Proving continuity – Serlo

Aus Wikibooks
Zur Navigation springen Zur Suche springen
UnderCon icon.svg

Diese Seite ist noch im Entstehen und noch nicht offizieller Bestandteil des Buchs. Gib der Autorin / dem Autor Zeit, die Seite anzupassen!

Overview[Bearbeiten]

There are several methods for proving continuity:

  • Concatenation Theorems: If the function can be written as a concatenation of continuous functions, it's continuous by the Concatenation Theorems.
  • Using the local nature of continuity: If a function looks like another well-known continuous function in a small neighborhood of a point, then it must also be continuous in this point.
  • Considering the left- and right-sided limits: If one can show that the left- and right-sided limits of a function are the same in some point, then the function is continuous in this point.
  • Showing the sequence criterion: Using the sequence criterion means that the limit can be pulled into the function, i.e. we can consider the limit of the arguments. For a sequence of arguments with limit value it must hold .
  • Showing the epsilon-delta criterion: For every it must be show that there exists some such that for all arguments with a distance smaller than from the point , the inequality is satisfied.

Composition of continuous functions[Bearbeiten]

Main article: Composition of continuous functions

General proof sketch[Bearbeiten]

Following the concatenation theorems, every composition of continuous functions is again continuous function. So if can be written as a concatenation of continuous functions, we can directly infer continuity of . A corresponding proof could be of the following form:

Let with . The function is a concatenation of the following functions:

...List of continuous functions, which serve as building bricks for ...

Since (Expression how is constructed out of those bricks) , we know that is a concatenation of continuous functions and hence continuous, as well.

In a lecture, this proof scheme may of course only be applied if the concatenation theorems have already been treated before. In any way, it is a very efficient method to characterize continuous functions.

Example Problem[Bearbeiten]

Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)

Show, using the epsilon-delta criterion, that the following function is continuous:

How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)

We need to show, that for any given , there is a , such that all with satisfy the inequality . So let us take a look at the target inequality and estimate the absolute from above. We are able to control the term . Therefor, would like to get an upper bound for including the expression . So we are looking for an inequality of the form

Here, is some expression depending on and . The second factor is smaller than and can be made arbitrarily small by a suitable choice of . Such a bound is constructed as follows:

Since , there is:

If we now choose small enough, such that , then we obtain our target inequality . But still depends on , so would have to depend on , too - and we required one choice of which is suitable for all . THerefore, we need to get rid of the -dependence. This is done by an estimate of the first factor, such that our inequality takes the form  :

We even made independent of , which would in fact not have been necessary. So we obtain the following inequality

We need the estimate , in order to fulfill the target inequality . The choice of is sufficient for that. So let us write down the proof:

Proof (Epsilon-Delta proof for the continuity of the Square Root Function)

Let with . Let and an arbitrary be given. We choose . For all with there is:

Hence, is a continuous function.

Using the Local Nature of Continuity[Bearbeiten]

If applicable, one can use the fact that continuity is a local property. Namely, when a function looks the same as another function in a certain point , i.e. on some small neighborhood of , and we know to be continuous, then must also be continuous at . For example, consider the function with for positive and for negative . Now we choose some random positive number . In a sufficiently small neighborhood of , is constant :

Die Funktion f ist in einer hinreichend kleinen Umgebung um positive Argumente konstant 1.

Since constant functions are continuous, is also continuous at the point . Similarly one can show that is also locally constant for negative number. Then is also continuous for negative numbers, and is therefore continuous for all real numbers. In the proof we write:

"For every there exists a neighborhood around where is either constant or constant . Since constant functions are continuous, must also be continuous at the point . Therefore, is continuous“.

Such an argument can often be applied to functions that are defined by a case differential. Our function is a good example of that. In conclusion, it's defined as:

However, the local nature of continuity can not be used as an argument for every function defined by a case differential! Let's consider the following function :

For all points that are not zero we can construct a proof like we've done earlier in this chapter to show that the function is continuous at these points. However, at the point we have to construct a different argument. For example, here we could consider the left- and right-sided limits.

Sequence Criterion[Bearbeiten]

Main article: Sequential definition of continuity

Review: Sequence Criterion[Bearbeiten]

Definition (Sequence criterion of continuity)

A function with is continuous, if for all and all sequences with and there is:

General Proof Structure[Bearbeiten]

In order to prove continuity of a function at some , we need to show that for each sequence of arguments converging to , there is . A proof for this could schematically look as follows:

Let be a function defined by and . In addition, let be any sequence of arguments satisfying . Then, there is:

In order to prove continuity for the function (for all arguments in its domain of definition), we need to slightly adjust that scheme:

Let be a function defined by and let be any element of the domain of definition for . In addition, let be any sequence of arguments satisfying . Then, there is:

Example Problem[Bearbeiten]

Exercise (Continuity of quadratic functions)

Show that the quadratic function is continuous.

Proof (Continuity of quadratic functions)

Let . Consider any sequence , converging to . There is

So we may pull the limit inside the brackets for the quadratic function and hence, it is continuous.

Epsilon-Delta Criterion [Bearbeiten]

Main article: Epsilon-delta definition of continuity

Review: Epsilon-Delta Criterion[Bearbeiten]

Definition (Epsilon-Delta-definition of continuity)

A function with is continuous at , if and only if for any there is a , such that holds for all with . Written in mathematical symbols, that means is continuous at if and only if

General Proof Structure[Bearbeiten]

In mathematical quantifiers, the epsilon-delta definition of continuity of a function at the point reads:

This technical method of writing the claim specifies the general proof structure for proving continuity using the epsilon-delta criterion:

Example Problem and General Procedure[Bearbeiten]

Exercise (Continuity of the quadratic function)

Prove that the function with is continuous.

How to get to the proof? (Continuity of the quadratic function)

For this proof, we need to show that the square function is continuous at any argument . Using the proof structure for the epsilon-delta criterion, we are given an arbitrary . Our job is to find a suitable , such that the inequality holds for all .

In order to find a suitable , we plug in the definition of the function into the expression which shall be smaller than :

The expression may easily be controlled by . Hence, it makes sense to construct an upper estimate for which includes and a constant. The factor appears if we perform a factorization using the third binomial formula:

The requirement allows for an upper estimate of our expression:

The we are looking for may only depend on and . So the dependence on in the factor is still a problem. We resolve it by making a further upper estimate for the factor . We will use a simple, but widely applied "trick" for that: A is subtracted and then added again at another place (so we are effectively adding a 0) , such that the expression appears:

The absolute is obtained using the triangle inequality. This absolute is again bounded from above by  :

So reshaping expressions and applying estimates, we obtain:

With this inequality in hand, we are almost done. If is chosen in a way that , we will get the final inequality . This is practically found solving the quadratic equation for . Or even simpler, we may estimate from above. We use that we may freely impose any condition on . If we, for instance, set , then which simplifies things:

So will also do the job. This inequality can be solved for to get the second condition on (the first one was ):

So any fulfilling both conditions does the job: and have to hold. Ind indeed, both are true for . This choice will be included into the final proof:

Proof (Continuity of the quadratic function)

Let be arbitrary and . If an argument fulfills then:

This shows that the square function is continuous by the epsilon-delta criterion.

Concatenated absolute function[Bearbeiten]

Exercise (Example for a proof of continuity)

Prove that the following function is continuous at :

How to get to the proof? (Example for a proof of continuity)

We need to show that for each given , there is a , such that for all with the inequality holds. In our case, . So by choosing for small enough, we may control the expression . First, let us plug into in order to simplify the inequality to be shown  :

The objective is to "produce" as many expressions as possible, since we can control . It requires some experience with epsilon-delta proofs in order to "directly see" how this is achieved. First, we need to get rid of the double absolute. This is done using the inequality . For instance, we could use the following estimate:

However, this is a bad estimate as the expression no longer tends to 0 as . To resolve this problem, we use before applying the inequality  :

A factor of can be directly extracted out of this with the third binomial fomula:

And we can control it by :

Now, the required must only depend on and . Therefore, we have to get rid of the -dependence of . This can be done by finding an upper bound for which does not depend on . As we are free to chose any for our proof, we may also impose any condition to it which helps us with the upper bound. In this case, turns out to be quite useful. In fact, or an even higher bound would do this job, as well. What follows from this choice?

As before, there has to be . As , we now have and as , we obtain and . This is the upper bound we were looking for:

As we would like to show , we set . And get that our final inequality holds:

So if the two conditions for are satisfied, we get the final inequality. In fact, both conditions will be satisfied if , concluding the proof. So let's conclude our ideas and write them down in a proof:

Proof (Example for a proof of continuity)

Let be arbitrary and let . Further, let with . Then:

Step 1:

As , there is . Hence and . It follows that and therefore .

Step 2:

Hence, the function is continuous at .

Hyperbola[Bearbeiten]

Exercise (Continuity of the hyperbolic function)

Prove that the function with is continuous.

How to get to the proof? (Continuity of the hyperbolic function)

The basic pattern for epsilon-delta proofs is applied here, as well. We would like to show the implication . Forst, let us plug in what we know and reshape our terms a bit until a appears:

By assumption, there will be , of which we can make use:

The choice of may again only depend on and so we need a smart estimate for in order to get rid of the -dependence. To do so, we consider .

Why was and not chosen? The explanation is quite simple: We need a -neighborhood inside the domain of definition of . If we had simply chosen , we might have get kicked out of this domain. For instance, in case, the following problem appears:

The biggest -value with is and the smallest one is . However, is not inside the domain of definition as . In particular, is element of that interval, where cannot be defined at all.

A smarter choice for , such that the -neighborhood doesn't touch the -axis is half of the distance to it, i.e. . A third of this distance or other fractions smaller than 1 would also be possible: , or .

As we chose and by , there is . This allows for an upper bound: and we may write:

So we get the estimate:

Now, we want to prove . Hence we choose . Plugging this in, our final inequality will be fulfilled.

So again, we imposed two conditions for  : and . Both are fulfilled by , which we will use in the proof:

Proof (Continuity of the hyperbolic function)

Let with and let be arbitrary. Further, let be arbitrary. We choose . For all with there is:

Hence, the function is continuous at . And as was chosen to be arbitrary, the whole function is continuous.

Concatenated square root function[Bearbeiten]

Exercise (Epsilon-Delta-Beweis für Stetigkeit einer Wurzelfunktion)

Beweise mit der Epsilon-Delta-Definition der Stetigkeit, dass folgende Funktion stetig ist:

How to get to the proof? (Epsilon-Delta-Beweis für Stetigkeit einer Wurzelfunktion)

Wir müssen zeigen, dass für jedes ein existiert, so dass alle mit die Ungleichung erfüllen. Hierzu betrachten wir zunächst die Zielungleichung und schätzen den Betrag geschickt nach oben ab. Da wir den Term kontrollieren können, schätzen wir so nach oben ab, dass wir den Betrag erhalten. Wir suchen also eine Ungleichung der Form

Dabei ist irgendein von und abhängiger Term. Der zweite Faktor ist kleiner als und kann damit durch eine geschickte Wahl von beliebig klein gemacht werden. Eine solche Abschätzung ist folgende:

Wegen ist:

Wenn wir so klein wählen, dass ist, folgt die Zielungleichung . Jedoch hängt von ab und diese Abhängigkeit würde sich auf vererben und wir dürfen nicht in Abhängigkeit von wählen. Deswegen müssen wir die Abhängigkeit des ersten Faktors von eliminieren. Dies erreichen wir, indem wir den ersten Faktor nach oben so abschätzen, dass wir eine Ungleichung der Form erreichen. Eine solche Umformung ist:

Wir haben sogar unabhängig von gemacht, was nicht nötig gewesen wäre. Somit haben wir die Ungleichung

Wir brauchen nun die Abschätzung , damit die Zielungleichung erfüllt ist. Die Wahl von ist hierfür ausreichend.

Proof (Epsilon-Delta-Beweis für Stetigkeit einer Wurzelfunktion)

Sei mit . Sei und beliebig. Wir wählen . Für alle mit gilt:

Damit ist eine stetige Funktion.

Discontinuity of the topological sine function[Bearbeiten]

Exercise (Discontinuity of the topological sine fucntion)

Prove the discontinuity at for the topological sine function:

How to get to the proof? (Discontinuity of the topological sine fucntion)

In this exercise, discontinuity has to be shown for a given function. This is done by the negation of the epsilon-delta criterion. Our objective is to find both an and an , such that and . Here, may be chosen depending on , while has to be the same for all . For a solution, we may proceed as follows

Step 1: Simplify the target inequality

First, we may simplify the two inequalities which have to be fulfilled by plugging in and . We therefore get: and .

Step 2: Choose a suitable

We consider the graph of the function . It will help us finding the building bricks for our proof:

Graph of the topological sine function

We need to find an , such that there are arguments in each arbitrarily narrow interval whose function values have a distance larger than from . Visually, no matter how small the width of the --rectangle is chosen, there will always be some points below or above it.

Taking a look at the graph, we see that our function oscillates between and . Hence, may be useful. In that case, there will be function values with in every arbitrarily small neighborhood around . We choose . This is visualized in the following figure:

The topological sine function oscillates between -1 and 1 nea the origin. Hence, in each neighborhood around 0, there will be arguments x , where f(x) has a distance larger than 1/2 to 0.

After has been chosen, an arbitrary will be assumed, for which we need to find a suitable . This is what we will do next.

Step 3: Choice of a suitable

We just set . Therefore, has to hold. So it would be nice to choose an with . Now, is obtained for any with . The condition for such that the function gets 1 is therefore:

So we found several , where . Now we only need to select one among them, which satisfies for the given . Our depend on . So we have to select a suitable , where . To do so, let us plug into this inequality and solve it for  :

So the condition on is . If we choose just any natural number above this threshold , then will be fulfilled. Such a has to exist by Archimedes' axiom (for instance by flooring up the right-hand expression). So let us choose such a and define via . This gives us both and . So we got all building bricks together, which we will now assemble to a final proof:

Proof (Discontinuity of the topological sine fucntion)

Choose and let be arbitrary. Choose a natural number with . Such a natural number has to exist by Archimedes' axiom. Further, let . Then:

In addition:

Hence, the function is discontinuous at .

Example Problems[Bearbeiten]

Sequence Criterion: Absolute Value Function[Bearbeiten]

Exercise (Continuity of the absolute function)

Prove continuity for the absolute function.

Proof (Continuity of the absolute function)

Let with be the absolute function. Let be a real number and a sequence converging to it . In chapter „Grenzwertsätze: Grenzwert von Folgen berechnen“ we have proven the absolute rule, stating that , whenever there is . Hence:

This proves continuity of the absolute function by the sequence criterion.

Discontinuity of the topological sine function[Bearbeiten]

Exercise (Discontinuity of the topological sine function)

Prove discontinuity of the following function:

How to get to the proof? (Discontinuity of the topological sine function)

Discontinuity of means that this function has at least one argument where is discontinuous. For each , is equal to the functionn in a sufficiently small neighbourhood of . Since is just a composition of continuous functions, it is continuous itself and therefore, must be continuous for all , as well. So we know that the discontinuity may only be situated at .

In order to prove that is discontinuous at , we need to fincd a sequence of arguments converging to but with . To find such a function, let us take a look at the graph of the function  :

Graph of the topological sine function f

In this figure, we recognize that takes any value between and infinitely often in the vicinity of . So, for instance, we may just choose such that is always . This guarantees that - and actually any other real number between and in place of would do the job. But we need to make a specific choice for , and is a very simple one. In addition, we will choose to converges to zero from above.

The following figure also contains the sequence elements beside our function . We may clearly see that for the sequence of function values converges to , which is different from the function value  :

Graph of the topological sine function f together with the function values for our sequence of arguments, which converges to 0 from the right and always takes function value 1.

But what are the exact values of these for which we would like to have To answer this question, let us resolve the equation for  :

So for each with , we have . In order to get positive converging to zero from above, we may for instance choose . In that case:

And we have seen that . So we found just a sequence of arguments , which proves discontinuity of at .

Proof (Discontinuity of the topological sine function)

Let with for and . We consider the sequence defined by . For this sequence:

And there is:

Hence, although . This proves that is discontinuous at and therefore it is a discontinuous function.

Epsilon-Delta Criterion: Linear Function[Bearbeiten]

Exercise (Continuity of a linear function)

Prove that a linear function with is continuous.

How to get to the proof? (Continuity of a linear function)

Graph of a function with . Considering the graph , we see that this function is continuous everywhere.

To actually prove continuity of , we need to check continuity at any argument . So let be an arbitrary real number. Now, choose any arbitrary maximal error . Our task is now to find a sufficiently small , such that for all arguments with . Let us take a closer look at the inequality  :

That means, has to be fulfilled for all with . How to choose , such that implies  ?

We use that the inequality contains the distance . As we know that this distance is smaller than . This can be plugged into the inequality  :

If is now chosen such that , then will yield the inequality which we wanted to show. The smallness condition for can now simply be found by resolving for :

Any satisfying could be used for the proof. For instance, we may use . As we now found a suitable , we can finally conduct the proof:

Proof (Continuity of a linear function)

Let with and let be arbitrary. In addition, consider any to be given. We choose . Let with . There is:

This shows , and establishes continuity of at by means of the epsilon-delta criterion. Since was chosen to be arbitrary, we also know that the entire function is continuous.

Epsilon-Delta Criterion: Concatenated Absolute Value Function[Bearbeiten]

Exercise (Example for a proof of continuity)

Prove that the following function is continuous at :

How to get to the proof? (Example for a proof of continuity)

We need to show that for each given , there is a , such that for all with the inequality holds. In our case, . So by choosing for small enough, we may control the expression . First, let us plug into in order to simplify the inequality to be shown  :

The objective is to "produce" as many expressions as possible, since we can control . It requires some experience with epsilon-delta proofs in order to "directly see" how this is achieved. First, we need to get rid of the double absolute. This is done using the inequality . For instance, we could use the following estimate:

However, this is a bad estimate as the expression no longer tends to 0 as . To resolve this problem, we use before applying the inequality  :

A factor of can be directly extracted out of this with the third binomial fomula:

And we can control it by :

Now, the required must only depend on and . Therefore, we have to get rid of the -dependence of . This can be done by finding an upper bound for which does not depend on . As we are free to chose any for our proof, we may also impose any condition to it which helps us with the upper bound. In this case,