REAL ANALYSIS II: PROBLEM SET 2

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1 REAL ANALYSIS II: PROBLEM SET 2 21st Feb, 2016 Exercise 1. State and prove the Inverse Function Theorem. Theorem Inverse Function Theorem). Let f be a continuous one to one function defined on an interval, and suppose that f is differentiable at f 1 b), with derivative f f 1 b)) 0. Then f 1 is differentiable at b, and f 1 ) b) = 1 f f 1 b)). Proof. Let b = fa). What we wish to calculate is f 1 b + h) f 1 b) f 1 b + h) a. h 0 h h 0 h Furthermore, for every h, there exists a unique k such that b + h = fa + k). As f is continuous and one to one at a, then f 1 is continuous at b. This allows us to liberally switch between the it as h 0 and k 0. Then, f 1 b + h) a h 0 b + h b f 1 fa + k) a k 0 fa + k) fa) k k 0 fa + k) fa) As fa) = f f 1 b)), and we know that f f 1 b)) 0, this gives us that ) k k 0 fa + k) fa) = f f 1 b) + k) f f 1 1 b)) = f f 1 b) )) 1. k 0 k Putting everything together yields that f 1 is differentiable at b and has derivative f 1 ) b) = 1 f f 1 b)). Exercise 2. Let f be continuous on [a, b], g differentiable on [a, b], and ga) = 0. Prove that if there exists a λ 0 such that then gx) = 0 on [a, b]. Date: 21st Feb, gx)fx) + λg x) gx) for x [a, b],

2 Proof. Assume that f be continuous on [a, b], g differentiable on [a, b], and ga) = 0. Furthermore, let λ 0 such that gx)fx) + λg x) gx) for x [a, b]. 1) Since f and g are continuous functions on the compact set [a, b], both functions are bounded. Let M, N R such that fx) M and gx) N for all x [a, b]. Turning our attention to rewriting 1) and utilizing the reverse triangle inequality, a bound we just described, and that λ 0, we get that for all x [a, b], gx)fx) + λg x) gx) λg x) gx)fx) λg x) gx)fx)) gx) gx) λg x) gx)fx) gx) gx) g x) 1 + M g x) gx) + gx) fx) 1 + M) gx) gx) We will use the following equation, which holds for all x [a, b]. g x) 1 + M gx) 2) Now, we will recursively construct a sequence of that will force the evaluation of a point under g to be zero. Let [c, d] [a, b] such that d c < 1 and gc) = 0. By the Mean Value Theorem, 2 1+M there exists x 1 c, d) such that g x 1 ) = gd) gc) d c = gd) d c. 3) By evaluating 2) at x = x 1 and utilizing 3), we have the following which we will rewrite as well. g x 1 ) = gd) d c 1 + M gx 1 ) gd) d c) 1 + M gx 1 ) < M gx 1 = M 2 gx 1) Then, for [c, x 1 ] [a, b], we know that x 1 c < d c < 1 2 tells us there exists x 2 c, x 1 ) such that g x 2 ) = gx 1) gc) x 1 c 1+M Again, by evaluating 2) at x = x 2 and utilizing 4), we get that gd) < gx 1) < gx 2).. The Mean Value Theorem = gx 1) x 1 c. 4) Continuing in this manner gives us a sequence of distinct points between c and d that satisfy the following inequality. gd) < gx 1) < gx 2) < < 1 2 n gx n) <. 5)

3 Utilizing the bound of N for g, we can rewrite 5) as the following inequality. gd) < N < N < < 1 2 n N <. 6) The sequence { } N is a convergent with it of 0. This forces gd) = 0, as well as 2 n n=1 gx n ) = 0 for any n N +. Now, to conclude our argument, the interval [a, b] has finite length and can therefore be covered by a finite number of closed intervals of length less than 1. Furthermore, for 2 1+M any point x [a, b], we can build a finite number of closed intervals of length less than 1 2 which will end with an interval that ends at x. So, gx) = 0 and since this choice of x was arbitrary, we can conclude that the function g is identically 0 on [a, b]. 1+M Exercise 3. Suppose f is a real function with domain R which has the intermediate value property: If fa) < c < fb), then fx) = c for some x between a and b. Suppose also, for every rational r, that the set of all x with fx) = r is closed. Prove that f is continuous. Proof. Let f be a real valued function with domain R that satisfies the intermediate value property and the inverse image of every rational number under f is a closed set. Let x n x be a convergent sequence and assume, by way of contradiction, that fx n ) > r > fx) for some rational number rwe can find r using the density of the Rational numbers in the Real numbers). Since f satisfies the intermediate value property, there exists some t n between x n and x such that ft n ) = r, for each n. By construction, t n x and so x is a it point of the sequence {t n }. Because {t n : n N} f 1 r) and we know that f 1 r) is closed, by assumption, f 1 r) must contain all its it points. However, fx) r, so x / f 1 r), a contradiction. Therefore, fx n ) fx) as n and so f is continuous. Exercise 4. Definition. Assume that f is differentiable on an open interval I and that [a, b] I. We say that f is uniformly differentiable on [a, b], if for any ε > 0 there is a δ > 0 such that fx + h) fx) f x) h < ε for all x [a, b] and h < δ and x + h I. Prove that f is uniformly differentiable on [a, b] iff f is continuous on [a, b]. Proof. First, suppose that f is continuous on [a, b]. By the mean value theorem, for x [a, b], x + h I and some 0 < λ < 1, fx + h) fx) h f x) = f x + λh) f x).

4 Now, since f is continuous on a compact interval, f is uniformly continuous. Let ε > 0 and δ > 0 be given by uniform continuity. Then, for h < δ, x + λh x < δ which gives us that ε > f x + λh) f x) = fx + h) fx) f x) h and so f is uniformly differentiable on [a, b]. Now, assume that f is uniformly differentiable on [a, b]. Let {h n } n=1 be any sequence that converges to zero with h n 0 for all n N and x + h n I for x [a, b]. Claim 1. The sequence { fx+hn) fx) h n } n=1 converges uniformly to f on [a, b]. Since f is assumed to be uniformly differentiable, this claim is just a reading of the definition. { Since fx+hn) fx) is a sequence of continuous functions that converge uniformly to f h n } on [a, b], then f is continuous on [a, b]. Exercise 5 Bernoulli s Inequality). Let α R such that 0 α 1 be fixed. Prove that for any x R with x > 1 and x 0, a) 1 + x) α > 1 + αx if α < 0 or α > 1, b) 1 + x) α < 1 + αx if 0 < α < 1. Proof. Fix α R such that 0 α 1 and consider the function hx) = 1 + x) α αx 1 defined on the interval 1, ). Note that h0) = 0 and that h x) = α α 1) 1 + x) α 2. Then, for any x 1, ) with x 0, 1 + x) α 2 > 0 Case 1 α < 0). If α < 0, then α 1) < 0 and 0 < α α 1). So, for any x 1, ) with x 0, h x) > 0. This tells us that the point h0) = 0 is the absolute minimum on 1, ), so for every x not equal to 0 in this interval, hx) = 1 + x) α αx 1 > x) α > 1 + αx. Case 2 0 < α < 1). If 0 < α < 1, then α 1) < 0 and α α 1) < 0. So, for any x 1, ) with x 0, h x) < 0. This tells us that the point h0) = 0 is the absolute maximum on 1, ), so for every x not equal to 0 in this interval, hx) = 1 + x) α αx 1 < x) α < 1 + αx. Case 3 1 < α). If 1 < α, then 0 < α 1) and 0 < α α 1). So, for any x 1, ) with x 0, h x) > 0. This tells us that the point h0) = 0 is the absolute minimum on 1, ), so for every x not equal to 0 in this interval, hx) = 1 + x) α αx 1 > x) α > 1 + αx.

5 Exercise 6. Let f 1, f 2,..., f n and g 1, g 2,..., g n be continuous on [a, b] and differentiable on a, b). Suppose that for all k = 1,..., n that g k a) g k b). Prove that there exists c a, b) for which f kc) = g kc) f kb) f k a) g k b) g k a). Proof. Consider the function hx) = g k x) g k a)) f ) kb) f k a) g k b) g k a) f kx) f k a)). The function h is a finite sum of differentiable functions multiplied by constants and added to a finite number of constants. Thus h is continuous on [a, b] and differentiable on a, b). As such, h x) = g kx) f ) kb) f k a) g k b) g k a) f kx). 7) and, By construction, ha) = = = 0 hb) = = = = 0 g k a) g k a)) f kb) f k a) 0) f kb) f k a) g k b) g k a) 0) g k b) g k a)) f kb) f k a) ) g k b) g k a) f ka) f k a)) ) ) g k b) g k a) f kb) f k a)) f k b) f k a)) g ) kb) g k a) g k b) g k a) f kb) f k a)) f k b) f k a)) f k b) f k a))) By Rolle s theorem, there exists a c a, b) such that h c) = 0. By rewriting 7) and collecting terms on various sides, f kc) = g kc) f kb) f k a) g k b) g k a).

6 Exercise 7 L A TEX). Consider f : [ 1, 1] R defined by { x fx) = 4 e 1 4 x2 sin 8, if x 0 x 3 0, if x = 0. Show that f has a derivative everywhere on the interval, that f is bounded, and that f possesses no absolute) extreme values on [ 1, 1]. Proof. Our previous methods give us that the derivative at every point other than x = 0 is given by e 1 4 x2 4x x5) sin 8 x 3 ) 24 cos 8 x 3 )). So, we must calculate f 0) f 0) x 0 fx) f0) x 0 x 4 e 1 4 x2 sin 8 0 x 3 x 0 x 0 x 4 e 1 4 x2 sin 8 x 3 x 0 x To calculate this it, we will turn to the squeeze theorem, 1 sin 8 x 3 1 x 3 e 1 4 x2 x 3 e 1 4 x2 sin 8 x 3 x3 e 1 4 x2 x 0 x3 e 1 4 x2 x 3 e 1 4 x2 sin 8 x 0 x 3 x 0 x3 e 1 4 x2 ) x 0 x3 1 x2) x 0 e 4 x 3 e 1 4 x2 sin 8 ) x 0 x 3 x 0 x3 1 x2) x 0 e 4 0) 1) x 3 e 1 4 x2 sin 8 0) 1) x 0 x3 0 x 0 x 3 e 1 4 x2 sin 8 x x 0 x 3 e 1 4 x2 sin 8 x 3. x 0 x 3 e 1 4 x2 sin 8 x 3. Hence, f 0) = 0. In turn, this gives us that the following equation for the derivative of f f x) = { e 1 4 x2 4x x5) sin 8 x 3 ) 24 cos 8 x 3 )), if x 0 0, if x = 0. On every neighborhood of the origin, f takes values arbitrarily close to 24 and 24. We must now show that f is bounded by 24 everywhere else on the interval [ 1, 1]. Let 0 < h = x 1, then the following is the first of two inequalities we will utilize, Showing 8) requires the following lemma: 0 < e 1 4 x2 < h2 e 1 4 h2 < h2. 8) Lemma 1. For a < b, e a b a) < e b e a < e b b a). Proof of Lemma 1. Left to the reader.

7 The second inequality is given by 4x 3 12 ) ) ) 8 8 x5 sin 24 cos x 3 x 3 4x 3 12 ) ) ) 8 x5 sin + 8 x 3 24 cos 9) x 3 4x 3 1 ) 2 x ) So, for 0 < h 1, and using 8) and 12), f x) < 1 14 ) 24 h ) h3 4h h ) 4h h = h3. 12) = h5 6h h3 = h5 32 ) h2 9 2 h h3 < h h3 = h2 1 h) 24. Therefore, on the closed interval [ 1, 1], f has a supremum of 24 and an infimum of 24, but f does not assume either of these values on the interval [ 1, 1]. Exercise 8. Show that f defined on ]0, 2[ by setting { x fx) = 2, for rational x ]0, 2[ 2x 1, for irrational x ]0, 2[. is differentiable only at x = 1 and that f 1) 0. Is the inverse function differentiable at 1 = y = f1)? Proof. We shall first show that f is only continuous at x = 1. Let x n be a sequence of rational numbers in ]0, 2[, that converge to 1 but x n 1 for all n. Then, n fx n ) = n x 2 n = 1. Now, let {x n } be a sequence of irrational numbers in ]0, 2[ that converge to 1. Similarly, n fx n ) n 2x n + 1 = 1. These its equal the evaluation of f at x = 1, so f is continuous at 1. To see that f is not continuous at any other point in ]0, 2[, let x ]0, 1[ ]1, 2[. If x is a rational number, let {x n } be a sequence of irrational numbers in ]0, 1[ ]1, 2[ such that x n x as n. For continuity to hold, we need the following it to go to zero, n fx) fx n) n x 2 2x n + 1 n x 2 2x + 1 = x 1 2 > 0.

8 If x is an irrational number, let {x n } be a sequence of rational numbers in ]0, 1[ ]1, 2[ such that x n x as n. Similarly, for continuity to hold, the following it must go to zero, fx n) fx) xn 2 2x + 1 x 2 2x + 1 = x 1 2 > 0. n n n Hence, f is not continuous at any point but at x = 1. Now, we want to determine the derivative of f at x = 1. Let {x n } be any sequence of irrational numbers in ]0, 1[ ]1, 2[ with x n 1 as n. Then, fx n ) 1 2x n = 2. n x n 1 n x n 1 n Now, let {x n } be any sequence of rational numbers in ]0, 1[ ]1, 2[ with x n 1 as n. So, fx n ) 1 x 2 n 1 n x n 1 n x n 1 x n n n Hence, f 1) = 2. As we move toward investigating the differentiability of f 1, we need to verify that f is a one to one function to know if we even have an inverse function. Let x, y ]0, 2[ such that x y with r ]0, 2[ where, without loss of generality x + r = y. Case 4 x and y are irrational). Suppose x and y are both irrational. Then, Since r > 0, fx) fy). fy) = 2y 1 = 2x + r) 1 = fx) + 2r. Case 5 x and y are rational). Suppose x and y are both rational. Then, Since r, x > 0, fx) fy). fy) = y 2 = x + r) 2 = fx) + 2xr + r 2. Case 6 x is rational and y is irrational). Suppose, without loss of generality, that x is rational and y is irrational. Then, fx) = x 2 and as the product of two rational numbers is again rational. Also, fy) = 2y 1 but, since y is irrational, 2y 1 is also irrational. Hence, fx) fy). This gives us that f is one to one. We can therefore define f 1 on ]0, 3[. However, f 1 is not defined entirely on ]0, 3[, the inverse function cannot evaluate rational numbers whose square roots are irrational. As these are dense in ]0, 3[, the domain of f 1 has no interior points; f 1 is not continuous on ]0, 3[ and so we cannot define f 1 ) 1). The Inverse Function Theorem requires that f 1 be continuous on some open neighborhood of the point we are attempting to evaluate the derivative of f 1 at.