

*contrapositive*: $\(-q \to -p\)$ - logically equivalent*triangle inequality*: \(\|a+b\| \leq \|a\| + \|b\|\) (often use |a-b| = |(a-c)+(c-b)|)*axiom of completeness*- every nonempty set \(A \subseteq \mathbb{R}\) that is bounded above has a least upper bounddoesnâ€™t work for \(\mathbb{Q}\)

*supremum*= supA = least upper bound (similarly,*infimum*)supA is an upper bound of A

if \(s \in \mathbb{R}\) is another u.b. then \(s \geq supA\)

can be restated as \(\forall \epsilon > 0, \exists a \in A\) \(s-\epsilon < a\)

*nested interval property*- for each \(n \in N\), assume we are given a closed interval \(I_n = [a_n,b_n]=\{ x \in \mathbb{R} : a_n \leq x \leq b_n \}\) Assume also that each \(I_n\) contains \(I_{n+1}\). Then, the resulting nested sequence of nonempty closed intervals \(I_1 \supseteq I_2 \supseteq ...\) has a nonempty intersection use AoC with x = sup{\(a_n: n \in \mathbb{N}\)} in the intersection of all sets*archimedean property*\(\mathbb{N}\) is unbounded above (sup \(\mathbb{N}=\infty\))

\(\forall x \in \mathbb{R}, x>0, \exists n \in \mathbb{N}, 0 < \frac{1}{n} < x\)

\(\mathbb{Q}\) is dense in \(\mathbb{R}\) - for every \(a,b \in \mathbb{R}, a<b\), \(\exists r \in \mathbb{Q}\) s.t. \(a<r<b\)

pf: want \(a < \frac{m}{n} < b\)

by Archimedean property, want \(\frac{1}{n} < b-a\)

corollary: the irrationals are dense in \(\mathbb{R}\)

there exists a real number \(r \in \mathbb{R}\) satisfying \(r^2 = 2\)

pf: let r = \(sup \{ t \in \mathbb{R} : t^2 < 2 \}\). disprove \(r^2<2, r^2>2\) by considering \(r+\frac{1}{n},r-\frac{1}{n}\)

A ~ B if there exists f:A->B that is 1-1 and onto

A is

*finite*- there exists n \(\in \mathbb{N}\) s.t. \(\mathbb{N}_n\)~A*countable*= \(\mathbb{N}\)~A.uncountable - inifinite set that isnâ€™t countable

Q is countable

pf: Let \(A_n = \{ \pm \frac{p}{q}:\) where p,q \(\in \mathbb{N}\) are in lowest terms with p+q=n}

R is uncountable

pf: Assume we can enumerate \(\mathbb{R}\) Use NIP to exclude one point from \(\mathbb{R}\) each time. The intersection is still nonempty, so we didnâ€™t succesfully enumerate \(\mathbb{R}\)

\(\frac{x}{x^2-1}\) maps (0,1) \(\to \mathbb{R}\)

countable union of countable sets is countable

if \(A \subseteq B\) and B countable, then A is either countable or finite

if \(A_n\) is a countable set for each \(n \in \mathbb{N}\), then their union is countable

the open interval (0,1) = \(\{ x \in \mathbb{R} : 0 < x < 1 \}\) is uncountable

pf: diagonalization - assume there exists a function from (0,1) to \(\mathbb{R}\). List the decimal expansions of these as rows of a matrix. Complement of diagonal does not exist.

*cantorâ€™s thm*- Given any set A, there does not exist a function f:\(A \to P(A)\) that is ontoP(A) is the set of all subsets of A

## 3.3.2. ch 2 - sequences and seriesÂ¶

a sequence \((a_n)\)

*converges*to a real number if \(\forall \epsilon > 0, \exists N \in \mathbb{N}\) such that \(\forall n\geq N, \|a_n-a\| < \epsilon\)otherwise it

*diverges*

if a limit exists, it is unique

a sequence \((x_n)\) is

*bounded*if there exists a number M > 0 such that \(\|x_n\|\leq M \forall n \in \mathbb{N}\)every convergent sequence is bounded

*algebraic limit thm*- let lim \(a_n = a\) and lim \(b_n\) = b. Thenlim(\(ca_n\)) = ca

lim(\(a_n+b_n\)) = a+b

lim(\(a_n b_n\)) = ab

lim(\(a_n/b_n\)) = a/b, provided b \(\neq\) 0

pf 3: use triangle inequality, \(\|a_nb_n-ab\|=\|a_nb_n-ab_n+ab_n-ab\|=...=\|b_n\|\|a_n-a\|+\|a\|\|b_n-b\|\)

pf 4: show \((b_n) \to b\) implies \((\frac{1}{b_n}) \to \frac{1}{b}\)

*order limit thm*- Assume lim \(a_n = a\) and lim \(b_n\) = b.If \(a_n \geq 0\) \(\forall n \in \mathbb{N}\), then \(a \geq 0\)

If \(a_n \leq b_n\) \(\forall n \in \mathbb{N}\), then \(a \leq b\)

If \(\exists c \in \mathbb{R}\) for which \(c \leq b_n\) \(\forall n \in \mathbb{N}\), then \(c \leq b\)

pf 1: by contradiction

*monotone*- increasing or decreasing (not strictly)*monotone convergence thm*- if a sequence is monotone and bounded, then it converges*convergence of a series*define \(s_m=a_1+a_2+...+a_m\)

\(\sum_{n=1}^\infty a_n\) converges to A \(\iff (s_m)\) converges to A

*cauchy condensation test*- suppose \(a_n\) is decreasing and satisfies \(a_n \geq 0\) for all \(n \in \mathbb{N}\). Then, the series \(\sum_{n=1}^\infty a_n\) converges iff the series \(\sum_{n=1}^\infty 2^na_{2^n}\) converges*p-series*\(\sum_{n=1}^\infty 1/n^p\) converges iff p > 1

### 3.3.2.1. 2.5Â¶

let \((a_n)\) be a sequence and \(n_1<n_2<...\) be an increasing sequence of natural numbers. Then \((a_{n_1},a_{n_2},...)\) is a

*subsequence*of \((a_n)\)subsequences of a convergent sequence converge to the same limit as the original sequence

can be used as divergence criterion

*bolzano-weierstrass thm*- every bounded sequence contains a convergent subsequencepf: use NIP, keep splitting interval into two

### 3.3.2.2. 2.6Â¶

\((a_n)\) is a

*cauchy sequence*if \(\forall \epsilon > 0, \exists N \in \mathbb{N}\) such that \(\forall m,n\geq N, \|a_n-a_m\| < \epsilon\)*cauchy criterion*- a sequence converges \(\iff\) it is a cauchy sequencecauchy sequences are bounded

overview: AoC \(\iff\) NIP \(\iff\) MCT \(\iff\) BW \(\iff\) CC

### 3.3.2.3. 2.7Â¶

*algebraic limit thm*- let \(\sum_{n=1}^\infty a_n\) = A, \(\sum_{n=1}^\infty b_n\) = B\(\sum_{n=1}^\infty ca_n\) = cA

\(\sum_{n=1}^\infty a_n+b_n\) = A+B

*cauchy criterion for series*- series converges \(\iff\) \((s_m)\) is a cauchy sequence

if the series \(\sum_{n=1}^\infty a_n\) converges then lim \(a_n=0\)

*comparison test**geometric series*- \(\sum_{n=0}^\infty a r^n = \frac{a}{1-r}\)\(s_m = a+ar+...+ar^{m-1} = \frac{a(1-r^m)}{1-r}\)

*absolute convergence test**alternating series test*decreasing

lim \(a_n\) = 0

then, \(\sum_{n=1}^\infty (-1)^{n+1} a_n\) converges

rearrangements: there exists one-to-one correspondence

if a series converges absolutely, any rearrangement converges to same limit

## 3.3.3. ch 3 - basic topology of RÂ¶

### 3.3.3.1. 3.1 cantor setÂ¶

C has small length, but its cardinality is uncountable

discussion of dimensions, doubling sizes leads to 2^dimension sizes

Cantor set is about dimension .631

### 3.3.3.2. 3.2 open/closed setsÂ¶

A set O \(\subseteq \mathbb{R}\) is

*open*if for all points a \(\in\) O there exists an \(\epsilon\)-neighborhood \(V_{\epsilon}(a) \subseteq O\)\(V_{\epsilon}(a)=\{ x \in R : \|x-a\| < \epsilon\)}

the union of an arbitrary collection of open sets is open

the intersection of a finite collection of open sets is open

a point x is a

*limit point*of a set A if every \(\epsilon\)-neighborhood \(V_{\epsilon}(x)\) of x intersects the set A at some point other than xa point x is a limit point of a set A if and only if x = lim \(a_n\) for some sequence (\(a_n\)) contained in A satisfying \(a_n \neq x\) for all n \(\in\) N

*isolated point*- not a limit pointset \(F \subseteq \mathbb{R}\)

*closed*- contains all limit points*closed*iff every Cauchy sequence contained in F has a limit that is also an element of F

density of Q in R - for every \(y \in \mathbb{R}\), there exists a sequence of rational numbers that converges to y

*closure*- set with its limit pointsclosure \(\bar{A}\) is smallest closed set containing A

iff set open, complement is closed

R and \(\emptyset\) are both open and closed

the union of a finite collection of closed sets is closed

the intersection of an arbitrary collection of closed sets is closed

### 3.3.3.3. 3.3Â¶

a set K \(\subseteq \mathbb{R}\) is

*compact*if every sequence in K has a subsequence that converges to a limit that is also in K*Nested Compact Set Property*- intersection of nested sequence of nonempty compact sets is not emptylet A \(\subseteq \mathbb{R}\).

*open cover*for A is a (possibly infinite) collection of open sets whose union contains the set A.given an open cover for A, a

*finite subcover*is a finite sub-collection of open sets from the original open cover whose union still manages to completely contain A*Heine-Borel thm*- let K \(\subseteq \mathbb{R}\). All of the following are equivalentK is compact

K is closed and bounded

every open cover for K has a finite subcover

## 3.3.4. ch 4 - functional limits and continuityÂ¶

### 3.3.4.1. 4.1Â¶

dirichlet function: 1 if r \(\in \mathbb{Q}\) 0 otherwise

### 3.3.4.2. 4.2 functional limitsÂ¶

def 1. Let f:\(A \to R\), and let c be a limit point of the domain A. We say that

*\(lim_{x \to c} f(x) = L\)*provided that for all \(\epsilon\) > 0, there exists a \(\delta\) > 0 s.t. whenever 0 < |x-c| < \(\delta\) (and x \(\in\) A) it follows that |f(x)-L|< \(\epsilon\)def 2. Let f:\(A \to R\), and let c be a limit point of the domain A. We say that \(lim_{x \to c} f(x) = L\) provided that for every \(\epsilon\)-neighborhood \(V_{\epsilon}(L)\) of L, there exists a \(\delta\)-neighborhood \(V_{\delta}(\)c) around c with the property that for all x \(\in V_{\delta}(\)c) different from c (with x \(\in\) A) it follows that f(x) \(\in V_{\epsilon}(L)\).

*sequential criterion for functional limits*- Given function f:\(A \to R\) and a limit point c of A, the following 2 statements are equivalent:\(lim_{x \to c} f(x) = L\)

for all sequences \((x_n) \subseteq\) A satisfying \(x_n \neq\) c and \((x_n) \to c\), it follows that \(f(x_n) \to L\).

*algebraic limit thm for functional limits**divergence criterion for functional limits*

### 3.3.4.3. 4.3 continuous functionsÂ¶

a function f:\(A \to R\) is

*continuous at a point*c \(\in\) A if, for all \(\epsilon\)>0, there exists a \(\delta\)>0 such that whenever |x-c|<\(\delta\) (and x\(\in\) A) it follows that \(\|f(x)-f( c)\|<\epsilon\). F is*continous*if it is continuous at every point in the domain Acharacterizations of continuouty

criterion for discontinuity

algebraic continuity theorem

if f is continuous at c and g is continous at f( c) then g \(\circ\) f is continuous at c

### 3.3.4.4. 4.4 continuous functions on compact setsÂ¶

*preservation of compact sets*- if f continuous and K compact, then f(K) is compact as well*extreme value theorem*- if f if continuous on a compact set K, then f attains a maximum and minimum value. In other words, there exist \(x_0,x_1 \in K\) such that \(f(x_0) \leq f(x) \leq f(x_1)\) for all x \(\in\) Kf is

*uniformly continuous on A*if for every \(\epsilon\)>0, there exists a \(\delta\)>0 such that for all x,y \(\in\) A, \(\|x-y\| < \delta \implies \|f(x)-f(y)\| < \epsilon\)a function f fails to be uniformly continuous on A iff there exists a particular \(\epsilon_o\) > 0 and two sequences \((x_n),(y_n)\) in A sastisfying \(\|x_n - y_n\| \to 0\) but \(\|f(x_n)-f(y_n)\| \geq \epsilon_o\)

a function that is continuous on a compact set K is

*uniformly continuous on K*

### 3.3.4.5. 4.5 intermediate value theoremÂ¶

*intermediate value theorem*- Let f:[a,b]\( \to R\) be continuous. If L is a real number satisfying f(a) < L < f(b) or f(a) > L > f(b), then there exists a point c \(\in (a,b)\) where f( c) = La function f has the

*intermediate value property*on an inverval [a,b] if for all x < y in [a,n] and all L between f(x) and f(y), it is always possible to find a point c \(\in (x,y)\) where f( c)=L.

## 3.3.5. ch 5 - the derivativeÂ¶

### 3.3.5.1. 5.2 derivatives and the intermediate value propertyÂ¶

let g: A -> R be a function defined on an interval A. Given c \(\in\) A, the

*derivative of g at c*is defined by gâ€™( c) = \(\lim_{x \to c} \frac{g(x) - g( c)}{x-c}\), provided this limit exists. Then g is differentiable at c. If gâ€™ exists for all points in A, we say g is*differentiable*on Aidentity: \(x^n-c^n = (x-c)(x^{n-1}+cx^{n-2}+c^2x^{n-3}+...+c^{n-1}\))

differentiable \(\implies\) continuous

*algebraic differentiability theorem*adding

scalar multiplying

product rule

quotient rule

*chain rule*: let f:A-> R and g:B->R satisfy f(A)\(\subseteq\) B so that the composition g \(\circ\) f is defined. If f is differentiable at c in A and g differentiable at f( c) in B, then g \(\circ\) f is differnetiable at c with (g\(\circ\)f)â€™( c)=gâ€™(f( c))*fâ€™( c)*interior extremum thm*- let f be differentiable on an open interval (a,b). If f attains a maximum or minimum value at some point c \(\in\) (a,b), then fâ€™( c) = 0.*Darbouxâ€™s thm*- if f is differentiable on an interval [a,b], and a satisfies fâ€™(a) < \(\alpha\) < fâ€™(b) (or fâ€™(a) > \(\alpha\) > fâ€™(b)), then there exists a point c \(\in (a,b)\) where fâ€™( c) = \(\alpha\)derivative satisfies intermediate value property

### 3.3.5.2. 5.3 mean value theoremsÂ¶

*mean value theorem*- if f:[a,b] -> R is continuous on [a,b] and differentiable on (a,b), then there exists a point c \(\in\) (a,b) where \(f'( c) = \frac{f(b)-f(a)}{b-a}\)*Rolleâ€™s thm*- f(a)=f(b) -> fâ€™( c)=0if fâ€™(x) = 0 for all x in A, then f(x) = k for some constant k

if f and g are differentiable functions on an interval A and satisfy fâ€™(x) = gâ€™(x) for all x \(\in\) A, then f(x) = g(x) + k for some constant k

*generalized mean value theorem*- if f and g are continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c \(\in (a,b)\) where |f(b)-f(a)|gâ€™( c) = |g(b)-g(a)|fâ€™( c). If gâ€™ is never 0 on (a,b), then can be restated \(\frac{f'( c)}{g'( c)} = \frac{f(b)-f(a)}{g(b)-g(a)}\)given g: A -> R and a limit point c of A, we say that

*\(lim_{x \to c} g(x) = \infty\)*if, for every M > 0, there exists a \(\delta\)> 0 such that whenever 0 < |x-c| < \(\delta\) it follows that g(x) â‰¥ M*Lâ€™Hospitalâ€™s Rule: 0/0*- let f and g be continuous on an interval containing a, and assume f and g are differentiable on this interval with the possible exception of the point a. If f(a) = g(a) = 0 and gâ€™(x) â‰ 0 for all x â‰ a, then \(lim_{x \to a} \frac{f'(x)}{g'(x)} = L \implies lim_{x \to a} \frac{f'(x)}{g'(x)} = L\)*Lâ€™Hospitalâ€™s Rule: \(\infty / \infty\)*- assume f and g are differentiable on (a,b) and gâ€™(x) â‰ 0 for all x in (a,b). If \(lim_{x \to a} g(x) = \infty \), then \(lim_{x \to a} \frac{f'(x)}{g'(x)} = L \implies lim_{x \to a} \frac{f'(x)}{g'(x)} = L\)

## 3.3.6. ch 6 - sequences and series of functionÂ¶

### 3.3.6.1. 6.2 uniform convergence of a sequence of functionsÂ¶

for each n \(\in \mathbb{N}\) let \(f_n\) be a function defined on a set A\(\subseteq R\). The sequence (\(f_n\)) of functions

*converges pointwise*on A to a function f if, for all x in A, the sequence of real numbers \(f_n(x)\) converges to f(x)let (\(f_n\)) be a sequence of functions defined on a set A\(\subseteq\)R. Then (\(f_n\))

*converges unformly*on A to a limit function f defined on A if, for every \(\epsilon\)>0, there exists an N in \(\mathbb{N}\) such that \(\forall n â‰¥N, x \in A , \|f_n(x)-f(x)\|<\epsilon\)*Cauchy Criterion*for uniform convergence - a sequence of functions \((f_n)\) defined on a set A \(\subseteq\) R converges uniformly on A iff \(\forall \epsilon > 0 \exists N \in \mathbb{N}\) s.t. whenever m,n â‰¥N and x in A, \(\|f_n(x)-f_m(x)\|<\epsilon\)

*continuous limit thm*- Let (\(f_n\)) be a sequence of functions defined on A that converges uniformly on A to a function f. If each \(f_n\) is continuous at c in A, then f is continuous at c

### 3.3.6.2. 6.3 uniform convergence and differentiationÂ¶

*differentiable limit theorem*- let \(f_n \to f\) pointwise on the closed interval [a,b], and assume that each \(f_n\) is differentiable. If \((f'_n)\) converges uniformly on [a,b] to a function g, then the function f is differentiable and fâ€™=glet (\(f_n\)) be a sequence of differentiable functions defined on the closed interval [a,b], and assume \((f'_n)\) converges uniformly to a function g on [a,b]. If there exists a point \(x_0 \in [a,b]\) for which \(f_n(x_0)\) is convergent, then (\(f_n\)) converges uniformly. Moreover, the limit function f = lim \(f_n\) is differentiable and satisfies fâ€™ = g

### 3.3.6.3. 6.4 series of functionsÂ¶

*term-by-term continuity thm*- let \(f_n\) be continuous functions defined on a set A \(\subseteq\) R and assume \(\sum f_n\) converges uniformly on A to a function f. Then, f is continuous on A.*term-by-term differentiability thm*- let \(f_n\) be differentiable functions defined on an interval A, and assume \(\sum f'_n(x)\) converges uniformly to a limit g(x) on A. If there exists a point \(x_0 \in [a,b]\) where \(\sum f_n(x_0)\) converges, then the series \(\sum f_n(x)\) converges uniformly to a differentiable function f(x) satisfying fâ€™(x) = g(x) on A. In other words, \(f(x) = \sum f_n(x)\) and \(f'(x) = \sum f'_n(x)\)*Cauchy Criterion for uniform convergence of series*- A series \(\sum f_n\) converges uniformly on A iff \(\forall \epsilon > 0 \exists N \in N\) s.t. whenever n>mâ‰¥N, x in A \(\|f_{m+1}(x) + f_{m+2}(x) + f_{m+3}(x) + ...+f_n(x)\| < \epsilon\)*Wierstrass M-Test*- For each n in N, let \(f_n\) be a function defined on a set A \(\subseteq\) R, and let \(M_n > 0\) be a real number satisfying \(\|f_n(x)\| â‰¤ M_n\) for all x in A. If \(\sum M_n\) converges, then \(\sum f_n\) converges uniformly on A

### 3.3.6.4. 6.5 power seriesÂ¶

power series f(x) = \(\sum_{n=0}^\infty a_n x^n = a_0 + a_1 x_1 + a_2 x^2 + a_3 x^3 + ...\)

if a power series converges at some point \(x_0 \in \mathbb{R}\), then it converges absolutely for any x satisfying |x|<|\(x_0\)|

if a power series converges pointwise on the set A, then it converges uniformly on any compact set K \(\subseteq\) A

if a power series converges absolutely at a point \(x_0\), then it converges uniformly on the closed interval [-c,c], where c = |\(x_0\)|

*Abelâ€™s thm*- if a power series converges at the point x = R > 0, the the series converges uniformly on the interval [0,R]. A similar result holds if the series converges at x = -R

if \(\sum_{n=0}^\infty a_n x^n\) converges for all x in (-R,R), then the differentiated series \(\sum_{n=0}^\infty n a_n x^{n-1}\) converges at each x in (-R,R) as well. Consequently the convergence is uniform on compact sets contained in (-R,R).

can take infinite derivatives

### 3.3.6.5. 6.6 taylor seriesÂ¶

*Taylorâ€™s Formula*\(\sum_{n=0}^\infty a_n x^n = a_0 + a_1 x_1 + a_2 x^2 + a_3 x^3 + ...\)centered at 0: \(a_n = \frac{f^{(n)}(0)}{n!}\)

*Lagrangeâ€™s Remainder thm*- Let f be differentiable N+1 times on (-R,R), define \(a_n = \frac{f^{(n)}(0)}{n!}.....\)not every infinitely differentiable function can be represented by its Taylor series (radius of convergence zero)

## 3.3.7. ch 7 - the Riemann IntegralÂ¶

### 3.3.7.1. 7.2 def of Riemann integralÂ¶

*partition*of [a,b] is a finite set of points from [a,b] that includes both a and b*lower sum*- sum all the possible smallest rectanglesa partition Q is a

*refinement*of a partition P if \(P \subseteq Q\)if \(P \subseteq Q\), then L(f,P)â‰¤L(f,Q) and U(f,P)â‰¥U(f,Q)

a bounded function f on the interval [a,b] is

*Riemann-integrable*if U(f) = L(f) = \(\int_a^b f\)iff \(\forall \epsilon >0\), there exists a partition P of [a,b] such that \(U(f,P)-L(f,P)<\epsilon\)

U(f) = inf{U(f,P)} for all possible partitions P

if f is continuous on [a,b] then it is integrable

### 3.3.7.2. 7.3 integrating functions with discontinuitiesÂ¶

if f:[a,b]->R is bounded and f is integrable on [c,b] for all c in (a,b), then f is integrable on [a,b]

### 3.3.7.3. 7.4 properties of IntegralÂ¶

assume f: [a,b]->R is bounded and let c in (a,b). Then, f is integrable on [a,b] iff f is integrable on [a,c] and [c,b]. In this case we have \(\int_a^b f = \int_a^c f + \int_c^b f.\)F

*integrable limit thm*- Assume that \(f_n \to f\) uniformly on [a,b] and that each \(f_n\) is integarble. Then, f is integrable and \(lim_{n \to \infty} \int_a^b f_n = \int_a^b f\).

### 3.3.7.4. 7.5 fundamental theorem of calculusÂ¶

If f:[a,b] -> R is integrable, and F:[a,b]->R satisfies Fâ€™(x) = f(x) for all x \(\in\) [a,b], then \(\int_a^b f = F(b) - F(a)\)

Let f: [a,b]-> R be integrable and for x \(\in\) [a,b] define G(x) = \(\int_a^x g\). Then G is continuous on [a,b]. If g is continuous at some point \(c \in [a,b]\) then G is differentiable at c and Gâ€™(c) = g(c).

## 3.3.8. overviewÂ¶

convergence

sequences

series

functional limits

normal, uniform

sequence of funcs

pointwise, uniform

series of funcs

pointwise, uniform

integrability

sequential criterion - usually good for proving discontinuous

limit points

functional limits

continuity

absence of uniform continuity

algebraic limit theorem ~ scalar multiplication, addition, multiplication, division

limit thm

sequences

series - canâ€™t multiply / divide these

functional limits

continuity

differentiability

~integrability~

limit thms

*continuous limit thm*- Let (\(f_n\)) be a sequence of functions defined on A that converges uniformly on A to a function f. If each \(f_n\) is continuous at c in A, then f is continuous at c*differentiable limit theorem*- let \(f_n \to f\) pointwise on the closed interval [a,b], and assume that each \(f_n\) is differentiable. If \((f'_n)\) converges uniformly on [a,b] to a function g, then the function f is differentiable and fâ€™=gconvergent derivatives almost proves that \(f_n \to f\)

let (\(f_n\)) be a sequence of differentiable functions defined on the closed interval [a,b], and assume \((f'_n)\) converges uniformly to a function g on [a,b]. If there exists a point \(x_0 \in [a,b]\) for which \(f_n(x_0) \to f(x_0)\) is convergent, then (\(f_n\)) converges uniformly

*integrable limit thm*- Assume that \(f_n \to f\) uniformly on [a,b] and that each \(f_n\) is integarble. Then, f is integrable and \(lim_{n \to \infty} \int_a^b f_n = \int_a^b f\).

functions are continuous at isolated points, but limits donâ€™t exist there

uniform continuity: minimize \(\|f(x)-f(y)\|\)

derivative doesnâ€™t have to be continuous

integrable if finite amount of discontinuities and bounded