h(x) lim H(x) = lim Since h is nondecreasing then h(x) 0 for all x, and if h is discontinuous at a point x then H(x) > 0. Denote

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1 Real Variables, Fall 4 Problem set 4 Solution suggestions Exercise. Let f be of bounded variation on [a, b]. Show that for each c (a, b), lim x c f(x) and lim x c f(x) exist. Prove that a monotone function (and hence a function of bounded variation) can have only a countable number of discontinuities. (a) Fix c (a, b). Since f is of bounded variation we can write f f f where f and f are nondecreasing functions. Since f is nondecreasing then lim x c f (x) sup x<c f (x) and lim x c f (x) inf x>c f (x), which both exist since f is bounded in a neighborhood of c. So both one-side limits exist for f, and similar reasoning can be done for f. Thus lim x c f(x) and lim x c f(x) exist. (b) Let h be a monotone function. We can assume that h is nondecreasing because if h is nonincreasing we can replace it by g, which is then nondecreasing. We show that h can have only a countable number of discontinuities. At each x (a, b) denote A as the set of discontinuity points of h and H(x) lim y x h(x) lim y x h(x). Since h is nondecreasing then h(x) for all x, and if h is discontinuous at a point x then H(x) >. Denote A k {x (a, b) : H(x) > k } for all k N. Note that A A k. We want to show that each A k is finite, implying that A must be countable. Fix k N and assume that x,..., x n A k so that x <... < x n. When i,,.., n choose rational numbers x, y Q so that x i < y < x i < x < x i. Such points exist since Q is dense in R. Since h is nondecreasing then f(x i ) f(x i ) f(x) f(y) H(x i ) > k.

2 This implies that f(b) f(a) n (f(x i ) f(x i )) i n i H(x i ) > n k. n H(x i ) i So n < (f(b) f(a))k +. We have shown that if we choose any collection of points from A k, then the number of points in this collection has a finite upper bound independent of the collection that we chose. So the set A k is finite. Since A A k then A is countable as a subset of a countable set. Exercise. Construct a monotone function on [, ] that is discontinuous at each rational point. Enumerate the rationals on R by Q {q n : n N}, and define f : R R so that The series k f(x) q n x n. is absolutely convergent so f is well defined. Since k Q is dense in R then for any x < y we find rational points from the interval (x, y) and thus f(x) < f(y). So f is monotone. Moreover if q k Q is fixed, we have f(x) f(q k ) > k for any x < q k, so lim x qk f(x) f(q k ). So f is discontinuous at each rational point. Exercise. Let (f n ) n be a sequence of functions that converges at each point of [a, b] to a function f. Show that T b a(f) lim inf T b a(f n ). Fix ε > and let {x k } N k be a partition of [a, b]. Since (f n) n converges pointwise to f, for each k,..., N there exists n k N so that f n (x k ) f(x k ) < ε N for all n n k. Choose n ε : max{n k : k,..., N}.

3 By triangle inequality we have for all n n ε that f(x k ) f(x k ) f(x k ) f(x k ) + f n (x k ) f n (x k ) + f n (x k ) f n (x k ) f(x k ) f n (x k ) + ε + N ε N + f(x k ) f n (x k ) + N ε N + f n (x k ) f n (x k ) f n (x k ) f n (x k ) ε + Ta(f b n ). So it follows that f n (x k ) f n (x k ) f(x k ) f(x k ) ε + inf n n ε T b a(f n ). Taking supremum over all partitions of [a, b] it follows that T b a(f) ε + lim inf T b a(f n ). Since the choice of ε > was arbitrary then T b a(f) lim inf T b a(f n ). Exercise 4. (a) Let f be defined by f() and f(x) x sin( x ) for x. Is f of bounded variation on [, ]? We show that f is not of bounded variation by showing that the total variation of f is bounded below by a divergent series. Since f( x) f(x) for all x [, ] it suffices to show that the total variation of f in [, ] is not finite. To make a suitable partition of [, ], note first that sin(x) for all x π +πk with k N, and

4 sin(x) for all x π +πk with k N. So sin(x) for all x π(+k) with k N. Hence sin( ) for all x x π(+k) for k N, and the value is negative if k is odd and positive if k is even. So for each N N we can take a partition {x k } N k of [, ] so that x k for all k,..., N. Now f(x π(+k) k) and f(x k ) always have the opposite sign, so it follows that T (f) f(x k ) f(x k ) ( π(k + ) + ) π(k ) π(k + ) π k. This inequality holds for all N N, so we get that T (f) π k. So f does not have bounded variation on [, ]. (b) Let g() and g(x) x sin( ) for x. Is g of bounded x variation on [, ]? We first establish the following fairly simple lemma, which is useful particularly in this exercise but also good to know for basic testing on bounded variations: Lemma Let h : [a, b] R be a function that is M-Lipschitz for some finite constant M >. Then h is of bounded variation. (Recall that h being M-Lipschitz means that h(x) h(y) M x y for all x, y [a, b].) Proof of Lemma: 4

5 Let {x k } N k be a partition of the interval [a, b]. Then h(x k ) h(x k ) M x k x k M(x N x ) M(b a). Taking supremum over all partitions of [a, b] it follows that T b a(h) M(b a) <, so h is of bounded variation. Going back to the exercise, we will then show that g is of bounded variation and this observation is based on the fact that g is M- Lipschitz for some constant M > that we will find. Note that for x we have by usual differentiation rules that and g () lim h g(h) g() h lim h sin(h) h g (x) x sin( x ) cos( x ). lim h h sin( h ) h lim h sin( h h ) lim sin( ) h h h Thus the derivative of g exists for all x [, ] and it is bounded by g (x) x sin( x ) cos( x ) x sin( x ) + cos( x ) + for all x [, ]. Since g(x) as x then g is continuous, so the standard trick to find the Lipschitz constant of g is to use the mean value theorem. Indeed, for any x < y in [, ] there exists by mean value theorem a point ξ (x, y) so that f(x) f(y) f (ξ) x y. Now since the derivative of g is bounded from above by, then this in particular implies that f(x) f(y) x y for all x, y [, ]. Hence f is -Lipschitz and thus of bounded variation by the above Lemma. Exercise 5. Let f be of bounded variation on [a, b]. Show that b a f (x) dx T b a(f). 5

6 Denote g(x) Pa x (f), h(x) Na x (f) and q(x) Ta x for all x [a, b] as in the proof of Theorem 5 in Chapter 5 at Royden s rd edition. We know that f(x) g(x) h(x) + f(a) and q(x) g(x) + h(x) for all x. Since f is of bounded variation then f is measurable and exists almost everywhere, and moreover we have f (x) g (x) h (x) for almost every x [a, b]. Since g and h are nondecreasing functions then f g + h q by triangle inequality whenever the derivative is defined. Thus b a f (x) dx b a q (x) dx q(b) q(a) T b a(f), where we used the fact that q(x) is nondecreasing. So we have shown that b a f (x) dx T b a(f). Exercise 6. Let f f(x, t) be a function of two variables on the square Q {(x, t) : x, t } [, ], and which is measurable as a function of x for each t. Suppose that for each x, lim t f(x, t) f(x), and that for all t we have f(x, t) g(x), where g(x) is an integrable function on [, ]. Then show that lim f(x, t) dx f(x) dx. t Show also that if the function f(x, t) is continuous in t for each x, then h(t) f(x, t) dx is a continuous function. (a) Pick a sequence (t n ) n [, ] so that t n as n and define f n (x) f(x, t n ) for all n N. Each f n is measurable, (f n ) n converges to f pointwise, and f n g for all n N. 6

7 Since g is integrable, we can use dominated convergence theorem to get lim f(x, t) dx lim f n (x) dx lim f t n(x) dx f(x) dx. (b) For the second statement, assume that f(x, t) is continuous in t for each x and denote h(t) f(x, t) dx for all t [, ]. Let t [, ] be fixed and choose (t n ) n [, ] so that t n t. Denote f n (x) f(x, t n ) for all x [, ]. Since f(x, t) was continuous in t, then lim f n (x) f(x, t). In other words, lim f(x, t n) f(x, t), and we also have by triangle inequality that f(x, t n ) f(x, t) g(x) for all x [, ]. Since g is integrable, we can use dominated convergence theorem to get lim h(s) h(t) lim h(t n ) h(t) s t lim lim f(x, t n ) dx f(x, t) dx f(x, t n ) f(x, t) dx lim f(x, t n ) f(x, t) dx lim f(x, t n) f(x, t) dx. So lim s t h(s) h(t) for all t [, ], which shows that h is a continuous function. Exercise 7. Let f(x) be the Cantor function on [, ]. Compute xn f(x) dx for n,,,... 7

8 Denote I n x n f(x) dx for all n N. For n, note that since f(x) f( x) then by change of variable f(x) dx f( x) dx f(u) du, f( x) dx and thus f, so f(x) dx. For n we note that the construction of f implies that f( x) f(x) for all x [, ], f(x + ) + f(x) for all x [, ], and f(x) for all x (, ). Then by change of variable I n x n f(x) dx x n f(x) dx + ( ( x n + x + x n f(x) dx + x n dx + ) n ) f(x) dx + x n f(x) dx + ( x + x n dx x n f(x) dx ) n ( + f(x) ) dx ( x n ( + x ) n ) ( x ) + f n dx + n + (x n + ( + x) n n+ )f(x) dx + n+ n+ (n + ) ( n ( n ) I n+ n + ) n k x k f(x) dx + I n + k k ( n ( n ) I n+ n + ) n k n+ I k + k n+ (n + ), k ( n+ ) n+ n+ (n + ) where we used the formula (a + b) n n k ( n k) a n k b k with a and b x. Multiplying both sides by n+ and subtracting I n we get n ( ) n ( n+ )I n n k I k + n+ k n +. k 8

9 Hence I n n ( ) n n k I n+ k + k n +. k This is a recursive relation that gives I n for any n and we know that I. 9