Math for Non-Geeks/ Exercises: Continuity

oben Lipschitz continuous functions are uniformly continuous Let be Lipschitz continuous with Lipschitz constant . That is for all . Prove that is uniformly continuous. 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 . Let be arbitrary. We choose . Then, for all with : Continuity at the origin Prove that the following function is continuous at the origin : with a real number 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 . Let . We choose . Let be any real number with . Then: This shows that is continuous at . Extreme value theorem Maximum and minimum of a function Prove that the function defined on attains a maximum, but not a minimum. 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 . 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 be for some , which is a contradiction ). How often is a value attained #1 # Show that there is no continuous function , which attains each of its function values exactly twice. Is there a continuous function , which attains each of its function values exactly three times? 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 Zero of a function Prove that the function has exactly one zero inside the interval . 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 . has exactly one zero is strictly monotonically increasing on . The function is also monotonically 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 monotonically increasing (it would either have to stay constant or go "down again" between and ). Solution of an equation Let with . Prove that the equation Has at least three solutions. 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. Solution of an equation Let be continuous with . Prove that there is a with . 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 This means , or equivalently So we have found a solution to . 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 . Existence of exactly one zero Let be a natural number. We define the function . Prove that has exactly one positive 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 monotonically increasing functions for . So we may assume that is also strictly monotone, 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 monotone in the sense that for . If there were two zeros , we would have although there is . This would contradict being monotone and in hence excluded. Therefore, can have at most one zero (as all strictly monotone 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 with . 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 Continuity of the inverse function 1 Let be defined by Prove that is continuous, strictly monotone and injective. Prove that is subjective (so it is a 1-to-1-map from to an inverse function exists). Why is the inverse function continuous, monotonically increasing and bijective? Explicitly determine . Part 1: is continuous on since it is the quotient of the continuous polynomials and . Note that for all . Let with . Then, strict monotonicity 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 monotonically 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: 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 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. Continuity of the inverse function 2 Let Prove that is injective. Determine the range of all attained values. Why is the inverse function continuous? Part 1: is continuous, as it is composed by the continuous functions , , and on . The logarithm is strictly monotonically increasing (and its inverse is decreasing): for with , there is: Now, for . Since in addition, is strictly monotonically decreasing on , we have So the -term is also strictly monotonically 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 monotonically 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. It tells us that is indeed continuous. unten