Problem set 5, Real Analysis I, Spring, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, 1 x (log 1/ x ) 2 dx 1

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1 Problem set 5, Real Analysis I, Spring, 25. (5) Consider the function on R defined by f(x) { x (log / x ) 2 if x /2, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, R f /2 /2 /2 /2 log 2 ( u x (log / x ) 2 dx x (log / x ) 2 dx x(log x) dx 2 u du 2 ) log 2 (u log x, du x dx) 2 log lim p p log 2 < So f is integrable. (b) Establish the inequality f (x) c x log / x for some c > and all x /2, to conclude that the maximal function f is not locally integrable.

2 2 Solution: Compute for x (, /2] f (x) sup f(y) dy (I an interval) I x I I x f(y) dy (choose I [, x]) x x log x u 2 du x ( log x ) x log /x (u log y, as above) Since x is even, a similar computation holds for x [, ). 2 It is also straightforward to check f (), as the computation above shows f () for all x (, /2]. x log /x This shows f (x), for all x /2. x log / x To show f is not locally integrable, compute /2 f (x) dx 2 2 log 2 x log /x dx x log x dx u du ( log u ) log 2 log log 2 + lim p log p. (u log x, du x dx) Thus f is not locally integrable. (2) Consider the function F (x) x 2 sin(/x 2 ), x, with F (). Show that F (x) exists for every x, but F is not integrable on [, ].

3 Solution: For x, it is clear that F (x) exists. For x, compute F () F (h) F () lim h h h 2 sin(/h 2 ) lim h h lim h sin(/h 2 ). h The last equality is by the squeeze theorem, as h h sin(/h 2 ) h. To show F is not integrable, compute for x F (x) x sin(/x 2 ) (2/x) cos(/x 2 ). Now 2x sin(/x 2 ) is clearly integrable over [, ]. Therefore, F is integrable if and only if the second term is. Compute (2/x) cos(/x 2 ) dx 4 4 (/x) cos(/x 2 ) dx cos u [ /(2u) du] (/u) cos u du for the substitution u /x 2. Now estimate cos u /2 if u I k [kπ π/3, kπ + π/3] for k a positive integer. For u I k, /u /(kπ + π/3). Therefore, (/u) cos u du k kπ + π/3 2. This sum is infinite by using the limit comparison test to the series k. So F is not integrable on [, ]. k (32) Let f : R R. Prove that if f satisfies the Lipschitz condition f(x) f(y) M x y for some M and all x, y R, if and only if f satisfies the following two properties (i) f is absolutely continuous. (ii) f (x) M for a.e. x. Hint to show that if f is Lipschitz with constant M, then f (x) M for almost every x: First prove that f is absolutely continuous, and conclude that f is differentiable almost 3

4 everywhere. Then apply the definition of the derivative and the Lipschitz inequality to show that f (x) M at each x where f is differentiable. Solution: First assume f satisfies the Lipschitz condition. To prove f is absolutely continuous, let ɛ > and δ ɛ/m. Then if {I n (a n, b n )} N n are a disjoint set of intervals in R with N (b n a n ) < δ, n then compute using the Lipschitz condition N N N f(b n ) f(a n ) M(b n a n ) M (b n a n ) < Mδ ɛ. n n Thus f is absolutely continuous. To prove (ii) assuming f is Lipschitz, note since f is absolutely continuous, it is differentiable almost everywhere. If x is a point where f (x) exists, compute n f f(x) f(y) (x) lim. y x x y The Lipschitz condition then implies for y x, the difference quotient f(x) f(y) [ M, M]. x y Therefore, the limit f (x) [ M, M] as well. This shows f (x) M for a.e. x. Now assume (i) and (ii). Since f is absolutely continuous, f exists almost everywhere and for a < b f(b) f(a) b a f (x) dx Condition (ii) then implies b b f(b) f(a) f (x) dx f (x) dx M(b a). a This is the Lipschitz condition for a < b. The remaining cases a b and a > b follow easily. () For f : R R which is continuous with continuous derivative, define f C f C + f C. Let C (R) be the set of all such functions f so that f C <. Show that C (R) is a Banach space: a 4

5 5 (a) Show that C is a norm. Solution: Let α R and f C (R). Then αf C αf C + (αf) C α f C + αf C α f C + α f C α f C. Now let f, g C (R). Then compute by the Triangle Inequality for C f + g C f + g C + (f + g) C f C + g C + f C + g C f C + g C Finally, assume f C. Then f C + f C, and so f and f are both identically. Clearly f in C (R). Thus C is a norm. (b) Let {f n } be a Cauchy sequence in C (R). Show that both {f n } and {f n} are Cauchy sequences in C (R). Solution: For all ɛ, there is an N so that if n, m N, then f n f m C + (f n f m ) C f n f m C < ɛ. This shows f n f m C and f n f m C are both < ɛ, and so {f n } and {f n} are Cauchy sequences in C (R). (c) Conclude that there are uniform limits f, g C (R) of f n and f n respectively. Solution: Since C (R) is complete, then by part (b), f n and f n are convergent to limits f, g respectively in C (R). This is the topology of uniform convergence. (d) For all x R, show that x f n(y) dy x g(y) dy as n. Solution: Since g C (R), g(y) M g C for all y R. We know f n g uniformly, and so there is an N so that if n N, then f n g C sup y R f n(y) g(y) <. For these n N, then sup y R f n(y) sup y R f n(y) g(y) + sup g(y) + M. y R So on each interval [, x] (or [x, ] for x < ), and n N, the functions f n are uniformly bounded by M +, which has integral (M + ) x <. Also f n g everywhere,

6 and the Dominated Convergence Theorem (or the Bounded Convergence Theorem) applies to show x f n(y) dy x g(y) dy. (e) Show that f g everywhere and that f n f in C (R). Solution: The fundamental theorem of calculus applies to the result of part (d) to show that f n (x) f n () G(x) G(), where G is an antiderivative of g. But since f n f in C, we also know f n (x) f n () f(x) f(). So f(x) G(x) + C, where C f() G() is a constant. Since G is an antiderivative of g, we see f (x) G (x) g(x). Therefore f n f C f n f C + f n f C as n, and C (R) is a Banach space. () For continuous f : R R, define f C, f C + sup x y f(x) f(y). x y Define C, (R) to be the set of continuous f so that f C, <. (a) For f C, (R), show that f C, f C + f L. Hint: use 32(ii). Solution: Problem 32 says that () f(x) f(y) M x y if and only if f is absolutely continuous and f (x) M for almost all x. So the quantity f(x) f(y) sup x y x y is bounded by M, and in fact is the infimum of all M satisfying (). Similarly, the L norm f L is the infimum of upper bounds for f (x). Problem 32 then says we are taking the infima over the same set, and so f(x) f(y) sup x y x y f L for f C, (R). This shows f C, f C + f L. (b) Repeat and modify the steps in problem () to show that C, (R) is a Banach space. Solution: Let {f n } be a Cauchy sequence in C, (R). Then part (a) and a simple computation as in (b) shows 6

7 that {f n } is a Cauchy sequence in C (R), and {f n} is a Cauchy sequence in L (R). Since C (R) and L (R) are both Banach spaces, there are limits f n f in C (R) and f n g in L (R). We would like to show f n f in C, (R), which involves two parts. First, we need to verify that f g almost everywhere, as by part (a), this will verify that f n f C,. Second, we need to verify that f C, (R). To show f g almost everywhere, we note that by Problem 32, each f n is absolutely continuous, and so f n (x) f n () x f n(y) dy. Now let n. Since f n f in C, f n (x) f(x) and f n () f(). Similarly, since f n g in L, we may use the Bounded Convergence Theorem as in (d) above to show that lim n This shows x f n(y) dy f(x) f() x x g(y) dy. g(y) dy. Since g L (R), it is locally integrable, and thus f is absolutely continuous and we may apply the Fundamental Theorem of Calculus to conclude that f (x) g(x) for almost every x. Finally, to verify that f C, (R), we know since f C (R) that f C <. The remaining term is taken care of by Problem 32: We know that f is both absolutely continuous and that its derivative f g is uniformly bounded almost everywhere (since g L ). So Problem 32 shows that f is Lipschitz, and so f C, (R). (c) Show that the identity map is an isometry from C (R) to C, (R). Hint: Use the corresponding fact about L and C. Solution: For f C (R), by part (a) and the fact that the L norm is the same as the C norm for continuous functions, f C, f C + f L f C + f C f C. 7

8 (d) Find, with proof, a function in C, (R) C (R). Solution: Consider for example for x <, f(x) x for x, for x >. Then f C, and it is easy to check that for x y, f(x) f(y) x y is < if at least one of x, y / [, ]. If x, y [, ], then this quantity is. This shows f C, +, and so f C, (R). On the other hand, f is not differentiable at x,, and so f / C (R). 8