Existence Proofs

An existence proof shows that an object exists. In some cases, this means displaying the object, or giving a method for finding it.

Example. Show that there is a real number x such that $\sin x > 0$ but $\cos x < 0$ .

There are many possibilities; for example,

$$\sin \dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}, \quad\hbox{but}\quad \cos \dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}.\quad\halmos$$

In some cases, you can know that an object exists without having any way of finding it (or finding it exactly). By analogy:

You've seen results of this kind in calculus. One such result is the Intermediate Value Theorem:

Example. Show that there is a real number x such that $x = \cos x$ .

The assertion means that the graphs of $y = x$ and $y = \cos x$ intersect:

$$\hbox{\epsfysize=1.75in \epsffile{existence-proofs1.eps}}$$

It looks like they do. Note, however, that a picture is not a proof.

Let $f(x) = \cos x - x$ . Then

$$f(0) = \cos 0 - 0 = 1, \quad\hbox{while}\quad f(\pi) = \cos \pi - \pi = -1 - \pi.$$

Since $f(0)$ is positive and $f(\pi)$ is negative, and since f is continuous for all x, the Intermediate Value Theorem implies that there is an x between 0 and $\pi$ for which $f(x) = 0$ . Then $\cos x - x = 0$ , so $\cos x = x$ .

Notice that the Intermediate Value Theorem doesn't tell you what x is, or how to find it. (It's approximately 0.73909.)

To say that there is an x satisfying a certain property does not mean that there is only one x satisfying the property. If that is what is meant, it has to be stated explicitly.

Example. The Mean Value Theorem says that if f is function which is continuous on the closed interval $a \le x \le b$ and differentiable on the open interval $a < x < b$ , then

$$f'(c) = \dfrac{f(b) - f(a)}{b - a}$$

for some number c such that $a < c < b$ .

The result says that there is a number c. This doesn't mean that you might not have several c's that work.

For example, suppose $f(x) = x^3$ and the interval is $1 \le x \le 3$ . Then

$$\dfrac{f(3) - f(1)}{3 - 1} = 13.$$

Now $f'(x) = 3x^2$ , so setting $3c^2 = 13$ , I find that $c = \pm\sqrt{\dfrac{13}{3}}$ . Both of these values satisfy the conclusion of the Mean Value Theorem.

The definition of the limit is an example of an existence assertion.

Let f be a function from the real numbers to the real numbers, and let c be a real number. The statement

$$\lim_{x\to c} f(x) = L$$

means:

\item{} For every $\epsilon > 0$ , there is a $\delta > 0$ , such that if $\delta > |x - c| > 0$ , then $\epsilon > |f(x) - L|$ .

Think of $\delta$ as a thermostat, $f(x)$ as the actual temperature in a room, and L as the ideal temperature. Someone challenges you to make the actual temperature $f(x)$ fall within a certain tolerance $\epsilon$ of the ideal temperature L. You must do that by setting your $\delta$ -thermostat appropriately (so that x is sufficiently close to c).

Moreover, note that it says "for every $\epsilon > 0$ ". It's isn't enough for you to say what you'd do if you were challenged with $\epsilon = 0.1$ or $\epsilon = 0.000004$ . You must prove that you can meet the challenge no matter what $\epsilon$ you're challenged with.

Finally, note the stipulation "$|x - c| > 0$ ". This implies that $x \ne c$ , since $x = c$ gives $|x - c| = 0$ . Thus, the conclusion "$\epsilon > |f(x) - L|$ " must hold only for x's close to c, but not necessarily for $x = c$ . (It may hold for $x = c$ , but it doesn't have to.)

What does this mean? It's a precise way of saying that the value of the limit of $f(x)$ as x approaches c does not depend on what $f(x)$ does at $x = c$ --- over even whether $f(c)$ is defined.

For example, consider the functions whose graphs are shown below.

$$\hbox{\epsfysize=1.75in \epsffile{existence-proofs2.eps}}$$

In both cases,

$$\lim_{x\to 3} f(x) = 4.$$

In the first case, $f(3) = 2$ : The value of the function at $x = 3$ is different from the value of the limit.

In the second case, $f(3)$ is undefined.

The fact that $\displaystyle \lim_{x\to 3} f(x) \ne f(3)$ means that f is not continuous at $x = 3$ .

Example. Use the $\epsilon-\delta$ definition of the limit to prove that

$$\lim_{x\to 2} (5x + 4) = 14.$$

In this case, $c = 2$ , $f(x) = 5x + 4$ , and $L = 14$ . So here is what I need to prove.

Suppose $\epsilon > 0$ . I must find a $\delta > 0$ such that if $\delta > |x - 2| > 0$ , then $\epsilon > |(5x + 4) - 14|$ .

Note that at this point $\epsilon$ is fixed --- given --- but all you can assume is that it's some positive number. Since it is given, however, I can use it in finding an appropriate $\delta$ .

I'll show how to find $\delta$ by working backwards; then I'll write the proof "forwards", the way you should write it.

I want

$$\epsilon > |(5x + 4) - 14|, \quad\hbox{or}\quad \epsilon > |5x - 10|, \quad\hbox{or}\quad \dfrac{\epsilon}{5} > |x - 2|.$$

It looks like I should set $\delta = \dfrac{\epsilon}{5}$ .

All of this has been on "scratch paper"; now here's the real proof.

Suppose $\epsilon > 0$ . Let $\delta = \dfrac{\epsilon}{5}$ . If $\delta > |x - 2| > 0$ , then

$$\dfrac{\epsilon}{5} > |x - 2|, \quad\hbox{so}\quad \epsilon > |5x - 10|, \quad\hbox{or}\quad \epsilon > |(5x + 4) - 14|.$$

Thus, if $\delta = \dfrac{\epsilon}{5}$ and $\delta > |x - 2| > 0$ , then $\epsilon > |(5x + 4) - 14|$ . This proves that $\displaystyle \lim_{x\to 2} (5x + 4) = 14$ .

Example. Let

$$f(x) = \cases{3x + 4 & if $x < 1$ \cr 9 - 2x & if $x \ge 1$ \cr}.$$

Use the $\epsilon-\delta$ definition of the limit to prove that

$$\lim_{x\to 1} f(x) = 7.$$

Let $\epsilon > 0$ . I must find $\delta > 0$ such that if $\delta > |x - 1| > 0$ , then $\epsilon > |f(x) - 7|$ .

Here's my scratch work. First, for $x < 1$ ,

$$\epsilon > |f(x) - 7|, \quad \epsilon > |(3x + 4) - 7|, \quad \epsilon > |3x - 3|, \quad \dfrac{\epsilon}{3} > |x - 1|.$$

It looks like I should take $\delta = \dfrac{\epsilon}{3}$ .

For $x > 1$ ,

$$\epsilon > |f(x) - 7|, \quad \epsilon > |(9 - 2x) - 7|, \quad \epsilon > |2 - 2x| = |2x - 2|, \quad \dfrac{\epsilon}{2} > |x - 1|.$$

It looks like I should take $\delta = \dfrac{\epsilon}{2}$ .

In order to ensure that both the $x < 1$ and $x > 1$ requirements are satisfied, I'll take $\delta$ to be the smaller of the two: $\delta = \dfrac{\epsilon}{3}$ .

Now here's the proof written out correctly.

Suppose $\epsilon > 0$ . Let $\delta = \dfrac{\epsilon}{3}$ , and assume that $\delta > |x - 1| > 0$ .

If $x < 1$ , then

$$\dfrac{\epsilon}{3} > |x - 1|, \quad\hbox{so}\quad \epsilon > |3x - 3| = |(3x + 4) - 7| = |f(x) - 7|.$$

Now consider the case $x > 1$ . Since $\dfrac{\epsilon}{3} > |x - 1|$ , and since $\dfrac{\epsilon}{2} > \dfrac{\epsilon}{3}$ , I have $\dfrac{\epsilon}{2} > |x - 1|$ . Therefore,

$$\epsilon > |2x - 2| = |2 - 2x| = |(9 - 2x) - 7| = |f(x) - 7|.$$

(The case $x = 1$ is ruled out because $|x - 1| > 0$ .)

Thus, taking $\delta = \dfrac{\epsilon}{3}$ guarantees that if $\delta > |x - 1| > 0$ , then $\epsilon > |f(x) - 7|$ . This proves that $\displaystyle \lim_{x\to 1} f(x) = 7$ .

Example. Use the $\epsilon-\delta$ definition of the limit to prove that

$$\lim_{x\to 2} x^2 = 4.$$

Let $\epsilon > 0$ . I want to find $\delta > 0$ such that if $\delta > |x - 2| > 0$ , then $\epsilon > |x^2 - 4|$ .

I start out as usual with my scratch work:

$$\epsilon > |x^2 - 4| = |x - 2||x + 2|.$$

Now I have a problem. I can use $\delta$ to control $|x - 2|$ , but what do I do about $|x + 2|$ ?

The idea is this: Since I have complete control over $\delta$ , I can assume $\delta \le 1$ . When I finally set $\delta$ , I can make it smaller if necessary to ensure that this condition is met.

Now if $\delta \le 1$ , then $|x - 2| < 1$ , so $1 < x < 3$ , and $3 < x + 2 < 5$ . In particular, the biggest $|x + 2|$ could be is 5. So now

$$\epsilon > |x - 2||x + 2| \quad\hbox{becomes}\quad \epsilon > |x - 2|\cdot 5, \quad\hbox{so}\quad \dfrac{\epsilon}{5} > |x - 2|.$$

This inequality suggests that I set $\delta = \dfrac{\epsilon}{5}$ --- but then I remember that I needed to assume $\delta \le 1$ . I can meet both of these conditions by setting $\delta$ to the smaller of 1 and $\dfrac{\epsilon}{5}$ : that is, $\delta = \min\left(1,\dfrac{\epsilon}{5}\right)$ .

That was scratchwork; now here's the real proof.

Let $\epsilon > 0$ . Set $\delta = \min\left(1,\dfrac{\epsilon}{5}\right)$ . Suppose $\delta > |x - 2| > 0$ .

Since $\delta \le 1$ , I have $|x - 2| < 1$ , so $1 < x < 3$ , or $3 < x + 2 < 5$ . Therefore, $5 > |x + 2|$ .

Now $\delta \le \dfrac{\epsilon}{5}$ , so $\dfrac{\epsilon}{5} > |x - 2|$ .

Now multiply the inequalities $5 > |x + 2|$ and $\dfrac{\epsilon}{5} > |x - 2|$ :

$$\epsilon = \dfrac{\epsilon}{5}\cdot 5 > |x - 2||x + 2| = |x^2 - 4|.$$

Thus, if $\delta = \min\left(1,\dfrac{\epsilon}{5}\right)$ and $\delta > |x - 2| > 0$ , then $\epsilon > |x^2 - 4|$ . This proves that $\displaystyle\lim_{x\to 2} x^2 = 4$ .

Example. Prove that $\displaystyle \lim_{x\to 2} \dfrac{x^2 + 11}{x + 3} = 3$ .

Let $\epsilon > 0$ . I must find $\delta$ such that if $\delta > |x - 2| > 0$ , then $\epsilon > \left|\dfrac{x^2 + 11}{x + 3} - 3\right|$ .

I'll start with some scratchwork.

$$\left|\dfrac{x^2 + 11}{x + 3} - 3\right| = \left|\dfrac{x^2 + 11 - 3(x + 3)}{x + 3}\right| = \left|\dfrac{x^2 - 3x + 2}{x + 3}\right| = \left|\dfrac{(x - 2)(x - 1)}{x + 3}\right| = |x - 2|\left|\dfrac{x - 1}{x + 3}\right|.$$

I can use $\delta$ to control $|x - 2|$ directly. I need to control the size of $\left|\dfrac{x - 1}{x + 3}\right|$ . It's important to think of this as $|x - 1|\cdot \left|\dfrac{1}{x + 3}\right|$ , not as $|x - 1|$ and $|x + 3|$ !

Assume $1 \ge \delta$ . Then $1 > |x - 2|$ , so $1 < x < 3$ .

For $x - 1$ , $0 < x - 1 < 2$ , so $|x - 1| < 2$ .

For $\left|\dfrac{1}{x + 3}\right|$ , $4 < x + 3 < 6$ , so $\dfrac{1}{4} > \dfrac{1}{x + 3} > \dfrac{1}{6}$ , and $\left|\dfrac{1}{x + 3}\right| < \dfrac{1}{4}$ .

Since all the number involved are positive, I can multiply the inequalities to obtain

$$2\cdot \dfrac{1}{4} > |x - 1|\cdot \left|\dfrac{1}{x + 3}\right|, \quad\hbox{or}\quad \dfrac{1}{2} > |x - 1|\cdot \left|\dfrac{1}{x + 3}\right|.$$

Thus, I'll get $\epsilon > |x - 2|\left|\dfrac{x - 1}{x + 3}\right|$ if I have $\epsilon > |x - 2|\cdot \dfrac{1}{2}$ , or $2\epsilon > |x - 2|$ . Here's the proof.

Let $\epsilon > 0$ . Set $\delta = \min(2\epsilon, 1)$ . Suppose $\delta > |x - 2| > 0$ .

Since $1 \ge \delta$ , $1 > |x - 2|$ , and $1 < x < 3$ .

First, $0 < x - 1 < 2$ , so $|x - 1| < 2$ .

Next, $\left|\dfrac{1}{x + 3}\right|$ , $4 < x + 3 < 6$ , so $\dfrac{1}{4} > \dfrac{1}{x + 3} > \dfrac{1}{6}$ , and $\left|\dfrac{1}{x + 3}\right| < \dfrac{1}{4}$ .

Hence,

$$2\cdot \dfrac{1}{4} > |x - 1|\cdot \left|\dfrac{1}{x + 3}\right|, \quad\hbox{or}\quad \dfrac{1}{2} > |x - 1|\cdot \left|\dfrac{1}{x + 3}\right|.$$

In addition, $2\epsilon \ge \delta > |x - 2|$ . Therefore,

$$\epsilon > |x - 2|\cdot \dfrac{1}{2} > |x - 2|\cdot |x - 1|\cdot \left|\dfrac{1}{x + 3}\right| = \left|\dfrac{x^2 + 11}{x + 3} - 3\right|.$$

This proves that $\displaystyle \lim_{x\to 2} \dfrac{x^2 + 11}{x + 3} = 3$ .

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