Math 104: Introduction to Analysis Summer 2019
Lecture 14: Bijective functions and continuity
Lecturer: Michael Christianson 18 July Aditya Sengupta
14.1
Bounded functions
Definition 14.1. Let f : S → R be a function. We say f is bounded if its image f (S) = {f (s) | s ∈ S} is a bounded set, i.e. ∃M ∈ R such that |f (s)| ≤ M ∀s ∈ S.
Theorem 14.1. (Extreme Value Theorem) Let a < b, and let f : [a, b] → R be a continuous function. Then f is bounded, and f assumes its maximum and minimum values on [a, b], i.e. ∃l, u ∈ [a, b] such that f (l) ≤ f (x) ≤ f (u) for all x ∈ [a, b].
Example 14.1. Consider f (x) = 1
x on (0, 1).
Example 14.2. Consider g(x) = x2 on (−1, 1). For all 0 < a < 1, g(√a) = a, but g(x) < 1 ∀x.
Proof. We claim f is bounded and prove it by contradiction. Suppose f is not bounded. Then ∀n ∈ Z>0,
∃xn ∈ [a, b] s. t. |f (xn)| > n. Then (xn) is bounded, so the Bolzano-Weierstrass theorem implies that there
exists a subsequence (xnk) of (xn) that converges to some y ∈ R. Therefore xnk∈ [a, b]∀k =⇒ y ∈ [a, b].
f is continuous at y, so f (y) = lim
k→+∞f (xnk), so (f (xnk)) is bounded. Therefore (|f (xnk)|) is bounded. But
lim
k→+∞|f (xnk)| =n→+∞lim |f (xn)| = +∞
This is a contradiction, therefore f is bounded. Let
M = sup{f (s) | s ∈ [a, b]} ∈ R
We find some u ∈ [a, b] such that f (u) = M . For all n ∈ Z>0, M − n1 is not an upper bound on f ([a, b]).
Lecture 14: Bijective functions and continuity 61 −1 n < f (yn) − M ≤ 0 < 1 n |f (yn) − M | < 1 n So (yn) is a sequence in [a, b] such that lim
n→+∞f (yn) = M . yn ∈ [a, b]∀n implies that (yn) is bounded, so
there exists a convergent subsequence (ynk). Write u = lim ynk; then u ∈ [a, b] be3cause ynk ∈ [a, b]∀k. f is
continuous at u, therefore by continuity,
f (u) = lim f (ynk) =n→+∞lim f (yn) = M
Repeat this argument with −f in place of f for the minimum. We get that ∃l ∈ [a, b] such that −f assumes its maximum at l.
−f (l) ≥ −f (x)∀x ∈ [a, b] f (l) ≤ f (x)∀x ∈ [a, b]
So f assumes its minimum value at l.
Definition 14.2. A subset I ⊆ R is an interval if ∃a < b in R ∪ {±∞} such that I has one of the following forms: (a, b), [a, b], [a, b), (a, b].
Theorem 14.2. (Intermediate Value Theorem) Let I ⊆ R be an interval, and let f : I → R be a continuous function. For all a < b in I and any y lying between f (a) and f (b), i.e. f (a) < f (y) < f (b) or f (b) < f (y) < f (a), there exists x ∈ I such that a < x < b and f (x) = y.
Proof. We prove the statement for f (b) < y < f (a). Define
S = {x ∈ [a, b] | f (x) > y}
a ∈ S implies S 6= ∅, i.e. sup S = x0 is a real number. Since S ⊆ [a, b], x0∈ [a, b] ⊆ I. We show f (x0) = y.
First, we show that f (x0) ≥ y.
For all n ∈ Z>0, x0−n1 is not an upper bound on S. Therefore, there exists sn∈ S such that x0−n1 < sn≤ x0,
i.e. |sn− x0| <n1, so lim
n→+∞sn= x0.
Then, f is continuous at x0, so f (x0) = lim
n→+∞f (sn) ≥ y. sn∈ S∀n, so f (sn) > y.
Next, we show that f (x0) ≤ y. For all n ∈ Z>0, tn = min{b, x0+n1}. Then, x0< b and x0< x0+n1, so
x0−
1
n < x0< tn ≤ x0+ 1 n
Lecture 14: Bijective functions and continuity 62
So |tn− x0| < n1, i.e. lim
n→+∞tn= x0.
For all n, tn> x0 =⇒ tn ∈ S, so f (t/ n) ≤ y. So continuity of f at x0 gives us
f (x0) = lim
n→+∞f (tn) ≤ y.
Therefore f (x0) = y.
Definition 14.3. Let f : S → T be a function.
1. f is injective if ∀s, s0 ∈ S s. t. f (s) = f (s0), we have s = s0.
2. The image of f is f (S) = {f (s) | s ∈ S}.
3. f is surjective or onto if f (S) = T , ki.e. ∀t ∈ T, ∃s ∈ S s. t. f (s) = t. 4. f is bijective if it is injective and surjective.
Corollary 14.3. Lert I ⊆ R be an interval and f : I → R be a continuous function. Then f (I) is either an interval or a single point.
Proof. If sup f (I) = inf f (I), then f (I) = {sup f (I)}. For all y ∈ f (I), inf f (I) ≤ y ≤ sup f (I). If instead inf f (I) < sup f (I), then I is an interval with end points inf f (I) and sup f (I). For all y such that inf f (I) < y < sup f (I), there exists y0, y1∈ f (I) such that y0< y < y1. By the intermediate value theorem,
there exists x ∈ I such that f (x) = y =⇒ y ∈ f (I). So
(inf f (I), sup f (I)) ⊆ f (I) ⊆ [inf f (I), sup f (I)]
Example 14.3. Let f : [0, 1] → [0, 1] be a continuous function. We show f has a fixed point, i.e. ∃a ∈ [0, 1] s. t. f (a) = a.
Let g(x) = f (x) − x on [0, 1]. Then
g(0) = f (0) − 0 = f (0) ≥ 0 g(1) = f (1) − 1 ≤ 1 − 1 = 0
If g(0) 6= 0 and g(1) 6= 0, then g(0) > 0 > g(1), so ∃x ∈ (0, 1) s. t. g(x) = 0 by the intermediate value theorem, i.e. f (x) = x.
Definition 14.4. Let f : S → R.
2. f is strictly decreasing if ∀x1< x2∈ S, f (x1) > f (x2).
Theorem 14.4. Let I ⊆ R be an interval and f : I → R a continuous function. The following are equivalent: 1. f is continuous and injective.
2. f (I) is an interval, and f is either strictly increasing or strictly decreasing.
Proof. In the forward direction, f being continuous and injective implies that f (I) is an interval or a point. But f being injective means it must be an interval.
We claim that ∀a < b < c in I, f (b) lies between f (a) and f (c) (strict inequalities because the function is injective.) Suppose this is false; then either f (b) > max{f (a), f (c)}, or f (b) < min{f (a), f (c)}.
If f (b) > max{f (a), f (c)}, pick y ∈ R such that f (b) > y > max{f (a), f (c)}. By IVT on [a, b] and [b, c], ∃x1 ∈ (a, b) and x2 ∈ (b, c) such that f (x1) = f (x2) = y. But this implies x1 = x2 because f is injective,
which is a contradiction.. Therefore f (b) < max{f (a), f (c)}.
If f (b) < min{f (a), f (c)}, pick y ∈ R such that min{f (a), f (c)} < y < f (b). By IVT on [a, b] and [b, c], ∃x1 ∈ (a, b) and x2 ∈ (b, c) such that f (x1) = f (x2) = y. But this implies x1 = x2 because f is injective,
which is a contradiction.. Therefore f (b) > min{f (a), f (c)}.
In the backward direction, f strictly increasing or strictly decreasing implies that f is injective, because x1< x2 =⇒ f (x1) < f (x2) or x1 < x2 =⇒ f (x1) > f (x2) both imply that x1 < x2 =⇒ f (x1) 6= f (x2).
Therefore it suffices to prove continuity. We prove the case where f is strictly decreasing, and assume the proof for the case f is strictly increasing follows similarly. (see Ross section 18).
Let a ∈ I and let > 0. Roughly, choose sufficiently small that it is in the image of the function around a then finish the proof.
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