Chapter 4 Flashcards
Definition: continuous at a point a in D_f, if
We say that a function f:R->R is continuous at a point a in D_f is limit as x tends to a of f(x) exists and equals f(a). We say that f is continuous on a set S subset of D_f if it is continuous at every point of S.
Theorem 4.1.1: for a function continuous, Sequence and epsilon-delta criterion
For a function f:R->R having domain D_f, the following statements are equivalent.
1) f is continuous at a in D_f
2) Given ant sequence (x_n) with x_n in D_f for all n in the naturals, such that
limit as n tends to infinity of x_n =a, we have that
limit as n tends to infinity of f(x_n) =f(a)
3) Given any ε>0 there exists δ> 0 such that whenever x∈D_f with 0 less than |x-a| < δ, we have | f(x) -f(a)| less than epsilon
Or
Given any epsilon , there exists δ>0 such that whenever x∈D_f with x∈(a-δ, a+δ) we have f(x)∈ (f(a)-ε,f(a)+ ε)
An open neighbourhood of a in the reals
Is the open interval (a-δ, a+δ)
Theorem 4.1.2 the algebra of limits revisited (continuity)
Suppose that f,g:R->R are both continuous at a∈D_f ∩ D_g. The following functions are also continuous at a:
a) f+g
b) fg
c) αf, for all α∈R
d) f/g, provided g(a) ≠ 0
Proof: direct consequence if thm 3.2.1
Theorem 4.1.3 for composition of continuous functions
Let f and g be functions from R to R. If g is continuous at a, g(a)∈D_f, and f is continuous at g(a), then f •g is continuous at a.
(Composite)
Proof:problem 57z
Theorem 4.3.1 the intermediate value theorem
Let f be continuous on [a,b] with f(a) bigger than and
f(b) less than 0,
Or f(a) less than 0 and f(b) bigger than 0.
Then there exists c∈(a,b) such that f(c) =0
Proof:
Case: f(a) bigger than 0, f(b) less than 0
Long proof
Corollary of the intermediate value theorem
Showing all values are attained in image
Let f be continuous on [a,b] with f(a) < f(b). Then for each γ∈ (f(a), f(b)) there exists c∈(a,b) with f(c) = γ
Proof: problem 66
2nd Corollary of the intermediate value theorem
Every polynomial of odd degree has at least one real root
Proof: writing ( with a_m bigger than 0, case less than 0 similar)
p(x) = a_m x^m + a_(m-1) x^(m-1) +…+ a₁x +a₀
= x^m( a_m + (a_(m-1) / x ) +…+ (a₁/(x^(m-1))+ a₀/x^m).
Then as shown in probes 54….
We show the limit of p(x) as x tends to infinity an as x tends to neg infinity are infinity and neg infinity do there exist a and b are between these (by the definition of divergence) such that p(a) is less than 0 and p(b) is bigger than 0. As p is continuous on the interval by the IVT there exists our c st p(c) =0z
DOESNT WORK FOR EVEN DEGREE POLYNOMIAL
Theorem 4.3.4 the boundedness
If f is continuous on [a,b] then it is bounded on [a,b] and it attains both of its bounds there.
Proof: long but using contradiction to show f is bounded, sequence and infimum along with Bolzano-weirstrass theorem etc
• 1/x us continuous on interval (0,1) but not bounded as divergent at 0
•x is bounded on (0,1) but doesn’t attain its bounds
HENCE BOUNDEDNESS USES CLOSED INTERVALS
Eg nothing about functions on R
Corollary 4.3.5 of bounded was, continuous functions and images
If f:R->R is continuous and non constant in [a,b] subset of D_f,
Then there exists m less than M st
f([a,b]) = [m,M]
(Image is also bounded on Interval)
This tells us that the image of the interval [a,b] under the continuous functions contains the interval [f(a), f(b)] which is a subset of f([a,b)].
Prop 4.3.6: mapping inverses
The mapping f:A->B is invertible if and only if it is bijective.
Theorem 4.3.7 the inverse function theorem
If f:R->R is continuous and strictly increasing (or strictly decreasing) on [a,b] then f is invertirle and
f^(-1) is strictly increasing on [f(a),f(b)] ( strictly decreasing on [f(b), f(a)] and continuous on (f(a),f(b)) (respectively, on (f(b), f(a)))
Proof:long
Using corollary of bounded image and showing injective and surjective
Discontinuity
A function f:R->R is said to have a discontinuity at a ∈D_f if it fails to be continuous there. Ie f is discontinuous at a.
Eg indicator function 1_[a,b] is discontinuous at a and at b but is continuous on R{a,b}. To show that a function is discontinuous at a it is sufficient to find a sequence (x_n) in D_f{a} st limit as n tends to infinity of x_n =a, but limit as n tends to infinity of f(x_n) is NOT equal to f(a).
Investigate discontinuity by using left and right limits
Theorem 4.2.1 continuous and left right limits
A function f:R-> R is continuous at a ∈D_f
If and only if
It is both right and left continuous there.
Proof: direct consequence of theorem 3.3.2
Jump discontinuity
If f is discontinuous at a but both limits:
Limit if x increasing to a of f(x)
And
Limit of x decreasing to a of f(x) exist
(Real numbers)
And are unequal we say that f has a jump discontinuity at a.
In this case the jump at a is defined as
J_f (a) =limit_ (x↓a) of f(x) - limit_(x↑a) of f(x)
Eg indicator function has J_f (a)=1 and J_f (b) =1
Continuity examples- continuous function
Fix c in reals then constant function f(x) =c is continuous on R.
To see this let a in R be arbitrary and let (x_n) be any sequence converging to a. Then f(x_n) =c for all n in the naturals and so the sequence (f(x_n)) clearly converges to c=f(a).
The linear function f(x)=x is also continuous on R. Again given any sequence (x_n) converging to a, f(x_n) =x_n for all n in N . And so the sequence (f(x_n)) clearly converges to f(a) =a.
Examples and the algebra of limits revisited (continuity)
Using the algebra of limits repeatedly we can show f(x) =x^n is continuous on R for all N.
Etc for polynomials continuous on R
Etc for rational functions
Examples- composite functions continuity
We can use theorem for the composite of two continuous functions being continuous for continuity of functions such as f(x) = sin( (2x-1)/(x^2 +1))
Eg g(x)= xsin(1/x) with domain R\{0} by the theorems algebra of limits and composition g is continuous at every point in the domain.
If we define a new function such that it's an extension that is continuous on the whole of R we can use the function is 0 if x=0
Define: given two functions from R to R an extension of the other
f_2 is an extension of f_1 ( f_1 is a restriction of f_2) if
D_f_1 is a subset of D_f_2
And
f_1(x) =f_2(x) for all x in the domain of f_1
If f_1 is continuous on D_f_1 and f_2 is continuous on D_f_2 we say that f_2 is a continuous extension of f_1
We can find a continuous extension if the limit exists o/w no continuous extensions of function to R.
Eg indicator function
Eg dirichlets function
Right/left continuous?
The function 1_Q is discontinuous at every point in R. Then f(x) = if x is rational and 0 if irrational.
F is discontinuous at every point a in Q.
We see this by considering sequence whose nth term is a+ 1/n. By theorem 1./2 we can find irrational b_n a less than b_n less than a+1/n for all n I. The naturals.
By the sandwich rule the limit as n tends to i finity of b_n is a. But the limit as n tends to infinity of f(b_n) is equal to 0 which isn’t equal to f(a), which is 1.
Suppose that a is irrational then a +1/n is irrational and by theorem 1.4.5 there exists a rational number c_n so that a< c_n < a+1/n for all n in the naturals. Then by the sandwich rule limit of c_n is a but limit as n tends infinity of f(c_n) =1 NOT 0=f(a) as a is irrational.
Example: let f(x) =x^n for n in N.
Inverses…
We can prove that f is strictly increasing on every [a,b] subset of [0, infinity ). f is continuous and hence the inverse exists by the theorem and is continuous and strictly monotone increasing on (a^n, b^n)
And hence on any finite interval.
Then the maximal Domain of f-1 is 0 to infinity and it’s continuous on that interval ( right continuous on 0)
So f-1 = x^1/n and we have proven the existence of the positive roots of real
Surjective objective bijective
For arbitrary sets A and B st f: A to B mapping with D_f =A
- surjective if R_f =B ( for all y in B there exists an x in A st f(x) =y
- injective if f(x₁) = f(x₂) implies x₁ = x₂ (for x_1 and x_2 in A)
•bijective if both, that is each b is mapped to uniquely by an a in A
Invertíble mapping
If there exists a mapping f^-1 :B to A with D _f-1 =B, the inverse of f
For which
f^-1(f(x) =x for all x in A and
f(f^-1(x)) =y for all y in B
Monotone vs strictly
Increasing decreasing
FUNCTIONS
if for x,y in D_f with x less than y we have:
• f(x) ≤ f(y) monotonic increasing
•f(y) ≤ f(x) monotoníc decreasing
Strictly decreasing or increasing if no equality
Eg f(x) = 1/x is strictly decreasing on R\{0}
We consider the invertibility of monotone functions that are continuous on closed intervals
Extension of inverse function theorem
f-1 is right continuous at f(a) and left continuous at f(b)
Example: inverse function theorem and f(x) =e^x
We prove monotonic increasing and we deduce that inverse is continuous and monotonic increasing on R.
Showing us every positive real number has a natural logarithm.
