Assignment 1 Solutions MATH 243, Analysis 2
Assignment 1 Solutions MATH 243, Analysis 2 Spencer Frei April 9, 2013 1. Section 5.4: # 2. Show that the function f (x) := 1/x2 is uniformly continuous on A = [1, ∞), but that it is not uniformly continuous on B = (0, ∞). Proof. To show that f is uniformly continuous on A, we need to show that for every ε > 0, there exists some δ > 0 such that for any x, y ∈ A satisfying |x − y| < δ, we have that |f (x) − f (y)| < ε. So let ε > 0 be given, and let us consider the difference |f (x) − f (y)| to see if we can squeeze out an |x − y| term, as this is what we will be able to bound. We have that 1 1 (by defn) |f (x) − f (y)| = 2 − 2 x y 2 y − x2 = 2 2 x y (x − y)(x + y) . = x2 y 2
Now we have that x − y term that we would like, and we need only deal with the term xx+y 2 y 2 . To do this, let us note that A = [1, ∞), and so for any x, y ∈ A, we certainly have that x, y ≥ 1, and hence we have the inequalities 1 ≤ 1, x
In particular, this means that
1 xy 2
and
1 ≤ 1, y 1 x2 y
1 ≤ 1, x2
1 ≤1. y2
are both ≤ 1. Therefore, we have that
(x − y) (x − y) (x − y) (x − y) x ≤ + x y + y x2 y 2 x2 y 2 x2 y 2 x2 y 2 (x − y) (x − y) + = xy 2 x2 y 1 1 ≤ 2 |x − y| + 2 |x − y| xy x y
(triangle ineq.)
(since |ab| ≤ |a||b|)
≤ |x − y| + |x − y|
(since
1 1 , ≤ 1) xy 2 x2 y
= 2|x − y| . Therefore, we see that if we are given ε > 0, then choosing δ = of x, y ∈ A, since we have that |f (x) − f (y)| ≤ 2|x − y| < 2 1
ε =ε 2
ε 2
will do the trick for any choice
for any x, y ∈ A .
As our choice of δ is independent of the points x, y ∈ A, f is uniformly continuous over A. As for showing that f is not uniformly continuous on B = (0, ∞), we can adapt the proof shown directly following section 5.4.2 in Bartle and Sherbert. Take sequences xn =
1 , n2
yn =
1 . 1 + n2
Both (xn ) and (yn ) are sequences in B. Then we have that 1 + n2 1 n2 = |xn − yn | = 2 − −→ 0 as n → ∞ . 2 2 2 n (1 + n ) n (1 + n ) n2 (1 + n2 )
But notice that
1 1 |f (xn ) − f (yn )| = 1 − 1 n4 (1+n2 )2 4 = n − (1 + n2 )2 = n4 − 1 − n4 − 2n2 = −2n2 − 1 > 1 for all n ∈ N . Therefore f is not uniformly continuous on (0, ∞). [N.B.: when writing proofs, it is better practice to use fewer symbols, and more words, as I have in this proof. If you ever get stuck with a proof, always write out exactly what you would like to prove—in this case, write what it means for a function to be uniformly continuous on A, or for what it means to not be uniformly continuous on B—and then write out what you have, and think about how you can get from the starting point to the end point.]
2. Section 5.4: # 4. Show that the function f (x) = 1/(1 + x2 ) is uniformly continuous on R. Proof. We need to show that for any ε > 0, there is some δ > 0 such that for any x, y ∈ R, we have that |x − y| < δ implies that 1 1 1 + y 2 − (1 + x2 ) x2 − y 2 |f (x) − f (y)| = = − = 2 2 2 2 2 2 1+x 1+y (1 + x )(1 + y ) (1 + x )(1 + y ) is less than ε. Now, we have that (x − y)(x + y) |f (x) − f (y)| = (1 + x2 )(1 + y 2 ) |x + y| ≤ |x − y| (1 + x2 )(1 + y 2 ) |x| |y| . ≤ |x − y| + 1 + x2 1 + y2
(since x2 − y 2 ≤ |x − y||x + y|)
The last line follows since 1 + x2 , 1 + y 2 are both ≥ 1 (and hence equal to their absolute value), and by the same argument in the previous question, their reciprocals are ≤ 1. Now, notice that we have the inequalities |x| ≤ 1 + x2 , |y| ≤ 1 + y 2 .
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The inequality is trivially true for |x| ≥ 1 (draw the parabola and |x|), and if |x| < 1, we have that 0 ≤ |x|2 < 1 and hence 1 + x2 > |x|. Therefore, we can continue from above to get that 1 + y2 1 + x2 = 2|x − y| . + |f (x) − f (y)| ≤ |x − y| 1 + x2 1 + y2 Thus, given ε > 0, we can choose δ =
ε 2
so that for any x, y ∈ R, we have that
|f (x) − f (y)| ≤ 2|x − y| < 2 ·
ε =ε. 2
Since our choice of δ was independent of the points x, y ∈ R, we have that f is uniformly continuous on R. 3. Section 5.4: # 12. Show that if f is continuous on [0, ∞) and uniformly continuous on [a, ∞) for some a > 0, then f is uniformly continuous on [0, ∞). Proof. Since f is continuous on [0, ∞), it is continuous on [0, a + 1]. Since [0, a + 1] is a closed bounded interval, f is uniformly continuous on [0, a + 1]. Let ε > 0. We would like to find some δ > 0 such that for any x, y ∈ [0, ∞), whenever |x − y| < δ we have |f (x) − f (y)| < ε. Now, for any x, y ∈ [0, a + 1], we know by uniform continuity that there is δ1 > 0 such that |f (x) − f (y)| < ε for |x − y| < δ1 . For any x, y ∈ [a, ∞), we know again by uniform continuity there is δ2 > 0 such that |f (x) − f (y)| < ε for |x − y| < δ2 . So take δ = min{δ1 , δ2 , 1}. Let x, y ∈ [0, ∞) with |x − y| < δ. We must have one of the following three cases: (i) both points x and y are in [0, a + 1] (ii) both points x and y are in (a + 1, ∞] (iii) none of the above holds, i.e. one point is in [0, a + 1] and the other in (a + 1, ∞]. We claim that in all three cases |f (x) − f (y)| < ǫ. Case (i): Note that δ ≤ δ1 , hence |x − y| < δ1 , hence |f (x) − f (y)| < ǫ by definition of δ1 . Case (ii): Note that x, y > a + 1 > a, so in particular x, y ∈ [a, ∞). Also note that δ ≤ δ2 , hence |x − y| < δ2 , hence |f (x) − f (y)| < ǫ by definition of δ2 . Case (iii): By symmetry we may let x denote the point which is in [0, a + 1] and y denote the point which is in (a + 1, ∞]. Note that δ ≤ 1, hence in particular we know that |x − y| < 1. Hence x > y − 1 > a + 1 − 1 = a, so that both points are certainly in [a, ∞). Now recall again that δ ≤ δ2 , hence |f (x) − f (y)| < ǫ by definition of δ2 . Thus f is uniformly continuous on [0, ∞).
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Now we have that x − y term that we would like, and we need only deal with the term xx+y 2 y 2 . To do this, let us note that A = [1, ∞), and so for any x, y ∈ A, we certainly have that x, y ≥ 1, and hence we have the inequalities 1 ≤ 1, x
In particular, this means that
1 xy 2
and
1 ≤ 1, y 1 x2 y
1 ≤ 1, x2
1 ≤1. y2
are both ≤ 1. Therefore, we have that
(x − y) (x − y) (x − y) (x − y) x ≤ + x y + y x2 y 2 x2 y 2 x2 y 2 x2 y 2 (x − y) (x − y) + = xy 2 x2 y 1 1 ≤ 2 |x − y| + 2 |x − y| xy x y
(triangle ineq.)
(since |ab| ≤ |a||b|)
≤ |x − y| + |x − y|
(since
1 1 , ≤ 1) xy 2 x2 y
= 2|x − y| . Therefore, we see that if we are given ε > 0, then choosing δ = of x, y ∈ A, since we have that |f (x) − f (y)| ≤ 2|x − y| < 2 1
ε =ε 2
ε 2
will do the trick for any choice
for any x, y ∈ A .
As our choice of δ is independent of the points x, y ∈ A, f is uniformly continuous over A. As for showing that f is not uniformly continuous on B = (0, ∞), we can adapt the proof shown directly following section 5.4.2 in Bartle and Sherbert. Take sequences xn =
1 , n2
yn =
1 . 1 + n2
Both (xn ) and (yn ) are sequences in B. Then we have that 1 + n2 1 n2 = |xn − yn | = 2 − −→ 0 as n → ∞ . 2 2 2 n (1 + n ) n (1 + n ) n2 (1 + n2 )
But notice that
1 1 |f (xn ) − f (yn )| = 1 − 1 n4 (1+n2 )2 4 = n − (1 + n2 )2 = n4 − 1 − n4 − 2n2 = −2n2 − 1 > 1 for all n ∈ N . Therefore f is not uniformly continuous on (0, ∞). [N.B.: when writing proofs, it is better practice to use fewer symbols, and more words, as I have in this proof. If you ever get stuck with a proof, always write out exactly what you would like to prove—in this case, write what it means for a function to be uniformly continuous on A, or for what it means to not be uniformly continuous on B—and then write out what you have, and think about how you can get from the starting point to the end point.]
2. Section 5.4: # 4. Show that the function f (x) = 1/(1 + x2 ) is uniformly continuous on R. Proof. We need to show that for any ε > 0, there is some δ > 0 such that for any x, y ∈ R, we have that |x − y| < δ implies that 1 1 1 + y 2 − (1 + x2 ) x2 − y 2 |f (x) − f (y)| = = − = 2 2 2 2 2 2 1+x 1+y (1 + x )(1 + y ) (1 + x )(1 + y ) is less than ε. Now, we have that (x − y)(x + y) |f (x) − f (y)| = (1 + x2 )(1 + y 2 ) |x + y| ≤ |x − y| (1 + x2 )(1 + y 2 ) |x| |y| . ≤ |x − y| + 1 + x2 1 + y2
(since x2 − y 2 ≤ |x − y||x + y|)
The last line follows since 1 + x2 , 1 + y 2 are both ≥ 1 (and hence equal to their absolute value), and by the same argument in the previous question, their reciprocals are ≤ 1. Now, notice that we have the inequalities |x| ≤ 1 + x2 , |y| ≤ 1 + y 2 .
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The inequality is trivially true for |x| ≥ 1 (draw the parabola and |x|), and if |x| < 1, we have that 0 ≤ |x|2 < 1 and hence 1 + x2 > |x|. Therefore, we can continue from above to get that 1 + y2 1 + x2 = 2|x − y| . + |f (x) − f (y)| ≤ |x − y| 1 + x2 1 + y2 Thus, given ε > 0, we can choose δ =
ε 2
so that for any x, y ∈ R, we have that
|f (x) − f (y)| ≤ 2|x − y| < 2 ·
ε =ε. 2
Since our choice of δ was independent of the points x, y ∈ R, we have that f is uniformly continuous on R. 3. Section 5.4: # 12. Show that if f is continuous on [0, ∞) and uniformly continuous on [a, ∞) for some a > 0, then f is uniformly continuous on [0, ∞). Proof. Since f is continuous on [0, ∞), it is continuous on [0, a + 1]. Since [0, a + 1] is a closed bounded interval, f is uniformly continuous on [0, a + 1]. Let ε > 0. We would like to find some δ > 0 such that for any x, y ∈ [0, ∞), whenever |x − y| < δ we have |f (x) − f (y)| < ε. Now, for any x, y ∈ [0, a + 1], we know by uniform continuity that there is δ1 > 0 such that |f (x) − f (y)| < ε for |x − y| < δ1 . For any x, y ∈ [a, ∞), we know again by uniform continuity there is δ2 > 0 such that |f (x) − f (y)| < ε for |x − y| < δ2 . So take δ = min{δ1 , δ2 , 1}. Let x, y ∈ [0, ∞) with |x − y| < δ. We must have one of the following three cases: (i) both points x and y are in [0, a + 1] (ii) both points x and y are in (a + 1, ∞] (iii) none of the above holds, i.e. one point is in [0, a + 1] and the other in (a + 1, ∞]. We claim that in all three cases |f (x) − f (y)| < ǫ. Case (i): Note that δ ≤ δ1 , hence |x − y| < δ1 , hence |f (x) − f (y)| < ǫ by definition of δ1 . Case (ii): Note that x, y > a + 1 > a, so in particular x, y ∈ [a, ∞). Also note that δ ≤ δ2 , hence |x − y| < δ2 , hence |f (x) − f (y)| < ǫ by definition of δ2 . Case (iii): By symmetry we may let x denote the point which is in [0, a + 1] and y denote the point which is in (a + 1, ∞]. Note that δ ≤ 1, hence in particular we know that |x − y| < 1. Hence x > y − 1 > a + 1 − 1 = a, so that both points are certainly in [a, ∞). Now recall again that δ ≤ δ2 , hence |f (x) − f (y)| < ǫ by definition of δ2 . Thus f is uniformly continuous on [0, ∞).
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