Continuous Functions

Definition

if a function is continuous on , it's saying that for any , s.t.:

Boundedness Theorem

If a function is continuous on a close-interval , then its bounded on that interval.

Proof with Bolzano-Weierstrass Theorem.

Suppose that: the function boundless on

If the function is boundless, loop this process, starts with :

  1. bisect the interval with
  2. select the boundless half or
  3. select a point making this half boundless to build a sequence
  4. continue this process with

And finally we have a infinite bounded sequence and an unbounded sequence s.t.:

  1. converge to some number (Bolzano Weierstrass Theorem)
  2. is unbounded, which contradicts to (Heine Theorem)

So the function is bounded on

Intermediate Value Theorem

IVT

If a function is continuous on a close-interval , then for any intermediate value , there's some value making that

Proof

Build nested intervals with this process:

  1. bisect the interval with
  2. if , stop.
  3. select the half making
  4. continue this process

So we get a nested interval , which locate exactly one real number , and (Nested Interval Theorem). As the function is continuous on , with Heine's Theorem:

So we have another set of nested intervals that locate a unique number, which can only be since . (Nested Interval Theorem)

Extreme Value Theorem

Extreme Value Theorem

If a function is continuous on , then there's a maxima and minima in .

As Boundedness Theorem said, the function is bounded, so the function has a supremum and infimum, w.r.t. Dedekind Completeness Theorem. If the supremum is .

Suppose . is also a continuous function, thus it's also bounded. Suppose is an upper bound of on :

so is also an upper bound for on , which contradicts to the setting that is the supremum. So there must be some value s.t. .

Mean Value Theorems

Fermat's Lemma

Fermat's Lemma

derivative on maxima / minima must be zero

Proof:

Suppose is a maxima.

It said that:

Rolle's MVT

Rolle's MVT

if function is continuous on and , then:

A function is continuous on , then with Extreme Value Theorem, there must be a maxima , and then with Fermat's Lemma,

Lagrange MVT

Lagrange MVT

If function is continuous on , then:

, which saids the derivative on some intermediate point concludes the change of endpoint.

Proof:

build a flat function containing to apply Rolle's MVT:

So that

So with Rolle's MVT,

Cauchy MVT

Cauchy MVT

if function are both continuous on , then:

construct a function whose derivative is in the form that fit the theorem:

so by Rolle's MVT:

MVT for Integrals

MVT for Integrals

If function is continuous on , then the mean integral is an intermediate value of

is continuous on , so with Boundedness Theorem, is bounded:

then with IVT:

Fundamental Theorem of Calculus

if is continuous at

accumulation function is one of the antiderivative

Integration is difference:

Proof:

Prove it by MVT for Integrals:

So any original function of , denoted by , can be represented by adding some constant to :

substituting with , we know the constant is

substituting with , then the integration of from to is the difference of the value of the endpoints of any antiderivative :

Prerequisite of differentiable

it must be continuous to be derivable, or equivalently, differentiable.

if is differentiable, then is also an infinitesimal -- which implies that is continuous.