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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 ...
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- Fall '10