MA10207: Exercise sheet 16

2013 MA10207: Exercise sheet 16
MA10207: Exercise sheet 16
Please hand in solutions to homework problems by Friday, 22nd March (!!!) 5:15pm.
(Monotonicity & inverse functions)
Warmup problems
Problem T 16.1. Recall what the inverse of a function is. How are the graphs of a function and its
inverse function related . . . and why
Problem T 16.2. Let I ⊂ R be an open interval, that is, inf I, sup I ∉ I. Suppose that f : I → R is
continuous and strictly increasing. Prove that f(I) ⊂ R is an open interval.
Problem T 16.3. Let f : D → R be a function. How do (strict) monotonicity and injectivity relate
Prove your claims.
Homework problems
Problem H 16.1. Let f : I → R be continuous and injective on an interval I.
(ii) Let A ⊂ I contain two elements, A = {x 1 , x 2 }. Use injectivity to argue that the restriction
f| A of f to A is strictly monotone.
(iii) Let A ⊂ I contain three elements, A = {x 1 , x 2 , x 3 }. Use the IVT to prove that f| A is strictly
monotone [Hint: proof by contradiction using (ii)].
(iv) Let A ⊂ I contain four elements, A = {x 1 , x 2 , x 3 , x 4 }. Deduce from (iii) that f| A is strictly
monotone [Hint: straight logic from (iii)].
(∞) Deduce from (iv) that f is strictly monotone.
Problem H 16.2. An increasing function f : D → R is continuous if f(D) is an interval (Thm 4.18).
(a) Fill in/elaborate any details in the proof below that you consider necessary and find any errors.
(b) What does the proof actually prove
(c) Can you “repair” the proof (i.e., make it prove the statement)
Proof. Suppose f was not continuous at some x ∈ D, that is,
∃(x n ) n∈N ⊂ D : x n → x but f(x n ) ↛ f(x).
Since f(x n ) ↛ f(x), ∃ε > 0∀N ∈ N∃n ≥ N : |f(x n ) − f(x)| ≥ ε.
Hence we can choose a subsequence (x nk ) k∈N of the sequence (x n ) n∈N so that
∀k ∈ N : f(x nk ) ≤ f(x) − ε or ∀k ∈ N : f(x) + ε ≤ f(x nk );
suppose, without loss of generality (wlog.), that ∀k ∈ N : f(x) + ε ≤ f(x nk ). Then
• f(x) + ε ∈ f(D) since f(x n1 ) ≥ f(x) + ε > f(x) and f(D) is an interval;
consequently, ∃δ > 0 : f(x + δ) = f(x) + ε;
• ∀k ∈ N : x nk ≥ x + δ since f(x nk ) ≥ f(x) + ε = f(x + δ).
As a consequence (convergence of subsequences) we obtain the desired contradiction:
x = lim k→∞ x nk ≥ x + δ > x.
Problem H 16.3. Use the functional identity e x+y = e x e y of the exponential function
exp : R → R, x ↦→ exp(x) = e x = ∑ ∞
n=0 xn
n!

2013 MA10207: Exercise sheet 16
to show that exp is strictly increasing. Conclude that ln = exp −1 : (0, ∞) → R is strictly monotone
and continuous.
Quiz questions
Problem Q 16.1. Let I be an interval and f : I → R some function (not necessarily continuous).
(i) x ∈ I is a minimum of f if ∀y ∈ I : f(x) ≤ f(y).
(ii) If I is compact then f has a maximum on I. [Hint]
(iii) If I is compact and f continuous then f(I) is a compact interval. [Hint]
(iv) If I is bounded, f continuous and f(I) is a compact interval, then I is a closed interval. [Hint]
(v) If f : R → R is continuous and satisfies lim x→±∞ f(x) = 0, then f is bounded. [Hint]
(vi) If f : R → R is continuous and satisfies lim x→±∞ f(x) = 0, then f has a minimum on R.
Evaluate
Problem Q 16.2. Let I be an interval and f : I → R continuous.
(i) If f is strictly monotone then f injects. [Hint]
(ii) If f injects then f must be strictly monotone. [Hint]
(iii) The previous statement is false, in general, if I is not an interval. [Hint]
(iv) If f injects then f −1 is continuous. [Hint]
Evaluate