Solutions to IVT Problems

(a) Since is continuous on a compact set ( it attains both its maximum and minimum values, so for all . Now using these poor estimates we have that . So if we consider the function we have that is continuous on the same interval as . Thus we just apply the intermediate value theorem to and the desired result follows.

(b) WLOG assume that . Let be given and define new function by . This is just the equation for the slope of the line connecting the points and and is continuous because is continuous. Now consider the set of points for . Now look at the values takes at these points. If for any then we have found a solution, so assume that for any chose of . So WLOG assume that , since we know that there must be some such that , call this . Then we there is a such that .

(c) For this one I will prove something a little stronger. All that we need to assume about is that it is integrable, and has what I shall call the extreme value property on our interval. Basically this property says that on the interval there exists a point such that and a point such that , where ranges over all the values in our interval. So clearly an increasing function satisfies these conditions. Now define a function by

Now is continuous and , thus by the intermediate value theorem we are done.

Some after thoughts: Problem (a) was pretty straight forward and didn’t take too much insight to actually solve. However, problem (b) was a little more challenging until I visualized what I was being asked to prove. This led me to the idea to look at the slope of the connecting line segment. Finally problem (c), I at first wondered why you were given that was increasing and not continuous, or even just integrable. As can be seen in my proof continuity imposes more conditions than are necessary, while integrability does not provide you with enough. This led me to think about what makes increasing functions special (well one of the things that makes them special).