Math 360: Uniform continuity and the integral D. DeTurck University of Pennsylvania October 25, 2017 D. DeTurck Math 360 001 2017C: Integral/functions 1 / 28 Integrals We've gotten pretty far with our \primitive" integral of functions of bounded variation. But not all continuous functions are BV (nor are all BV functions continuous), so we should define a more robust integral so that the set of integrable functions includes both cases. The setup is pretty much the same as for the integral we already have: To calculate b f (x) dx ˆa we partition the interval [a; b] with points x0 = a < x1 < x2 < ··· < xn = b { the new ingredient is how we choose to \sample" the function f in the interval [xi−1; xi ]. When our functions were monotonic, we chose to evaluate f either the left or right endpoint, since that would give the maximum or minimum value of f on the subinterval. D. DeTurck Math 360 001 2017C: Integral/functions 2 / 28 Riemann vs Darboux There are two approaches to sampling: • Riemann: In the \Riemann integral" you choose a random ∗ point xi in the interval [xi−1; xi ] for each i, and create the Riemann sum n X ∗ RS(f ; P) = f (xi )(xi − xi−1) i=1 If the limit of RS(f ; P) as P becomes finer and finer (n ! 1 and jPj ! 0) exists, then that is the Riemann integral. • Darboux: In the \Darboux integral" you make two sums: For the upper Darboux sum you sample [xi−1; xi ] by simply letting + fi be the supremum of f (x) on the interval [xi−1; xi ], and − you make the lower Darboux sum by letting fi be the infimum of f on the subinterval. D. DeTurck Math 360 001 2017C: Integral/functions 3 / 28 Setup for the integral { upper and lower sums This is going to look familiar (compare with slide set 4). Let f be any bounded function on [a; b], and P a partition of [a; b]. Definition: (upper and lower sums) The upper sum of f corresponding to the partition P is n X + U(f ; P) = fi (xi − xi−1) i=1 and the lower sum of f corresponding to the partition P is n X − L(f ; P) = f (xi − xi−1) i=1 Since f + = sup f (x) ≥ inf f (x) = f − for each i, i x2[xi−1;xi ] x2[xi−1;xi ] i we have U(f ; P) ≥ L(f ; P) D. DeTurck Math 360 001 2017C: Integral/functions 4 / 28 Setup for the integral { definition Since there's always a common refinement P for any partitions P1 and P2, we have L(f ; P1) ≤ L(f ; P) ≤ U(f ; P) ≤ U(f ; P2) where P is a common refinement for P1 and P2. So for any pair of partitions P1 and P2 we have L(f ; P1) ≤ U(f ; P2): Therefore sup L(f ; P) over all partitions P (which is called the lower Darboux integral is less than or equal to inf U(f ; P)(the upper Darboux integral). If these are equal then we say the function f is integrable and their common value is called the integral: b f (x) dx ˆa D. DeTurck Math 360 001 2017C: Integral/functions 5 / 28 Proving existence { regular partitions To show that the integral exists, it is sufficient to find, for any " > 0, a partition P such that U(f ; P) < L(f ; P) + ": We did this before for monotonic functions, we can do this by using sufficiently fine regular partitions { these are partitions having the b − a x 's evenly spaced (so x − x = for all i = 1;:::; n). i i i−1 n Theorem b If f :[a; b] ! R is monotonic, then f (x) dx exists. ˆa Proof: Given " > 0, choose n so large that the regular partition P f (b) − f (a) with n steps, will have U(f ; P) − L(f ; P) = < ". n D. DeTurck Math 360 001 2017C: Integral/functions 6 / 28 Functions of bounded variation Proposition (Greater generality) Then we extended the integral to functions of bounded variation: If f (x) can be written as the sum of two monotonic functions p(x) and q(x) with p increasing and q decreasing on [a; b] (such a function is called a function of bounded variation), then b b b f (x) dx exists and is equal to p(x) dx + q(x) dx. ˆa ˆa ˆa Now, the goal is to extend the whole thing to continuous functions. D. DeTurck Math 360 001 2017C: Integral/functions 7 / 28 Basic properties Because the upper and lower sums have these properties, it follows that the integral does: 1 Linearity: b b b αf (x) + βg(x) dx = α f (x) dx + β g(x) dx for ˆa ˆa ˆa constants α; β and f and g are integrable functions on [a; b]. 2 Monotonicity: If f (x) ≥ g(x) for all x 2 [a; b] then b b a f (x) dx ≥ a g(x) dx. 3 If´ f (x) is integrable´ on [a; b], then so is jf (x)j and b b f (x) dx ≤ jf (x)j dx. ˆa ˆa D. DeTurck Math 360 001 2017C: Integral/functions 8 / 28 More basic properties 4 If a < b < c then c b c f (x) dx = f (x) dx + f (x) dx: ˆa ˆa ˆb 5 If m ≤ infff (x) j x 2 [a; b]g and M ≥ supff (x) j x 2 [a; b]g then b (b − a)m ≤ f (x) dx ≤ (b − a)M: ˆa 6 Mean value theorem for integrals: If f is continuous and integrable (soon we can remove this as an assumption) on [a; b] then there is a c with a < c < b such that 1 b f (c) = f (x) dx: b − a ˆa D. DeTurck Math 360 001 2017C: Integral/functions 9 / 28 Fundamental Theorem { integrals of derivatives Fundamental theorem of calculus I: Integrals of derivatives Let F (x) be a differentiable function on [a; b] with derivative F 0(x), and suppose F 0 is an integrable function on [a; b]. Then b F 0(x) dx = F (b) − F (a): ˆa For any interval (xi−1; xi ) in the partition P we have 0− F (xi ) − F (xi−1) 0+ Fi ≤ ≤ Fi xi − xi−1 by the mean value theorem (for derivatives), since the middle quantity is F 0(x) for some x in the interval. D. DeTurck Math 360 001 2017C: Integral/functions 10 / 28 Fundamental Theorem { proof conclusion But then n 0 X 0− L(F ; P) = Fi (xi − xi−1) i=1 n X ≤ F (xi ) − F (xi−1) i=1 n X 0+ 0 ≤ Fi (xi − xi−1) = U(F ; P): i=1 But the middle sum telescopes to F (xn) − F (x0) = F (b) − F (a). Now F (b) − F (a) is trapped between sup L(F 0; P) and P b inf U(F 0; P), both of which are equal to the integral F 0(x) dx P ˆa because F 0 is integrable. D. DeTurck Math 360 001 2017C: Integral/functions 11 / 28 Second fundamental theorem { derivatives of integrals Fundamental theorem of calculus II: Derivatives of integrals Let f (x) be a continuous (and integrable) function on [a; b]. Define the function F (x) via x F (x) = f (t) dt: ˆa Then F is differentiable and F 0(x) = f (x). First, if h > 0 we have x+h F (x + h) − F (x) = f (t) dt = hf (c) ˆx for some c between x and x + h by properties 4 and 6 (mean value theorem for integrals). D. DeTurck Math 360 001 2017C: Integral/functions 12 / 28 Second fundamental theorem { proof conclusion Therefore the difference quotient F (x + h) − F (x) = f (c) where x < c < x + h h But since f is continuous, f (c) ! f (x) as h ! 0+. x For h < 0 we use that F (x + h) − F (x) = − f (t) dt = hf (c) ˆx+h for some c between x + h and x, and the proof goes through as for the h > 0 case. D. DeTurck Math 360 001 2017C: Integral/functions 13 / 28 Uniform continuity To show that continuous functions on closed intervals are integrable, we're going to define a slightly stronger form of continuity: Definition (uniform continuity): A function f (x) is uniformly continuous on the domain D if for every " > 0 there is a δ > 0 that depends only on " and not on x 2 D such that for every x; y 2 D with jx − yj < δ, it is the case that jf (x) − f (y)j < ". A uniformly continuous function is necessarily continuous, but on non-compact sets (i.e., sets that are not closed and bounded) a continuous function need not be uniformly so. D. DeTurck Math 360 001 2017C: Integral/functions 14 / 28 Not uniformly continuous To help understand the import of uniform continuity, we'll reverse the definition: Definition (not uniformly continuous): A function f (x) is not uniformly continuous on D if there is some " > 0 such that for every δ > 0, no matter how small, it is possible to find x; y 2 D with jx − yj < δ but jf (x) − f (y)j > ". For instance f (x) = x2 is not u.c. on the set [0; 1) because we can choose " = 1 and then for any δ > 0, we have (x + δ)2 − x2 = 2xδ + δ2 and we can choose x > 1=(2δ) so that 2xδ > 1.