4.5: Monotone Function - Mathematics LibreTexts

If a function f:A→E∗(A⊆E∗) is monotone on A, it has a left and a right (possibly infinite) limit at each point p∈E∗. Skiptomaincontent Theorem\(\PageIndex{1}\)Example\(\PageIndex{1}\)Theorem\(\PageIndex{2}\)Theorem\(\PageIndex{3}\) Afunction\(f:A\rightarrowE^{*},\)with\(A\subseteqE^{*},\)issaidtobenondecreasingonaset\(B\subseteqA\)iff \[x\leqy\text{implies}f(x)\leqf(y)\text{for}x,y\inB.\] Itissaidtobenonincreasingon\(B\)iff \[x\leqy\text{implies}f(x)\geqf(y)\text{for}x,y\inB.\] Notation:\(f\uparrow\)and\(f\downarrow(\text{on}B),\)respectively. Inbothcases,\(f\)issaidtobemonotoneormonotonicon\(B.\)If\(f\)isalsoonetooneon\(B\)(i.e.,whenrestrictedto\(B\)),wesaythatitisstrictlymonotone(increasingif\(f\uparrow\)anddecreasingif\(f\downarrow\)). Clearly,\(f\)isnondecreasingiffthefunction\(-f=(-1)f\)isnonincreasing.Thusinproofs,weneedconsideronlythecase\(f\uparrow\).Thecase\(f\downarrow\)reducestoitbyapplyingtheresultto\(-f.\) Theorem\(\PageIndex{1}\) Ifafunction\(f:A\rightarrowE^{*}\left(A\subseteqE^{*}\right)\)ismonotoneon\(A,\)ithasaleftandaright(possiblyinfinite)limitateachpoint\(p\inE^{*}\). Inparticular,if\(f\uparrow\)onaninterval\((a,b)\neq\emptyset,\)then \[f\left(p^{-}\right)=\sup_{ax_{0}\text{}(x\inB).\] Moreover,as\(f(x)\inf[B],\)wehave \[f(x)\leq\supf[B]=qp\},q=\inff[B].\quad\square\] Note1.ThesecondclauseofTheorem1holdsevenif\((a,b)\)isonlyasubsetof\(A,\)forthelimitsinquestionarenotaffectedbyrestricting\(f\)to\((a,b).\)(Why?)Theendpoints\(a\)and\(b\)maybefiniteorinfinite. Note2.If\(D_{f}=A=N\)(thenaturals),thenbydefinition,\(f:N\rightarrowE^{*}\)isasequencewithgeneralterm\(x_{m}=f(m),m\inN\)(see§1,Note2).Thensetting\(p=+\infty\)intheproofofTheorem1,weobtainTheorem3ofChapter3,§15.(Verify!) Example\(\PageIndex{1}\) Theexponentialfunction\(F:E^{1}\rightarrowE^{1}\)tothebase\(a>0\)isgivenby \[F(x)=a^{x}.\] Itismonotone(Chapter2,§§11-12,formula(1)),so\(F\left(0^{-}\right)\)and\(F\left(0^{+}\right)\)exist.Bythesequentialcriterion(Theorem1of§2),wemayuseasuitablesequencetofind\(F\left(0^{+}\right),\)andwechoose\(x_{m}=\frac{1}{m}\rightarrow0^{+}.\)Then \[F\left(0^{+}\right)=\lim_{m\rightarrow\infty}F\left(\frac{1}{m}\right)=\lim_{m\rightarrow\infty}a^{1/m}=1\] (seeChapter3,§15,Problem20). Similarly,taking\(x_{m}=-\frac{1}{m}\rightarrow0^{-},\)weobtain\(F\left(0^{-}\right)=1.\)Thus \[F\left(0^{+}\right)=F\left(0^{-}\right)=\lim_{x\rightarrow0}F(x)=\lim_{x\rightarrow0}a^{x}=1.\] (SeealsoProblem12of§2.) Next,fixany\(p\inE^{1}.\)Notingthat \[F(x)=a^{x}=a^{p+x-p}=a^{p}a^{x-p},\] weset\(y=x-p.\)(Whyisthissubstitutionadmissible?)Then\(y\rightarrow0\)as\(x\rightarrowp,\)soweget \[\lim_{x\rightarrowp}F(x)=\lima^{p}\cdot\lim_{x\rightarrowp}a^{x-p}=a^{p}\lim_{y\rightarrow0}a^{y}=a^{p}\cdot1=a^{p}=F(p).\] As\(\lim_{x\rightarrowp}F(x)=F(p),F\)iscontinuousateach\(p\inE^{1}.\)Thusallexponentialsarecontinuous. Theorem\(\PageIndex{2}\) Ifafunction\(f:A\rightarrowE^{*}\left(A\subseteqE^{*}\right)\)isnondecreasingonafiniteorinfiniteinterval\(B=(a,b)\subseteqA\)andif\(p\in(a,b),\)then \[f\left(a^{+}\right)\leqf\left(p^{-}\right)\leqf(p)\leqf\left(p^{+}\right)\leqf\left(b^{-}\right),\] andforno\(x\in(a,b)\)dowehave \[f\left(p^{-}\right)f(p),\)sotherightjumpisgreaterthan\(0.\)Since \[f(p)=f\left(p^{-}\right)=\lim_{x\rightarrowp^{-}}f(x),\] \(f\)isleftcontinuous(butnotrightcontinuous)at\(p\). Theorem\(\PageIndex{3}\) If\(f:A\rightarrowE^{*}\)ismonotoneonafiniteorinfiniteinterval\((a,b)\)containedin\(A,\)thenallitsdiscontinuitiesin\((a,b),\)ifany,are"jumps,"that is,points\(p\)atwhich\(f\left(p^{-}\right)\)and\(f\left(p^{+}\right)\)exist,but\(f\left(p^{-}\right)\neqf(p)\)or\(f\left(p^{+}\right)\neqf(p).\) Proof ByTheorem1,\(f\left(p^{-}\right)\)and\(f\left(p^{+}\right)\)existateach\(p\in(a,b)\). If,inaddition,\(f\left(p^{-}\right)=f\left(p^{+}\right)=f(p),\)then \[\lim_{x\rightarrowp}f(x)=f(p)\] byCorollary3of§1,sofiscontinuousat\(p\).Thusdiscontinuitiesoccuronlyif\(f\left(p^{-}\right)\neqf(p)\)or\(f\left(p^{+}\right)\neqf(p).\square\)