Assignment 2 Solutions MATH 243, Analysis 2
Assignment 2 Solutions MATH 243, Analysis 2 Spencer Frei February 18, 2013 1. 5.6–4. Show that if f and g are positive increasing functions on an interval I, then their product f g is increasing on I. Proof. By definition, the product of two functions f and g is their pointwise product; i.e., (f g)(t) = f (t) · g(t) ∀t ∈ I . Given two points x1 , x2 ∈ I satisfying x1 ≤ x2 , since f and g are both positive and increasing, we know that we have ( f (x1 ) ≤ f (x2 ) , g(x1 ) ≤ g(x1 ) . Therefore, we have that their product is also increasing, since (f g)(x1 ) = f (x1 ) · g(x1 ) ≤ f (x2 ) · g(x1 ) ≤ f (x2 ) · g(x2 ) = (f g)(x2 ) . This completes the proof. 2. 5.6–5. Show that if I = [a, b] and f : I → R is increasing on I, then f is continuous at a if and only if f (a) = inf{f (x) : x ∈ (a, b]} . (*) Proof. Suppose that (*) holds. We would like to show that for every ε > 0 and x ∈ (a, b], there is some δ > 0 such that if |x − a| < δ then |f (x) − f (a)| < ε. Now, since x > a and f is increasing, we know that f (x) ≥ f (a), and so we just need to show that there is some δ > 0 such that x − a < δ implies f (x) − f (a) < ε. So let ε > 0. Then f (a) + ε is not a lower bound for the set {f (x) : x ∈ (a, b]}, and hence there is some point x0 ∈ (a, b] such that f (a) ≤ f (x0 ) < f (a) + ε . So let δ = min{b − a, x0 − a}. Then for x − a < δ, since f is increasing we must have f (x) − f (a) ≤ f (x0 ) − f (a) < ε . Thus f is continuous at a. To prove the other direction, we will prove the contrapositive. That is, we will show that if (*) does not hold, then f is not continuous at a. If (*) doesn’t hold, then f (a) 6= inf{f (x) : x ∈ (a, b)}. Since f is increasing, this inequality means that f (a) is strictly less than inf{f (x) : x ∈ (a, b]}. Thus, we know that ε0 = inf{f (x) : x ∈ (a, b]} − f (a) > 0 . In particular, since inf{f (x) : x ∈ (a, b]} − f (a) = ε0 > 0, for any point x0 ∈ (a, b] we have that f (x0 ) − f (a) ≥ ε0 > 0. Thus, for any point x0 ∈ (a, b] and for any δ > 0, if |x0 − a| < δ then we will still have that |f (x0 ) − f (a)| = f (x0 ) − f (a) ≥ ε0 > 0. Thus, f is discontinuous at a, proving the contrapositive. 1
3. 5.6–10. Let I = [a, b] and let f : I → R be continuous on I. If f has an absolute maximum (minimum) at an interior point c ∈ (a, b), show that f is not injective on I. Proof. We will prove the case where f has an absolute maximum, since the case of an absolute minimum will follow by just applying our method to −f . Since f has an absolute maximum at an interior point c ∈ (a, b), we know that f (c) ≥ f (x) for all x ∈ [a, b]. Since c is in the interior (a, b), we can take δ > 0 small enough such that [c − δ, c + δ] ⊂ (a, b) . Now, if f (c − δ) = f (c + δ), then we immediately get that f is not injective on I. Otherwise, we suppose without loss of generality that f (c − δ) < f (c + δ). [Otherwise, we may just switch the sign of δ.] The interval [c − δ, c + δ] is closed and bounded, and hence f is uniformly continuous on this interval. In particular, f is uniformly continuous on [c − δ, c], and so by the intermediate value theorem, we know that f attains all values between f (c − δ) and f (c). Since f has an absolute maximum at c, we have that f (c − δ) ≤ f (c). If this is an equality, we have shown that f is not injective on I. If it is strict, then by intermediate value theorem f attains all values in the interval [f (c − δ], f (c)]. Since f (c − δ) < f (c + δ) ≤ f (c)], this means that there is some point x0 ∈ [c − δ, c] such that f (x0 ) = f (c + δ). Thus f is not injective on I.
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3. 5.6–10. Let I = [a, b] and let f : I → R be continuous on I. If f has an absolute maximum (minimum) at an interior point c ∈ (a, b), show that f is not injective on I. Proof. We will prove the case where f has an absolute maximum, since the case of an absolute minimum will follow by just applying our method to −f . Since f has an absolute maximum at an interior point c ∈ (a, b), we know that f (c) ≥ f (x) for all x ∈ [a, b]. Since c is in the interior (a, b), we can take δ > 0 small enough such that [c − δ, c + δ] ⊂ (a, b) . Now, if f (c − δ) = f (c + δ), then we immediately get that f is not injective on I. Otherwise, we suppose without loss of generality that f (c − δ) < f (c + δ). [Otherwise, we may just switch the sign of δ.] The interval [c − δ, c + δ] is closed and bounded, and hence f is uniformly continuous on this interval. In particular, f is uniformly continuous on [c − δ, c], and so by the intermediate value theorem, we know that f attains all values between f (c − δ) and f (c). Since f has an absolute maximum at c, we have that f (c − δ) ≤ f (c). If this is an equality, we have shown that f is not injective on I. If it is strict, then by intermediate value theorem f attains all values in the interval [f (c − δ], f (c)]. Since f (c − δ) < f (c + δ) ≤ f (c)], this means that there is some point x0 ∈ [c − δ, c] such that f (x0 ) = f (c + δ). Thus f is not injective on I.
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Graph of d/dx(g(x)) against x 200
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Graph of d/dx(g(x)) against x 2000
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