lacunary - Mathnotes

Limits of Functions

Note \@{note-2}

In the above definition, the symbols dX and dY refer to distances in X and Y, respectively.

Furthermore, while p∈X, p need not be a point of E. In fact, even if p∈E, it's possible that f(p)≠limx→pf(x).

Theorem \@{limit-of-a-function-characterized-by-limits-of-sequences}

Let X and Y be metric spaces; suppose E⊂X, f:E→Y, and p is a limit point of E. Then

limx→pf(x)=q(a)

if and only if

limn→∞f(pn)=q(b)

for every sequence {pn} in E such that

pn≠p,limn→∞pn=p.(c)

Proof contrapositive \@{proof-of-limit-of-a-function-characterized-by-limits-of-sequences}

Suppose that (a) is true and let {pn}⊂E be a sequence such that (c) holds. Let ϵ>0. Then, for some δ>0, if dX(x,p)<δ, then dY(f(x),q)<ϵ. Now, there is also some N such that dX(pn,p)<δ whenever n≥N, and thus, dY(f(pn),q)<ϵ whenever n≥N, so (b) holds.

Conversely, suppose that (a) is false. Then, for some ϵ>0, that for every δ>0 there exists a point x∈E such that dY(f(x),q)≥ϵ but 0<dX(x,p)<δ. If we let δn=1/n,n=1,2,3,…, and each xn a point such that 0<dX(xn,p)<δn, then {xn} is a sequence satisfying (c). However, since dY(f(xn),q)≥ϵ for all n, (b) does not hold.

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Corollary \@{limit-of-a-function-at-a-point-is-unique-if-it-exists}

If f has a limit at p, this limit is unique.

Proof \@{proof-of-limit-of-a-function-at-a-point-is-unique-if-it-exists}

This follows directly from the facts that limits of sequences are unique and that limits of functions are characterized by limits of sequences.

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Referenced by (2 direct)

Direct references:

Theorem \@{theorem-7}

Suppose E⊂X, a metric space, p is a limit point of E, f and g are complex functions on E, and

limx→pf(x)=A,limx→pg(x)=B.

Then:

limx→p(f+g)(x)=A+B.

limx→p(fg)(x)=AB.

limx→p(fg)(x)=AB,B≠0.

Proof \@{proof-of-theorem-7}

This follows directly from the that limits of functions are characterized by limits of sequences, and the algebraic properties of sequences of limits.

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Remark \@{theorem-7-remark}

If f and g map E into Rk, then (a) remains true, and (b) becomes

limx→p(f⋅g)(x)=A⋅B.

That is, the limit of an inner product of vector valued functions is the inner product of their limits.

(See A sequence in Rk converges iff its components converge).

Continuous Functions

Definition: Continuous \@{continuous}

Suppose X and Y are metric spaces, E⊂X,p∈E, and f:E→Y. Then f is said to be continuous at p if for every ϵ>0 there exists a δ>0 such that

dY(f(x),f(p))<ϵ

for all points x∈E for which dX(x,p)<δ.

If f is continuous at every point of E, then f is said to be continuous on E.

Referenced by (47 direct, 67 transitive)

Direct references:

Transitive (depth 1):

Transitive (depth 2):

Transitive (depth 3):

Note \@{note-11}

A more geometric way to view continuity is that for any ball B(f(p)) centered at f(p), there is some ball B(p) centered at p for which every point in B(p) (including p) is mapped by f to some point in B(f(p)).

Also, note that if p is an isolated point of E, then the definition of continuous implies that every function f which has E as its domain of definition is continuous at p. This is because for any ϵ>0 we choose, we can pick δ>0 so that the only point x∈E for which dX(x,p)<δ is x=p, and so

dY(f(x),f(p))=0<ϵ.

Also note that, unlike the definition of limit, the definition of continuous requires f to be defined at p in order to be continuous at p.

Theorem \@{function-is-continuous-at-point-iff-limit-at-point-equals-function-at-point}

Suppose X and Y are metric spaces, E⊂X,p∈E, with p a limit point of E, and f:E→Y. Then, f is continuous if and only if limx→pf(x)=f(p).

Proof \@{proof-of-function-is-continuous-at-point-iff-limit-at-point-equals-function-at-point}

Note that the definition of a function having a limit at point in a metric space is different from the definition of a function being continuous at a point in a metric space only in that the continuous definition requires the function to be defined at the point (and equal to the limit at the point.)

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Definition: Composition \@{composition}

Suppose X,Y,Z are metric spaces, E⊂X, f:E→Y, g:f(E)→Z, h:E→Z with

h(x)=g(f(x))(x∈E).

The function h is called the composition or the composite of f and g. The notation

h=g∘f

is frequently used.

Referenced by (1 direct, 1 transitive)

Direct references:

Transitive (depth 1):

Theorem \@{composition-of-continuous-functions-is-continuous}

Suppose X,Y,Z are metric spaces, E⊂X, f:E→Y, g:f(E)→Z, h:E→Z with

h(x)=g(f(x))(x∈E).

If f is continuous at a point p∈E and if g is continuous at the point f(p), then h is continuous at p.

Proof \@{proof-of-composition-of-continuous-functions-is-continuous}

Let ϵ>0. Since g is continuous at f(p), there exists η>0 such that

dZ(g(y),g(f(p)))<ϵ if dY(y,f(p))<η and y∈f(E).

Since f is continuous at p, there exists δ>0 such that

dY(f(x),f(p))<η if dX(x,p)<δ and x∈E.

It follows that

dZ(h(x),h(p))=dz(g(f(x)),g(f(p)))<ϵ

if dX(x,p)<δ and x∈E. Thus, h is continuous at p.

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Intuition \@{composition-of-continuous-functions-is-continuous-intuition}

Basically, since g is continuous, we can control how close its output is to g(f(p)) by controlling how close its input is to f(p), which we can certainly do, since f is also continuous, and we can control how close its output is to f(p) by controlling how close its input is to p.

Theorem \@{mapping-continuous-iff-inverse-images-of-open-sets-are-open}

A mapping f of a metric space X into a metric space Y is continuous on X if and only if f−1(V) is open in X for every open set V in Y (see inverse image.)

Proof \@{proof-of-mapping-continuous-iff-inverse-images-of-open-sets-are-open}

Assume f is continuous on X and V is an open set in Y. Suppose, for the sake of contradiction, that f−1(V) is not open. Then, some point p∈f−1(V) is not an interior point of f−1(V), which means there is no neighborhood of p that contains only points in f−1(V), that is, every neighborhood of p contains some point q that is not in f−1(V), i.e., f(q)∉V. Now, since V is open and f(p)∈V, there is some ϵ>0 for which Bϵ(f(p))⊂V, but, since all neighborhoods of p contain some q∉f−1(V), there is no δ>0 for which all points within δ of p are mapped to Bϵ(f(p)) by f, a contradiction, since f is continuous on X by hypothesis. Therefore, our assumption is incorrect and f−1(V) is open.

Conversely, suppose f−1(V) is open in X for every open set V in Y. Let V be an open set in Y. Assume, for the sake of contradiction that there is some p∈f−1(V) at which f is not continuous. Let ϵ>0. Then, there is no δ>0 for which Bδ(p) contains only points that are mapped by f to V, i.e. every neighborhood of p contains some point that is not in f−1(V), and therefore p is not an interior point of f−1(V), and f−1(V) is not open, a contradiction. Thus, our assumption must be incorrect, and there is no such p, and since entire space Y is an open subset of itself, f must be continuous on all of X.

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