Lecture 3. Limits of Functions and Continuity


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1 Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live without more definitions (unless you pln to go to grd school in mth) Briefly n open set in the plne hs fuzzy boundry like the open disc {x x < 1} A closed set is the complement of n open set So the boundry is hrd not fuzzy; eg, like the closed disc {x x 1} You cn use open sets to eliminte εδ nd limit from the definition of continuous function See Lng, Theorem 56, p 156 See Figure 1 for pictures of open nd closed sets So tht brings us to Lng, Chpter 7  the story of limits of functions in nd on normed vector spces Why re we interested in this question? We wnt to know when we cn interchnge limit nd integrl, limit nd derivtive We wnt to know whether series of functions such s Tylor series or Fourier series converges In fct, we need to know precisely wht we men by convergence of series of functions We wnt to think of the definite integrl s function on the spce C[, b] of continuous functions on finite closed intervl [, b] Wht sort of function is it? Liner? Continuous? In fct proving things bout limits of functions into/on normed vector spces is no hrder thn it ws for functions into/on the rel line We will bsiclly copy the proofs from Lecture 3 of Mth 142 Suppose tht V nd W re normed vector spces I will use the sme symbol for the norm on V nd the norm on W Suppose S V nd f : S W Before defining the limit lim f(x), we need to mke sure tht there re points which re rbitrrily close to V We mke the x sme sort of definition s before Definition 1 Let V be normed vector spce with norm nd S V We sy the point V is dherent to S iff for every δ > there is point such tht x < δ You should picture n dherent point to set s point sticking to the set See Figure 2 Exmples 1) Any point in S is dherent to S 2) If V = R 2, using ny of our fvorite norms, ny point in the closed bll { x R 2 x r } is dherent to the open bll S = { x R 2 x < r } Definition 2 Suppose S V nd f : S W Here V nd W re normed vector spces We will denote both norms by though they my be different Assume V is dherent to S Define lim f(x) = L W to men tht for every ε >, there exists δ > (depending on ε) such tht nd x < δ implies f(x) L < ε Exmples 1) Let V = C[, b], the spce of continuous relvlued functions on the finite intervl [, b] Define I : V R by I(f) = f(x)dx Suppose our norms re f 1 = f(x) dx on V nd the usul bsolute vlue on R Does Answer: Yes In fct, here δ = ε, since using the fct tht integrls preserve inequlities, we see tht lim I(f) =? f 1
2 Figure 1: Pictures of n open set (top) nd closed set (bottom) Figure 2: Picture of red dherent point to purple set 2
3 f 1 < ε implies I(f) I() = f(x)dx f(x) dx = f 1 < ε 2) Set f(x, y) = y2 x 2 y 2 +x, for (x, y) (, ) in R 2 Does lim f(x, y) =? Here you cn use ny of our fvorite 2 (x,y) (,) norms on R 2 nd ordinry bsolute vlue on R Figure 1 below shows the grph of z = f(x, y) Answer No Set y = kx Then f(x, kx) = k2 1 k 2 +1 Thus f(x, y) hs different vlue on vrious lines through the origin In prticulr, it is 1 on the xxis nd +1 on the yxis There is no limit s (x, y) pproches the origin There re more exmples in homework 2 from Mth 142A The properties of limits re similr to those stted in prt 3 of the lectures Properties of Limits in Normed Vector Spces We ssume tht V, W re normed vector spces, is dherent to S V, f, g : S W 1) Uniqueness Suppose lim 2) Linerity Suppose α, β R nd f(x) = L nd lim lim f(x) = M Then L = M f(x) = L nd lim g(x) = M Then 3) Composite Suppose we hve 3 normed vector spces V, W, Z nd functions f : S T, is dherent to the set S V nd the vector L is dherent to the set T W If then lim g (f(x)) = M 4) Inequlities Suppose f, g : S R nd f(x) g(x) for ll If lim 5) Independence of Which Equivlent Norm is Used Using equivlent norms on either V or W leds to the sme definition of lim (αf(x) + βg(x)) = αl + βm g : T Z, plus we know tht the vector lim f(x) = L nd lim g(y) = M, y L y T f(x) = L nd lim lim f(x) = L g(x) = M, then L M Some Proofs 1) L M = L f(x) + f(x) M L f(x) + f(x) M For ny ε > δ 1 such tht x < δ 1 implies L f(x) < ε 2 And δ 2 such tht x < δ 2 implies f(x) M < ε 2 Thus tking δ = min{δ 1, δ 2 } we see tht x < δ implies L M L f(x) + f(x) M < ε Since ε ws rbitrry this mens L M = (see Lng, p 3 if you don t believe this) By the first xiom of norms, this implies L M = 2) We know tht given ε > δ 1 such tht y L < δ 1 implies g(y) M < ε (αf(x) + βg(x)) (αl + βm) = αf(x) αl + βg(x) βm αf(x) αl + βg(x) βm = α f(x) L + β g(x) M 3
4 Figure 3: Picture of the h(x) vlues (red circles to the right of ) on the rel xis, long with the limiting vlue K ( blue str to the left of ) No wy cn this hppen since the red circles cn never get close to the blue str ε δ 1 so tht x < δ 1 implies f(x) L < 2(1+ α ) ε And δ 2 so tht x < δ 2 implies g(x) M < Then let δ = min{δ 1, δ 2 } so tht x < δ implies 2(1+ β ) (αf(x) + βg(x)) (αl + βm) α f(x) L + β g(x) M < ε α 2 (1 + α ) + ε β 2 (1 + β ) < ε 3) ε δ 1 such tht y L < δ 1 implies g(y) M < ε And δ 2 such tht x < δ 2 implies f(x) L < δ 1 Then x < δ 2 implies, setting y = f(x), g(y) M < ε since y L < δ 1 4) Define h(x) = g(x) f(x) Then h(x) for ll Let K = M L If K <, we cn get contrdiction We know from Property 1 of Limits tht K = M L = lim h(x) See Figure 3 The red circles re the vlues of h(x) to the right of nd the blue str is K, to the left of Tke ε = K K 2 Then δ such tht x < δ implies h(x) K < 2 Then, since K is negtive, h(x) K < K 2 nd, dding K to both sides, h(x) < K 2 <, contrdiction to h(x) We refer you to Lng, p1623 for the generl story of limits of products You get the joy of considering specil cse in Homework 2, problem 2 There re lots of other cses one could look t; eg, sclr vlued function times vector vlued function, mtrix vlued function times mtrix vlued function, If you hte εδ stuff, you will love the following theorem, which llows you to think bout limits of sequences insted Theorem 3 Sequentil Definition of Limits Suppose S V nd f : S W, where V nd W re normed vector spces Let V be dherent to S Then the existence of lim f(x) = L is equivlent to sying tht for every sequence of vectors {x n } in S such tht lim n =, we hve lim n) = L exists n n 4
5 Proof = We leve this prt s n extr credit exercise = Proof by Contrdiction Suppose {x n } in S st lim lim x n =, we hve lim f(x n) = L If, by contrdiction, n n f(x) does not equl L Then using the rules for negting sttement involving lots of, we see tht ε > st n Z +, x n S with x n < 1 n nd f(x n) L ε Since then lim x n =, this is contrdiction to n lim f(x n) = L n Mybe we should try to drw picture of the definition of limit in higher dimensions The problem is tht it is hrd to drw the grph of function unless it mps subset of the plne into the rels Here the grph of z = f(x, y) is 3dimensionl Just plot the points (x, y, f(x, y)) in 3spce Of course in the infinite dimensionl cse good luck drwing pictures Even drwing the grph of function from the plne to the plne requires 4 dimensionl pictures You cn still project them down to 2 dimensions s you would in the cse of rel vlued function of 2 vribles Or you cn mke movie of the grph being rotted Figure 1 shows 3D grph (drwn using Scientific Workplce) of the function z = f(x, y) = y2 x 2 y 2 +x 2 from our erlier exmple It is hrd to see how bdly the function fils to hve limit s (x, y) pproches (, ) z 2 1 x y D plot of z = f(x, y) = y2 x 2 y 2 +x Continuous Functions Suppose tht V, W re normed vector spces nd U is subset of V Definition 4 f : U W is continuous t c U iff lim x c f(x) = f(c) When V = W = R, we view continuity to men tht the grph of y = f(x) does not brek up t x = c When V = R 2, you cn think similr thing bout the surfce z = f(x, y) in 3spce But reclling Figure 1 of the function f(x, y) = y2 x 2 y 2 +x, 2 it is even hrd to see the brek up for function of 2 vribles It is esier to recll tht we sw tht f(x, y) hs different vlue on vrious lines through the origin It is 1 on the yxis nd 1 on the xxis, for exmple Weird or Perhps Ridiculous Fct Define c to be n isolted point of U to men tht there exists δ > such tht the bll of rdius δ bout c contins no points of U; ie the set {x U x c < δ } is empty If c is isolted thn ny function defined t c is continuous there If V = W = R, the point (c, f(c)) would be disconnected from the rest of the grph of y = f(x) This seems to be bd choice of terminology, but it ppers to be the usul one We cn use the properties of limits to deduce the following properties of continuous functions 5
6 Properties of Continuous Functions 1) Linerity Suppose tht f, g : U W, where U V nd V, W re normed vector spces Let α, β be (rel) sclrs Then f nd g continuous t c U implies tht (αf + βg) is continuous t c 2) Composition Suppose tht V, W, Z re normed vector spces with U V nd T W Let c U Suppose tht f : U T nd g : T Z Suppose f is continuous t c nd g is continuous t f(c) Then g f is continuous t c 3) Sequentil Version Assume V, W re normed vector spces with U V The function f : U W is continuous t c U iff sequence {v n } of vectors in V such tht lim n v n = c, we hve lim n f (v n) = f (c) For the proofs, you just hve to look t the corresponding properties of limits We leve it to you nd Serge Exmples (the sme s those in the section on limits) 1) Let V = C[, b], the spce of continuous relvlued functions on the finite intervl [, b] Define I : V R by I(f) = f(x)dx Suppose we use f 1 = f(x) dx on V nd the usul bsolute vlue on R We showed erlier tht the liner function I(f) is ctully continuous t f = Now I clim I(f) is continuous on V ; ie, continuous everywhere Why? Using properties of the integrl on continuous functions tht we proved lst qurter, I(f) I(g) = I(f g) I ( f g ) = f g 1 This mens tht given ε >, we cn tke δ = ε nd then f g 1 < ε implies I(f) I(g) < ε (which is the εδ definition of continuity t g (or f) In fct, since δ depends only on ε nd not on f or g, we hve proved tht the function I(f) is uniformly continuous  concept we re bout to define Extr Credit Is I(f) still continuous when we replce the norm f 1 on V with f 2? Explin your nswer 2) Look gin t the function f(x, y) = y2 x 2 y 2 +x 2, for (x, y) (, ) in R 2 We know tht this function cnnot be continuous t (, ) since it hs no limit s (x, y) (, ) There re more such exmples in the homework Definition 5 Suppose tht V, W re normed vector spces nd U is subset of V We sy tht f : U W is uniformly continuous t on U iff ε > δ > (with δ depending only on ε) such tht u, v U, u v < δ implies f(u) f(v) < ε The point here is tht δ does not depend on u, v U Exmple 1 just considered is n exmple of uniformly continuous function where in fct δ = ε If I were good person nd lectured on the section of Lng bout continuous functions on compct sets, we d hve mny more exmples of uniformly continuous functions, since Theorem 25 on pge 198 of Lng sys the following Theorem 6 Suppose K is compct subset of normed vector spce V nd W is ny normed vector spce A continuous function f : K W must be uniformly continuous on K Wht does "compct set" men? The definition cn be found in Lng, p 193 Let V be finite dimensionl normed vector spce like R n Then K R n is compct iff K is closed nd bounded (This is theorem not the definition of compct) So, for exmple, closed bll of rdius r ; ie, {x R n x r }is compct This is flse in infinite dimensions A bll of rdius r in infinite dimensions is not compct Very inconvenient Anywy this mens tht for f : {x R n x r } W continuity implies uniform continuity More Exmples 3) V =normed vector spce Let f(x) = x Then f is uniformly continuous on V For ll x, y V, the tringle inequlity implies (s it did for ordinry bsolute vlue in n erly homework problem from 142), x y x y 6
7 This sys we cn tke ε = δ gin 4) Suppose tht L : R n R m is liner Then tke e j to be the stndrd bsis vector e j = 1, with 1 in the jth row nd the rest of the entries being Every vector v R n, cn be written uniquely in the form: It follows from linerity of L, tht Write Le j = 1j j 1,j jj mj R m So we see tht v = Lv = v 1 v j 1 v j v n Lv = n j=1 v j = n v j e j j=1 n v j Le j j=1 1j j 1,j jj mj = Av, where we multiply the mtrix A whose entries re ij with the vector v According to homework 2, problem 4, the liner function L is uniformly continuous You cn see this by using the infinity norm on R n nd R m nd showing tht there is constnt C > so tht Lx C x The constnt C depends on the entries ij of the mtrix A If you tke the K = mx ij, then C = nk should work 3 Completeness of C[,b] with respect to the Norm In prt 2 of the Lectures we promised to prove the following Theorem 7 The normed vector spce C[, b] of continuous rel vlued functions on the finite intervl [, b] is complete with respect to the norm f = mx x [,b] f(x) Proof Recll tht V "complete" mens every Cuchy sequence in V converges to n element of V So let {f n } be Cuchy sequence in C[, b] using the norm f This mens for every x [, b], the sequence {f n (x)} of rel numbers is Cuchy; s ε N ε st f n (x) f m (x) f n f m < ε when n, m N ε (1) 7
8 We showed lst qurter tht Cuchy sequences of rel numbers converge to limit in R Thus x [, b] there is function f(x) = lim n f n(x) Now we need to show tht f n converges uniformly to f on [, b] Let ε > be given There is M = M(x, ε) N ε so tht m M implies f m (x) f(x) < ε Then for n N ε we hve the following sneky formul by dding nd subtrcting f m (x) nd using the tringle inequlity: f(x) f n (x) f(x) f m (x) + f m (x) f n (x) < ε + f n f m < 2ε (2) We chose N ε so tht formul (1) holds This implies f f n < 2ε for n N ε which is uniform convergence of f n to f on [, b] since N ε does not depend on x Next we need to show tht f is continuous on [, b] To see this, note tht for x, y [, b], using the tringle inequlity in sneky wy gin (this time dding nd subtrcting f n (x) f n (y)): f(x) f(y) f(x) f n (x) + f n (x) f n (y) + f n (y) f(y) f(x) f n (x) + f n (x) f n (y) + f n (y) f(y) We know tht for n N ε the 1st nd 3rd terms re < 2ε by formul (2) Since f n is continuous, there is positive δ, depending on n, ε nd y such tht x y < δ implies the middle term is lso < ε So the finl result is tht f(x) f(y) < 5ε Replce ε by ε/5, if you like 8
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