319HW9Solutions

319HW9Solutions - MAT 319/320 Solutions for HW9 Exercise 1....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAT 319/320 Solutions for HW9 Exercise 1. Section 5.3, #3. Proof. Start with any point, x0 = 1/2 for example. Then we know that there exists x1 such that f (x1) 1 2 f (x0) . If f (x1) = 0, then we are done; if not then necessarily x1 1 2 x0, and there is then another point x2, such that f (x2) sequence (xn) satisfying f (xn +1) 1 2 f (x1) . Continue this process and build a f (xn) . Since all these points are in a bounded interval, 1 2 α(n) 1 2 the sequence is bounded and by Bolzano-Weierstrass we can extract a sequence (xkn) converging to some c ∈ [a, b]. If we write kn +1 = kn + α(n), we get f (xkn+1) f (xkn) f (xkn) . (This just means that the extracted sequence satisfies the same inequality). Since (xkn) → c, and f is continuous, then we deduce that f (xkn) → f (c), but then the previous inequality says that: f (c) 1 2 f (c) so f (c) = 0 and we are done. Exercise 2. Section 5.3, #6. Proof. The hint says it all: g(0) = f (0) − f (1/2) and g (1/2) = f (1/2) − f (1) = − g (0). Now if g (0) = 0 we are done. If not, then apply the intermediate value theorem to g : you will get a zero 1 for g between 0 and 1/2. But g (x) = 0 ⇒ f (x) = f (x + 2 ). Exercise 3. Section 5.3, #11. Proof. This has been proved in class.Let me quickly give the argument: if f (w) < 0, then necessarily w < b (because f (b) > 0). By continuity of f there will be a small neighborhood of w on which the function is strictly negative: this contradicts the definition of w . If f (w ) > 0, then by continuity of f there is a small δ − neighborhood where the function is strictly positive: this is a contradiction, because between w − δ and w there should be a point in W. Exercise 4. Section 5.3, #13. Proof. Take ε = 1: then there exists an α > 0 such that for any x > α we have | f (x)| < 1, and there is a β < 0 such that for any x < β we have | f (x)| < 1. But now f is continuous on [ β , α] so it is bounded, say by M > 0. Putting everything together, we get that f is bounded on the entire line by max (1, M ). 1 Since it is bounded, we can consider L = supx ∈ R ( f (x)) and l = infx ∈ R ( f (x)). If L = l then 1 the function is constant. If L l, then one of them is nonzero. Assume it is L: pick ε = 2 L . Then there is an α > 0 such that for any x > α we have f (x) < ε, and there is a β < 0 such that for any x < β we have f (x) < ε. Now since again f is continuous on [ β , α], it reaches a maximum at a point X in that closed interval. I claim that f (X ) = L. Indeed if we have f (X ) < L then this would contradict the definition of L (because we know we can find points y ∈ R, such that f ( y) is arbitrarily close to L, and such points are necessarily in [ β , α] because outside of 1 this closed interval everybody has an image less than 2 .L). If inf f < 0, then similarly we obtain a global minimum for the function. Now it can happen that one of the two values inf f , sup f is zero, in which case it is possible 1 that the extremum is not reached: take for example f (x) = x2 + 1 (max is reached at zero, but infimum is zero, not reached). Exercise 5. Section 5.4, #2. Proof. Just notice that 1 x2 − y2 = x − y . 1, y ε : 2 1 x+y x2.y 2 . x+y x2.y 2 |x | + | y | x2.y 2 x2 + y 2 x2 y 2 1 y2 But observe now that for x 1 x2 1 we have + x2 1 2. So now given ε > 0, just pick δ = − 1 y2 then for any x, y such that x − y < δ we get: <2 x−y ε < 22 = ε. The key thing is that the first inequality is true for anybody in [1, + ∞) (it is “uniformly true”!). 1 It is not uniformly continuous on (0, ∞): pick the sequence (xn) = 2n , then notice that (xn +1 − xn) → 0 but that f (xn) − f (xn +1) = 22(n +1) − 22n = 22n.(3) > 3. Exercise 6. Section 5.4, #4. Proof. Same stuff: 1 1 + x2 − 1 + y2 = x − y . 1 x+ y (1 + x2).(1 + y2) x − y . 1 + x2 . 1 + y 2 |x | |y | x−y . The last inequality comes from the fact that for any x ∈ R we have x 1 + x |2. Indeed 2 2 2 either x ≤ 1 and then x 1 + x | , or x > 1 but then x < x 1+x . Now for a given ε > 0, just pick δ = ε: then for any pair x, y satisfying |x − y | < δ we get that 1 1 + x2 − 1 + y2 1 x − y < δ = ε. Exercise 7. Section 5.4, #9. Proof. Just notice 1 f (x) − f ( y) = f (x) − f ( y ) . 1 1 f (x).f ( y ) 1 k2 f (x) − f ( y ) . Now : f is uniformly continuous so for any ε > 0 there is a δ > 0 such that for any x, y satisfying x − y < δ we have f (x) − f ( y ) < k 2.ε. This will imply that for that given ε > 0, if we have |x − y | < δ we deduce 1 f (x ) − f ( y) < k2 k 2.ε = ε. 1 1 Exercise 8. Section 5.6, #10. Proof. Let c be the interior point where f attains a max. We have a < c < b. If either f (a) = f (c) or f (b) = f (c) then we are done (f will not be injective). If not, then pick any k between 2 max ( f (a), f (b)) and f (c). Then by the intermediate value theorem, you know that f will reach that value k once in [a, c], and once in [c, b] (observe that k is not reached at c, so we really get two different points having the same value), and this proves that f is not injective. 3 ...
View Full Document

Ask a homework question - tutors are online