College Algebra

Chapter 4: Continuity and the Fundamental Theorem of Algebra

4.1 Continuity

Intuitively, a function f(x) is said to continuous at b if functional values f(x) are as close as we would like to f(b) as soon as x is sufficiently close to b and at a place where f(x) is defined. For example, the top surface of a table defines a flat surface which is continuous. It is even continuous at the edge of the table because f(x) is not defined for points x beyond the edge of the table. On the other hand, we have a room containing only a table and consider the function consisting of the table top and the part of the floor which is not under the table, then this function is still continuous at all points except at the points corresponding to the edge of the table. Given such a point, there are points arbitrarily close to it where the functional value is defined by the table top and other points arbitrarily close to it where the functional value is defined by the floor level.

Example 1: i. Another example is the function consisting of all the pairs for non-zero x together with the single point (0, 1). This is usually written as

This function is continuous at all x except for x = 0. All the functional values for x near 0 are close to 0 whereas f(0) = 1.

ii. A less extreme case is

Again, this function is continuous at all x except x = 0.

iii. This function is not continuous anywhere:

iv. This monstrosity is continuous everywhere except at (0, 0):

Let's give a more careful definition: Let f(x) be a real valued function defined for certain real values x. If f is defined at b, we want to say that f is continuous at b if f(x) is as close as we wish to f(b) for all x sufficiently close to b where f is defined. The problem is to make sense of expressions such as "as close as we wish". One might wish for different degrees of closeness at various times. To cover the most stringent case, we will interpret this as meaning the distance between the two is less than any specified positive number. So, "f(x) is as close as we wish to f(b)" means that, if you are given a maximal distance , then one must have . The expression, "for all x sufficiently close to b" means that there is a positive number such that the assertion holds for all x satisfying . So, our careful definition is:

Definition 1: Let f be a function with domain a set of real numbers and with range space the set of all real numbers. We say that f is continuous at b in its domain if for every there is a such that for all x in the domain of f with .

Showing that a function is continuous can be a lot of work:

Example 2: i. The function is continuous at x = 0. In fact, if we are given , then we need to show that there is a such that for all x with . If we choose to be the smaller of and 1, then we can see that this works. In fact, if , then because . So . Since , we know that . So, as required.

ii. Now let's try and show that the same function is continuous at x = 1. For a given , we need to have for all x with , where is yet to be chosen. Now, . So, if we want this to be small by making |x - 1| small, we can do it provided that |x + 1| is not made large in the process. But if . So, if , we would have

where the last inequality holds provided we choose . This is precisely what we want provided that we also have .

So, given , let be any number smaller than both 1 and . Then, if , we have because

This proves that is continuous at x = 1.

iii. Now, let's try to show that is continuous at all real number b. Repeating the same kind of reasoning, we would want

Given an \epsilon > 0, we could choose so that it is less than 1 and Then, we would have for all x with .

We will need to study the continuity of real valued functions of two variables. The careful definition is almost the same as in the one variable case:

Definition 2: Let f be a function with domain a set of pairs of real numbers and with range space the set of all real numbers. We say that f is continuous at (a,b) in its domain if for every there is a such that for all (x,y) in the domain of f with .

All we really did was replace the absolute value with the distance function: the condition that is replaced with the distance from (x,y) to (a,b) be less than . We could have written Definition 1 in the same way, since absolute value gives the distance between two real numbers.

Example 3: i. Any constant function is continuous at every point in its domain. Suppose f(x) = c where c is real number for all x in the domain of f. Let b be in the domain of f. If , then we need

for all x in the domain of f with . But this holds for any choice of as long as it is a positive number.

ii. Any linear function f(x,y) = ax + by + c is continuous at all points (x, y) in its domain. In fact, let and (r, s) be in the domain of r. We need

for all (x, y) in the domain with .

For (x, y) with , one has:

because the square root function is increasing. Similarly . But then, we have:

If we choose our so that , then the right hand side will be smaller than as desired.

4.2 More Continuous Functions

Proving that a function is continuous using only the definition can be quite tedious. So, we will need to develop some results which make it easy to check that certain functions are continuous. Throughout this section we will be dealing with real valued functions of one or two variables. We will take their values at points z where, by point, we mean either a real number or a pair of real numbers depending on whether the function is of one or two variables. We will also use absolute value signs to indicate the distance to the origin, either 0 or (0, 0), depending on whether the function is of one or two variables.

Given two functions f and g, we can define sums, differences, products, and quotient functions by:

1. (f + g)(z) = f(z) + g(z)
2. (f - g)(z) = f(z) - g(z)
3. (fg)(z) = f(z)g(z)
4. (f/g)(z) = f(z)/g(z)

Proposition 1: Let f and g be real valued function of one or two variables. Let z be a point in the domain of f and the domain of g where both f and g are continuous.

1. f + g and f - g are continuous at z.
2. fg is continuous at z.
3. If , then f/g is continuous at z.

Proof: Let w be a point in the intersection of the domains of f and of g. Then

If is any positive number, then we can use the continuity of f and g at z to know that there is a and such that and for all points w in the domains of f and g such that (for f) and (for g). Choosing to be any number smaller than both and gives

as needed. You should check through the same proof using subtraction instead of addition of functions.

For products, one has

Let . Choose smaller than 1 and subject to another condition which we will specify below. By the continuity of f at z, we know that there is a such that for all w in the domain of f with . Similarly, by the continuity of g at z, we know that there is a such that for all w in the domain of g with . Choose any smaller than both and . Continuing our series of inequalities, we get for all w in the domains of both f and of g with that

where the last inequality follows because we add the condition to the list of conditions which is required to satisfy.

Finally, let's consider the case of quotients. We need to assume that z is in the domains of f and of g as well as . The inequality looks like

Now the numerator is like the one for products and the same sort of argument will be able to handle it. The new ingredient is the denominator. In order to get an upper bound on a quotient |a|/|b|, you need to either make the numerator larger or the denominator smaller. So, we need a lower bound on |g(w)|. But and so we can choose a such that for all w in the domain of g with , one has . Since g(w) is close to g(z), it cannot be too close to zero; more specifically,

where we have used:

Lemma 1: If a and b are real numbers, then .

Proof: This is an exercise. Let be chosen according to criteria which we will figure out below. Now, choose so and for all w in the domains of f and g such that . Then we can continue our inequalities:

where the last inequality would be true if we chose so that .

Corollary 1: i. Every polynomial with real coefficients is continuous at all real numbers. ii. If p(z) is a polynomial with complex coefficients and p(x, y) = r(x,y) + i s(x,y), then r(x,y) and s(x,y) are continuous at all pairs (x, y). iii. If p(z) is a polynomial with complex coefficients, then |p(x,y)| is continuous at all (x, y).

Proof: i. and ii. All of these are functions made up of a finite number of additions and multiplications of continuous functions. So the result follows by applying the Proposition a certain number of times. (More correctly, one can proceed by descent assuming that one has the function made up with the least number of multiplications and additions.)

For assertion iii, one needs

Lemma 2: i. If f(x, y) is a real valued function continuous at (a, b) and g(x) is a real valued function continuous at f(a, b), then g(f(x,y)) is continuous at (a, b)

ii. The square root function is continuous at all non-negative reals.

Proof: i. Let . Since g is continuous at f(a, b), there is a so that for all w in the domain of g such that . Further, since f is continuous at (a, b), one knows that there is a such that for all (x, y) in the domain of f where . But then, letting w = f(x,y), we have

ii. Let a > 0 and . One has for with where is a quantity to be determined:

So, if we choose so that , then one has .

Now consider the case where a = 0. If , choose smaller than 1 and . If x is non-negative and , then since the square root function is increasing, and so, the square root function is even continuous at zero.

3.3 Theorems of Bolzano and Weierstrass

The typical discontinuity is a point where the function makes a jump. Continuous functions cannot do this:

Proposition 2: ( Bolzano's Theorem) Let f(x) be a continuous function defined on the closed interval [0,1]. If f(0) and f(1) have different signs (i.e. one is positive and the other is negative), then f has a root c in (0, 1).

Proof: Let's try to find the number c by developing a binary decimal expansion using binary search. At each step, we will add one binary digit to the expansion.

Originally, we only know to look in the interval . So, the first approximation is . Now, consider the midpoint 1/2 of the interval . If f(1/2) = 0, we have found our . Otherwise, either f(0) and f(1/2) have different signs or else f(1/2) and f(1) have different signs. In the first case, let and . Otherwise, let and . We have replaced our original interval with one half as long without losing the property that the function has different signs at the endpoints.

Now, just repeat the process over and over again. Here are the details: Assuming that one has , already defined where and have different signs. Let be the midpoint of the interval. If , then is the sought after point. If not, then it could happen that and have different signs; in this case, let be with a 0 digit added on the right and let , , and . Otherwise, and f(b_k) have different signs; in this case, let be with a 1 digit added on the right and let , , and .

The define an infinite binary decimal, which agrees with each in the first digitis. Because we are in the real numbers, the binary decimal converges to a number which we also call . This is a root of f. Otherwise, would be non-zero and we could find a such that for all with . So, for all such , has the same sign as . But this is absurd because for large enough , the interval lies entirely within of c and and have different signs.

Corollary 2: (Intermediate Value Theorem) If f(x) is continuous on the closed interval [a, b], then for every real number d between f(a) and f(b), there is a c in (a, b) with f(c) = d.

Proof: Apply Bolzano's Theorem to the function f(a + (b-a)x) - d.

Definition 3: A point z is an absolute minimum of a function f defined on a set S if z is in the domain of f and for all w in the set S. A point z is an absolute maximum of a function f defined on a set S if z is in the domain of f and for all w in S.

Proposition 3: ( Weierstrass's Extreme Value Theorem) If f is a real valued function defined and continuous on a closed interval [a, b] (or a closed rectangle in the case of a function of 2 variables), then f has at least one absolute maximum and at least one absolute minimum on this interval (or rectangle).

Proof: We can use binary search here as well. Let's discuss the one variable case first, and then indicate what changes need to be made to handle the two variable one. First, note that it is enough to handle the case where because one can always replace the function with . Start with the interval , , and search for an absolute maximum. Let . If for every in , there is a number y in with , then let and (i.e. add a 0 digit on the right of . If not, then there is an x in with for every y in . In particular, for every y in , there is a number (viz. x) in such that . In this case, let and , i.e. add a 1 digit to the right of . We have assured that an absolute of f on will be an absolute maximum of f on and we have halved the length of the interval.

As in the previous proposition, we can repeat this process defining and given and . Each step gives an interval of half the length whose maximum would be a maximum over . Let c be the limit of the infinite binary decimal defined by the . Then c is an absolute maximum. In fact, if e is in [0, 1] with f(e) > f(c), then there is a k with e in and e not in . By the construction, there is a g in with . In fact, there are such g in for arbitrarily large j. In fact, g is different from c, and so there is a k with e in and g not in . Then there must be an h with h in and . You can now repeat the process with h in place of g.

But f is continuous at c and so there is a such that for all x with . Since lies within of c for all sufficiently large j, we have a contradiction.

The same sort of proof can be used to prove Weierstrass's Theorem in the case of functions of two variables. In this case, we have a rectangle which is just the cartesian product of two intervals. Instead of dealing with a nested set of intervals, we deal with rectangles. We can let and be the two midpoints. The crucial step is:

1. Divide the rectangle into and . Discard a half which does not contain a point p with f(p) > f(z) for all z in the other half. Suppose you keep .
2. Divide the new rectangle in half the other way, i.e. the halves are and . Discard a half which does not contain a point p with f(p) > f(z) for all z in the other half.

Remark: Although both Bolzano's and Weierstrass's Theorems were proved with binary search, examination shows that the proof of Weierstrass's Theorem really does not give us any effective means for finding the absolute maximum. We know it exists, but can't really find it. On the other hand, the proof of Bolzano's Theorem does allow one to actually approximate the desired root to any desired degree of accuracy. Since the size of the interval shrinks by a factor of 2 at each step, we get another decimal digit of accuracy every three or four steps.

4.4 The Proof of the Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states:

Theorem 1: ( Gauss) Every non-constant polynomial with complex coefficients has at least one complex root.

Lemma 3: ( D'Alembert's Lemma) Let f(z) be a non-constant polynomial with . Then there is a non-zero complex number c such that |f(cx)| < |f(0)| for all sufficiently small positive real values x.

Proof: Let . be a non-zero polynomial with complex coefficients. We assume that and that , i.e. the smallest positive degree term of degree k. Then .

We want to choose c so that . Since x is a positive, the factor does not affect the argument. Since arg converts products into sums, we can solve to get

One can choose any non-zero c with this argument. Then the last two terms of can be written in the form where d is a positive real number. The remaining terms are in absolute value at most for some positive e. So,

for all positive x small enough so that x <|a_n|r/(2e). This completes the proof.

Corollary 3 Let g(z) be any polynomial with complex coefficients and b be any complex number with . Then there are points c arbitrarily close to b where . In particular, b is not an absolute minimum of the function |g(z)|.

Proof: Let f(z) = g(b + z). Then and f is a polynomial with complex coefficients. The result now follows from Lemma 3.

Proof of the Fundamental Theorem of Algebra Suppose f(z) is a non-constant polynomial with complex coefficients and no complex roots. If d is the degree of g(z), then the triangle inequality shows that for z sufficiently large in absolute value, the highest degree term dominates and one has for some positive c and all z with |z| > M. In particular, one knows that there is a square centered at the origin which is guaranteed to contain in its interior all the absolute minima of the function |g(z)|. By the Weierstrass Extreme Value Theorem applied to a slightly larger square, there is at least one such absolute minimum, say z. By D'Alembert's Lemma, we must have f(z) = 0, which is a contradiction.