Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 10



Continuous functions

We denote the distance between two real numbers and by . For a given function

one may ask whether it is possible to control the distance in the image (the target) by controlling the distance in the domain. Let and let be its image. One would like that for points which are close to , also their image points are close to . Already linear functions with different slopes show that the measure for being close in the image and in the domain can not be the same. A reasonable question is whether for any desired exactness in the image there exists at all an exactness in the domain which ensures that the values of the function are lying inside the desired exactness. To make this intuitive idea more precise, let be given. This represents an accuracy for the target. The question is whether one can find a (representing an accuracy for the domain) such that for all fulfilling , also the relation holds. This idea leads to the concept of a continuous function.


Definition  

Let be a subset,

a function, and a point. We say that is continuous in the point , if for every , there exists a , such that for all fulfilling , the estimate holds. We say that continuous, if it is continuous in every point

stands for the set of definition of the function. Typically, is just , or an interval, or with finitely many points removed. Instead of working with the real numbers and , one might also work just with the unit fractions and .


Example

A constant function

is continuous. For every given , one can choose an arbitrary , since

holds anyway.

The identity

is also continuous. For every given , one can take , yielding the tautology: If

then

holds.



Example

We consider the function

given by

This function is not continuous in . For and every positive , there exists a negative number such that . But for such a number we have .

It is not possible to draw any continuous function, even if you allow arbitrary enlargements. The image shows an approximation of a so-called Weierstraß-function, which is continuous, but nowhere differentiable. For a continuous function one can control the extent of the fluctuation in the image by reducing the seize of the definition interval, but the number of the fluctuations (how often the direction of the graph is changing) can not be controlled.

The following statement relates continuity to convergent sequences.


Lemma

Let be a subset,

a function and

a point. Then the following statements are equivalent.
  1. is continuous in the point .
  2. For every convergent sequence in with , also the image sequence is convergent with limit .

Proof  

Suppose that (1) is fulfilled and let be a sequence in converging to . We have to show that

holds. To show this, let be given. Due to (1), there exists a fulfilling the estimation property (from the definition of continuity) and because of the convergence of to there exists a natural number such that for all the estimate

holds. By the choice of we have

so that the image sequence converges to .

Suppose now that (2) is fulfilled. We assume that is not continuous. Then there exists an such that for all there exist elements such that their distance to is at most , but such that the distance of their value to is larger than . This holds in particular for every , . This means that for every natural number , there exists a with

This sequence converges to , but the image sequence does not converge to , since the distance of its members to is always at least . This contradicts condition (2).



Rules for continuous functions

Lemma

Let and be subsets and let

and

denote functions with

. Then the following statements hold.
  1. If is continuous in and is continuous in , then also the composition is continuous in .
  2. If and are continuous, so is .

Proof  

The first statement follows from Lemma 10.4 . This implies also the second statement.



Lemma

Let be a subset and let

be continuous functions. Then also the functions

are continuous. For a subset such that has no zero in , also

is continuous.

Proof  

This follows from Lemma 10.4 and Lemma 8.1 .



Corollary

Polynomial functions

are

continuous.

Proof  

Due to Example 10. and Lemma 10.6 , the powers

are continuous for every . Hence, also the functions

are continuous for every and therefore, again due to Lemma 10.6 , the functions

are continuous.


Rational functions are continuous on their domain.

Corollary

Let be polynomials and let . Then the rational function

is continuous.

Proof  

This follows from Corollary 10.7 and Lemma 10.6 .



Limit of a function

Quite often, functions are not defined in certain points, e.g. if the used functional term is not defined there. However, it makes a huge difference whether only the functional term is not defined in a point, but has a useful (continuous) extension, or whether the function does not have a useful extension in this point, e.g. because it has a pole or an even more chaotic behavior. The following concept is in particular relevant for the definition of differentiability (if the difference quotient has a useful limit, then it is called differential quotient).


Definition  

Let denote a subset and a point. Let

be a function. Then is called limit of in , if for every there exists some such that for all fulfilling

the estimate

holds. In this case, we write

This concept is basically only useful if there exists at least some sequence within converging to . A typical situation is the following: Let denote a real interval, a point and let . The function is defined on but not in , and we are dealing with the question whether can be extended to a function defined on the whole . Here, should be determined by .


Lemma

Let denote a subset and a point. Let

be a function and

a point. Then the following statements are equivalent.
  1. We have
  2. For every sequence in which converges to , also the image sequence converges to .

Proof


This implies for a continuous function that it can be extended to a continuous function (by ) if and only if the limit of in equals .


Lemma

Let denote a subset and a point. Let and denote functions, such that the limits and

exist. Then the following statements hold.
  1. The sum has in the limit
  2. The product has in the limit
  3. Suppose that for all and . Then the quotient has in the limit

Proof  

This follows from Lemma 10.10 and from Lemma 8.1 .



Example

We consider the limit

where . For , this term is not defined, and from this term one can not read of directly whether the limit exists. It is however possible to multiply the numerator and the denominator by , then we get

Due to the rules for limits we can determine the limit in the numerator and in the denominator separately, where for the denominator we use the continuity of the square root according to Exercise 10.4 . Hence, the limit is .


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