0.5 Absolutely continuous functions Let I \u2282 R be an interval The function f I \u2192 R isabsolutely continuous(AC)if \u2200 \u03b5 > 0 \u2203\u03b4 > 0 \u2200 x 1 y 1

# 0.5 absolutely continuous functions let i ⊂ r be an

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0.5. Absolutely continuous functions.LetIRbe an interval. The functionf:IRisabsolutely continuous (AC)if(ε >0)(δ >0)(x1, y1, . . . , xn, ynI)nXi=1|xi-yi|< δnXi=1|f(xi)-f(y)i)|< εWe note that every Lipschitz continuous function is absolutely continuous and that everyabsolutely continuous function is uniformly continuous.0.5.1.Cantor function.The Cantor function is an example of the function that is uniformlycontinuous, but not absolutely continuous.Here we describe the analytical definition of the Cantor functionf: [0,1][0,1]. Setf(1) = 1. Forx[0,1)consider the representation ofxin digits with base3which does notend with infinite sequence2222222..... That is letx= 0.c1c2. . . cn. . .. To definef(x)do thefollowing. Let˜x=(xif for allncn6= 10.c1. . . ck00. . .otherwise, wherek= min{n:cn= 1}.We can write˜x= 0.˜c1˜c2. . .where˜ciare set by above definition of˜x. Letˆx= 0.b1b2. . .wherebn= 1if˜˜cn= 2orcn= 1, andbn= 0otherwise. We definef(x) = ˆxwhereˆxis interpreted asa binary number.This was discussed on December 1st.
405. CONTINUITY0.6. H¨older continuous functions.Let(X, dX)a metric space. A functionf:XRisH¨older continuous with exponentγ(0,1]if[f]γ:=sup{x1,x2X:x16=x2}|f(x2)-f(x1)|dX(x2, d1)γ<.We denote the set of H¨older continuous functions byC0(X).It is straightforward toshow that[·]γis a semi-norm (defined in subsection 0.0.2) onC0(X). Moreoverkfkγ:=[f]γ+ supxX|f(x)|is a norm on the set of bounded H¨older continuous functionsC0b(X).Note that H¨older continuity with exponentγ= 1is equivalent with Lipschitz continuity.Furthermore note that H¨older continuous functions are uniformly continuous, but not necessarilyabsolutely continuous. Finally absolutely continuous functions may fail to be H¨older continuous.0.7. Monotonic functions.LetAR. The functionf:ARis(i)increasingif for allx, yA, ifx < ythenf(x)< f(y).(ii)nondecreasingif for allx, yA, ifx < ythenf(x)f(y).(iii)nonincreasingif for allx, yA, ifx < ythenf(x)f(y).(iv)increasingif for allx, yA, ifx < ythenf(x)> f(y).A function is monotonic if it is nondecreasing or nonincreasing.LEMMA118.Letf: (a, b)Rbe monotonic. Letp(a, b)thenlimxp-f(x)andlimxp+exist. Consequently a monotonic function cannot have a type II discontinuity.The left and right limit used in the lemma were defined in subsection 0.1.1. This lemma isTheorem 4.29 in Rudin.LEMMA119.LetAbe an interval. A monotonic function can have at most countably manydiscontinuities.This lemma is Theorem 4.30 in Rudin.0.8. Infinite limits and limits at infinity.The meaning of statements likelimx→∞f(x) =is handled using the extended set of real numbersRext, as in section 0.0.1.Recall thatRext=R∪ {-∞,∞}and thatRextis homeomorphic to[-1,1].The limits at infinity (or minus infinity) are then considered as limits with the domainbeingRext, instead ofR. Using the definition of topology onRextone obtains the followingcharacterization.
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