Exercises: Continuity – Serlo

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Lipschitz continuous functions are uniformly continuous[Bearbeiten]

Exercise

Let be Lipschitz continuous with Lipschitz constant . That is

for all . Prove that is uniformly continuous.

How to get to the proof?

We need to show that for all , there is a , such that for all with there is . By our assumption, we have

In order for to hold, it suffices to have . We can reach this by taking .

Proof

Let be arbitrary. We choose . Then, for all with :

Continuity at the origin[Bearbeiten]

Exercise

Prove that the following function is continuous at the origin :

with a real number

How to get to the proof?

In order to establish continuity at the origin , we make use of the epsilon-delta criterion. That means, for all we have to find a such that the inequality holds at all with . The question now is how to find a suitable for any given . To answer this question, we take a look at the function around :

Since at , there is

This inequality also holds at , since . Visually, the inequality above means that the graph of the function fits inside the "double wedge" given by , where tells us, how much the wedge is "stretched" in -direction. So if we choose , then at any with , there is

Which we can use to carry through the proof (see below).

Note: We can also "move the double wedge" to any when investigating continuity at . If the graph fits inside the "moved double wedge" , then for any , there is

So any function fitting in such a double wedge is continuous. The converse does not hold true. There are functions which do not fit in any double wedge around but are continuous at . An example is the square root function around .

Proof

Let . We choose . Let be any real number with . Then:

This shows that is continuous at .

Extreme value theorem[Bearbeiten]

Exercise (Maximum and minimum of a function)

Prove that the function defined on attains a maximum, but not a minimum.

Solution (Maximum and minimum of a function)

Proof step: attains a maximum

The function is continuous on , since it is composed by continuous functions and the denominator is strictly positive (). The enumerator is also strictly positive, so for , , and

That means, there is an with for all . The extreme value theorem implies that attains a maximum on . One may even show that the maximum is even global. However, computing the maximum explicitly would require us to solve , which is quite a computational effort and can be even harder for other functions . The extreme value theorem just allowed us to quickly show that there is a maximum - and saved us from the tedious solution of .

Proof step: does not attain a minimum

There is on . And indeed, approaches 0 (see the previous proof step). But it does not attain 0. Since and by continuity, there can not be a minimum (if there was an attained minimum , then by , there would by for some , which is a contradiction ).

Exercise (How often is a value attained #1)

  1. Show that there is no continuous function , which attains each of its function values exactly twice.
  2. Is there a continuous function , which attains each of its function values exactly three times?

Exercise (How often is a value attained #2)

Let with . Show: There is no continuous function , which attains each of its function values exactly times.

Intermediate value theorem and zeros[Bearbeiten]

Exercise (Zero of a function)

Prove that the function

has exactly one zero inside the interval .

Solution (Zero of a function)

Proof step: has at least one zero

is continuous as it is composed by the continuous functions and . In addition,

and

By means of the intermediate value theorem, there must be an with .

Proof step: has exactly one zero

is strictly monotonously increasing on . The function is also monotonously increasing on , so is decreasing and again increasing. Hence there can be only one zero with , since a function with two zeros is never strictly monotonously increasing (it would either have to stay constant or go "down again" between and ).

Exercise (Solution of an equation)

Let with . Prove that the equation

Has at least three solutions.

Solution (Solution of an equation)

It is a powerful trick in mathematics, to transform the problem of finding solutions to as zeros of an auxiliary function (if , then and vice versa). In our case, the continuous auxiliary function is

When approaching and , this function goes to

and

Therefore, there must be two arguments with and ( is close to and close to ). By the intermediate value theorem, there must hence be a zero with . This zero is one solution of the above equation.

The same argument works between and . Since and , we can use the intermediate value theorem and get a zero with .This is the second solution we have been looking for.

The third solution follows by a similar argument. There is and . So the intermediate value theorem renders a with . The equation has therefore at least three solutions.

Exercise (Solution of an equation)

Let be continuous with . Prove that there is a with .

Solution (Solution of an equation)

We consider the following auxiliary function:

Finding a with now amounts to finding a zero of . Since is continuous, so is . In addition, at the endpoints of the interval, there is

and

Fall 1:

This means , or equivalently

So we have found a solution to .

Fall 2:

We will first consider the case . Since , there is . The intermediate value theorem now yields a with

This is a zero of and hence a desired solution for . The other case can be treated using exactly the same arguments.

So for any choice of , there is a with .

Exercise (Existence of exactly one zero)

Let be a natural number. We define the function . Prove that has exactly one positive zero.

Solution (Existence of exactly one zero)

We need to show two things: At first, we need to show that a zero exists inside the interval . Second, we need to assure that there is indeed only one such zero.

The function is a polynomial function and hence continuous. At the beginning of the interval , there is i.e. the graph of the function runs below the -axis. At infinity, there is , meaning that for large , the graph runs above the -axis. As is continuous, we can apply the intermediate value theorem and get a zero .

Now we need to show that there is at most one zero. Both and are strictly monotonously increasing functions for . So we may assume that is also strictly monotonous, there. We can prove this assumption be taking the first derivative:

For there is: .

One may show with a bit of effort that differentiable functions with positive derivative are strictly monotonous in the sense that for . If there were two zeros , we would have although there is . This would contradict being monotonous and in hence excluded. Therefore, can have at most one zero (as all strictly monotonous functions).

Note: We could also prove that has at most one zero, only using that is differentiable with :

Assume that, the function would have two zeros with . Since the function is differentiable and , we may use Rolle's theorem and get that some exists wit . But this is a contradiction to the first derivative of being strictly positive . So this is a second way to exclude the existence of two zeros.

Continuity of the inverse function[Bearbeiten]

Exercise (Continuity of the inverse function 1)

Let be defined by

  1. Prove that is continuous, strictly monotonous and injective.
  2. Prove that ist surjectiive (so it is a 1-to-1-map from to an inverse function exists).
  3. Why is the inverse function continuous, monotonously increasing and bijective? Explicitly determine .

Solution (Continuity of the inverse function 1)

Part 1: is continuous on since it is the quotient of the continuous polynomials and . Note that for all .

Let with . Then, strict monotony holds:

Therefore, is also injective.

Part 2: The function runs towards infinity at the end points of the open interval as follows:

and

Since is continuous, the intermediate value theorem ensures that for each there is a mapped onto it: . Therefore, is also surjective: .

Part 3: Since is bijective, the inverse map exists and is bijective, as well:

The theorem about continuity of the inverse function tells us that is continuous and strictly monotonously increasing. Now, let us compute . That means, we need to bring into the form - i.e. we need to get standing alone on the left side of the equation:

Fall 1:

Fall 2:

We can use the quadratic solution formula in order to solve for :

Since for , the only reasonable solution is . Putting all together, the full definition for the inverse function reads

Hint

The distinction of two cases above is not very convenient. We can avoid it using a little trick: In case enumerator and denominator of can be multiplied by a factor of :

Plugging in , we get that is described correctly. So we can use the definition above for all and avoid the case distinction.

Exercise (Continuity of the inverse function 2)

Let

  1. Prove that is injective.
  2. Determine the range of all attained values.
  3. Why is the inverse function continuous?

Solution (Continuity of the inverse function 2)

Part 1:

is continuous, as it is composed by the continuous functions , , and on .

The logarithm is strictly monotonously increasing (and its inverse is decreasing): for with , there is:

Now, for . Since in addition, is strictly monotonously decreasing on , we have

So the -term is also strictly monotonously increasing and so is :

Therefore, is also injective.

Part 2:

At the ends of the domain of definition, there is

and

this implies

is continuous on the interval . Hence, we can use a corollary of the intermediate value theorem, and get that is again an interval. Since is strictly monotonously increasing and , we can conclude

Part 3:

Since is an interval and in bijective, we can use the theorem about continuity of the inverse function. I tells us that

is indeed continuous.