math upper homework-Real analysis due tomorrow 9：00 am

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 61

4 Continuity

Lecture 15 - 11/5 - Functions, limits and continuity.

We are now ready to revisit calculus.

Functions. Given R and D two subsets of R, a function f : D ! R is an assignment f(x) 2 R to each x 2 D. Every point in D must have an image f(x), but not every point in R need be the image of some point. In particular, if one does not have specific information about the range R, it is safe to write f : D ! R. Example 25. 1. f : [0, 2] ! R given by f(x) = �x

2

+ 1 for x 2 [0, 2].

2. f : [�1, 1] ! R given by f(x) = ⇢

�1/4, if x  0, 1/2, if x > 0.

3. f : (0, 1] ! [1, 1) given by f(x) = 1 x .

4. f : R ! R given by f(x) = ⇢

1, if x 2 Q, 0, if x 2 R\Q. We write f = �Q, the characteristic function

of the rationals.

Definition 13. f : D ! R is one-to-one (or injective) if for all x, x0 2 D, f(x) = f(x0) implies x = x0. f : D ! R is onto (or surjective) if for every y 2 R, there exists x 2 D such that f(x) = y. The graph of f is the set of points {(x, f(x)), x 2 D} ⇢ D ⇥ R. Two functions f : Df ! R and g : Dg ! R are equal if Df = Dg and for all x 2 Df, f(x) = g(x).

As a motivating example, consider the function f(x) = x 2�1 x�1 defined for all x 6= 1 and g(x) =

x + 1 defined for all x. These functions are actually equal for all x 6= 1, yet they are di↵erent functions because their domains are di↵erent. Our intuition tells us that if f were to be extended to take some value at x = 1, we would like to define f(1) = 2. The reason for this choice is because the graph nearby tells us “the function wants to take the value” 2 as x approaches 1. This nearby behavior is what constitutes a “limit”.

Limits. To define continuity, heuristically, we would like to express the fact that we do not need to lift the pen when drawing the graph of f. Given a function f : D ! R and a a point of D, f takes the value f(a) there. If a also happens to be a limit point of D, there are sequences {xn}n of points in D which converge to a, and one may ask whether the sequences of images {f(xn)}n converge at all, and if their limits coincide with f(a). This question is at the heart of defining what a continuous function is. For simplicity of exposition, from now on we denote by I an interval, for which every point is also a limit point.

Definition 14. We say that f : I ! R has limit ` at a 2 I (in short, limx!a f(x) = `) if

8" > 0, 9� > 0, 8x 2 I, 0 < |x � a| < � =) |f(x) � `| < ".

In the definition above, � = �(a, ✏) is called the modulus of continuity and describes how small we must choose the interval (a � �, a + �) so that we can map every point of it, by f, into the interval (` � ✏, ` + ✏).

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 62

Example 26. 1. In proving the limits limx!1(2x + 1) = 3 or limx!1(5x + 1) = 6, you may find that the best � you can find is � = ✏

2

and � = ✏ 5

respectively. This tells you that � has something to do with the slope (if the latter is well-defined).

2. Other examples may yield non-linear relations between � and ✏, for instance, in limx!0 x2 = 0, we have � =

p ✏ (“slower” than linear shrinking) while when proving limx!0

p x = 0, we have

� = ✏2 (“faster” than linear shrinking).

3. Define f : [0, 3] ! R by f(x) = ⇢

x2 � 1, if x 6= 1, 1, if x = 1.

Let us show that limx!1 f(x) = 0, or in other words

8" > 0, 9� > 0, 8x 2 [0, 3], 0 < |x � 1| < � =) |x2 � 1| < ". Notice the preliminary bound |x2�1| = |x�1||x+1|  4|x�1| for any x 2 [0, 3]. In particular, if we can make |x�1| less than "/4, we can make |x2 �1| less than ". Therefore, given " > 0, set � = "

4

> 0. If x 2 [0, 3] is such that 0 < |x � 1| < � = " 4

, then |x2 � 1|  4|x � 1| < 4" 4

= ", hence the proof.

Similarly, you can show that limx!2 f(x) = 3.

In the example above, the fact that we have changed the value of the function at x = 1 does not impact its limit as x approaches 1.

Theorem 63. Let f : I ! R and a 2 I. Then, limx!a f(x) = ` if and only if for every sequence {xn}n in I\{a} converging to a, the sequence {f(xn)}n converges to `.

Proof. ( =) ) Suppose that limx!a f(x) = `, let {xn}n a sequence in I\{a} converging to a, and let us prove that limn!1 f(xn) = `. Let " > 0. Since limx!a f(x) = `, there exists � > 0 such that for every x 2 I, 0 < |x � a| < � implies |f(x) � `| < ". Since xn ! a as n ! 1, there exists N 2 N such that for every n � N, |xn � a| < �. Then for every n � N, |xn � a| < �, and thus, |f(xn) � `| < ". Thus, limn!1 f(xn) = `.

( (= ) By contraposition, suppose the negation of “limx!x0 f(x) = `”, that is to say, 9" > 0, 8� > 0, 9x 2 I, 0 < |x � a| < � and |f(x) � `| � ".

Using this sentence with � = 1 n > 0 for any n 2 N, there exists xn such that 0 < |xn � a| < 1n and

|f(xn) � `| � ". By squeezing, it is clear that xn ! a as n ! 1, and also that since |f(xn) � `| � " for every n, the sequence {f(xn)}n cannot converge to `. The theorem is proved.

In particular, to disprove that some function f(x) has a limit at x = a, one may find two distinct sequences {xn}n and {x0n}n, both converging to a, and such that the sequences {f(xn)}n and {f(x0n)}n converge to di↵erent limits (if they converge at all). The example below is an application of that.

Example 27. 1. For any a 2 R, the function �Q does not have a limit at a. In light of the theorem above, this is because for any a, by density of Q and R\Q, we can find a sequence of rationals {rn}n and a sequence of irrationals {xn}n both converging to a, and the correspond- ing sequences of images converge to distinct limits (1 and 0, respectively).

2. The function f(x) =

⇢ �1 x < 0 1 x � 0 does not have a limit at a = 0 because limn!1 f(�1/n) =

�1 6= 1 = limn!1 f(1/n).

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 63

One-sided limits. In the last example, the second function seems to have limits if x is forced to come from the left or the right side of 0. We can in fact define one-sided limits in this sense.

Definition 15. Let f : I ! R and a 2 I. We say that f has left-limit ` at a (in short limx!a� f(x) = `) if

8" > 0, 9� > 0, 8x 2 I, a � � < x < a =) |f(x) � `| < ".

We say that f has right-limit ` at a (in short limx!a+ f(x) = `) if

8" > 0, 9� > 0, 8x 2 I, a < x < a + � =) |f(x) � `| < ".

f can have a left-limit and a right-limit, and these limits might or might not agree.

Example 28. Define f : R ! R by f(x) = ⇢

0, if x  0, 1, if x > 0.

(sometimes called the Heaviside

function). Show that limx!0� f(x) = 0 and limx!0+ f(x) = 1.

As an exercise, you may prove the following:

Theorem 64. The limit limx!a f(x) exists if and only if limx!a� f(x) and limx!a+ f(x) both exist and are equal.

Continuity.

Definition 16. Let a function f : I ! R and a 2 I. We say that f is continuous at a if limx!a f(x) exists and equals f(a). By Theorem 63, f is continuous at a if and only if for every sequence {xn}n in I converging to a, we have limn!1 f(xn) = f(a). We say that f is continuous on I if f is continuous at every point of I

Combining this definition with Theorem 63, continous functions are those for which, when limn!1 xn exists and is in the domain of f, we can write

lim n!1

f(xn) = f ⇣ lim n!1

xn ⌘ , (sequential characterization of continuity) (13)

that is to say, we can interchange “f” and “limn!1” signs.

Example 29. 1. Any polynomial is continous on R. This is a direct consequence of the algebraic limit rules on sequences, which states that if P is a polynomial and if {xn} converges to x0, then the sequence {P(xn)}n converges to P(x0).

2. The function x 7! 1 x is continuous on (�1, 0) [ (0, 1).

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 64

Exercises for Lecture 15:

1. Prove Theorem 64.

2. Prove using the " � � definition of limit, (a) for every a > 0, limx!a 1x =

1

a . Can � be made independent of a ?

(b) for every a > 0, limx!a x3 = a3. Can � be made independent of a ?

3. Define the function f(x) = x�Q(x) (that is, f(x) = x if x 2 Q and f(x) = 0 if x 2 R\Q). Show that

(a) limx!0 f(x) = 0.

(b) For any a 6= 0, limx!a f(x) does not exist13. 4. Define the function f(x) = sin(x�1) for x 2 (0, 1). Show that the limit limx!0+ f(x) does

not exist. [Hint: use that sin(n⇡) = 0 and sin((2n + 1/2)⇡) = 1 for every n 2 N]

13This is an example of a function which is continuous at one point and one point only.