Continuity of Functions

Limit and Continuity

Limit on Metric Spaces

Let and be metric spaces and let be a function. For , we say that if for every there exists such that

Continuity of Functions

Let and be metric spaces and let be a function. Then is

• continuous at if .
• continuous on if it is continuous at every point of .

Neighbourhood

Let be a metric space and . A set is called a neighborhood of if there is an open ball with .

Theorem

Let and be metric spaces, let be a function, and let . The following statements are equivalent:

• is continuous at .
• For any sequence in with there holds .
• For any neighborhood of , is a neighborhood of .

Proof

Theorem

Let and be metric spaces and let be a function. The following are equivalent:

• is continuous on .
• For every open set in , is open in .
• For every closed set in , is closed in .

Theorem

Let be a metric space.

1. If are continuous then and are continuous and is continuous at all points where .
2. If ) is a normed vector space and are continuous then is continuous.

Theorem

Let be metric spaces. If is continuous at and is continuous at , then is continuous at .

Proof Let be any sequence with . By the continuity of at , we have . Next, by the continuity of at we have Therefore is continuous at .

Theorem

Let and be metric spaces and let be continuous. If is compact, then is compact.

Theorem

Let be a metric space and let be a continuous function. If is compact, then is bounded above and below and achieves its maximum and minimum values on .

Theorem

Continuous function on compact space must be uniformly continuous.

Lipschitz Continuity

Lipschitz Continuity

A function is Lipschitz continuous or just Lipschitz if there exists such that We say that is a Lipschitz constant for .

e.g. Let be a metric space. Suppose is non-empty. Then we can define the distance of from by setting . Then the function is Lipschitz with Lipschitz constant . Indeed, for all we have Hence . Similarly, . Thus $$d_{\R}(d(x,A),d(y,A))=|d(x,A)-d(y,A)|\leq d(x,y)$$$\square$

Proposition

A Lipschitz continuous function is continuous.

Proof Suppose is Lipschitz. Then for all and , let , then $$d_{X}(x,p)<\delta=\frac{\epsilon}{C} \implies d_{Y}(f(x),f(p))<C\cdot\frac{\epsilon}{C}=\epsilon$$$\square$

Topologically Equivalent Metrics

Lemma

Suppose that and are two metrics on . Then the following statements are equivalent:

• Every set that is open in is open in
• For any metric space , if is continuous from to then is continuous from to
• For any metric space , if is continuous from to then is continuous from into .

Theorem

Suppose that and are two metrics on . Then the following statements are equivalent:

• The open sets in and coincide
• For any metric space , a function is continuous from into if and only if is continuous from into
• For any metric space , a function is continuous from into if and only if is continuous from into .

Topologically Equivalence

Two metrics and on are called topologically equivalent, or just equivalent, if the open sets in and coincide.

Lipschitz Equivalence

Two metrics and on are called Lipschitz equivalent if there exist such that

Theorem

Let and be two metrics on that are Lipschitz equivalent on . Then and are topologically equivalent.

Corollary The metrics induced by equivalent norms are topologically equivalent.

Lemma If is a vector space and two norms and on induce topologically equivalent metrics then the norms are equivalent.

Uniform Continuity

Uniform Continuity

Let and be metric spaces and . A function is called uniformly continuous on if for any , there is a such that for any with :

Theorem

Let and be metric spaces. If is continuous and is compact, then is uniformly continuous on .