Continuity and Uniform Continuity
Definition 3. The function f is said to be uniformly continuous on S iff. ?? > 0 ?? > 0 ?x0 ? S ?x ? S.
Uniform continuity.pdf
Observation: Each uniformly continuous function is continuous. in the definition be the same tx?eA. -R be uniformly continuous.
Locally Uniformly Continuous Functions
is uniformly continuous every continuous function on Euclidean space (or more generally any locally compact metric space) is locally uniformly continuous
Bounded uniformly continuous functions
To study the basic properties of the C?-algebra of the bounded uniformly continuous functions on some metric space. Requirements. Basic concepts of analysis:
Uniformly Continuous Representations of Lie Groups
UNIFORMLY CONTINUOUS REPRESENTATIONS OF LIE GROUPS. BY I. M. SINGER. (Received July 30 1951). It is known that noncompact semisimple and nonabelian
A UNIFORMLY CONTINUOUS FUNCTION ON [01] THAT IS
uniformly continuous function on [01] has a positive infimum. Various applications to constructive mathematics are given. 0. Introduction.
global attractors and steady states for uniformly persistent dynamical
weaker sufficient conditions for the existence of interior global attractors for uniformly persistent dynamical systems and hence generalize the earlier
UNIFORMITY AND UNIFORMLY CONTINUOUS FUNCTIONS FOR
left uniform structures if (and only if) the sets of bounded complex-valued
f is uniformly continuous on S if: For all ? > 0 there is ? > 0 such that f
3 juin 2020 Hence f(sn) is Cauchy. ?. Page 6. 6. LECTURE 28: UNIFORM CONTINUITY (II). Note: This proof works precisely because f is uniformly continuous.
Uniform Continuity - Mathonline
It is obvious that a uniformly continuous function is continuous: if we can nd a which works for allx0 we can nd one (the same one) which worksfor any particularx0 We will see below that there are continuous functionswhich are not uniformly continuous Example 5 LetS=Randf(x) = 3x+ 7 Thenfis uniformly continuousonS Proof Choose" >0 Let ="=3
Chapter 11 Uniform Continuity - University of Kentucky
Why is uniform continuity important? Oneof the reasons for studying uniform continuityis its application to the integrability of contin-uous functions on a closed interval i e provingthat a continuous function on a closed intervalis integrable
LECTURE 28: UNIFORM CONTINUITY (II)
from Example 2 is NOT uniformly continuous on (0;1] because fhas no continuous extension on [a;b] Proof: The proof is magical! We’ll do some wishful thinking that actually works STEP 1: Suppose fis uniformly continuous on (a;b) Since on (a;b) f~(x) =: f(x) is continuous all we really need to do is de ne f~(a) and
Continuity and Uniform Continuity - University of Washington
Uniform continiuty is stronger than continuity that is Proposition 1Iffis uniformly continuous on an intervalI then it is continuous onI Proof: Assumefis uniformly continuous on an intervalI To provefis continuous at every point onI letc2Ibe an arbitrary point Let >0 be arbitrary
Searches related to uniformly continuous PDF
approaches that x0 whereas uniform continuity ask what happens if two sequences approach each other 2 De?nition: A function f : D ? Ris uniformly continuous if whenever {u n} and {v n} are sequences in D with {u n ?v n} ? 0 we must have {f(u n)?f(v n)} ? 0 Note: Uniform continuity is de?ned on the domain D not at a point
What is the definition of uniform continuity?
By the definition of uniform continuity, a function is uniformly continuous if we are given any , then we can find a so much so that for any , if we have that the distance between and is less than then we will have that the distance between and will be less than . We should note that this should work for all and that are within from each other.
Is 1 x uniformly continuous?
This is a stronger condition than continuity, and often very useful. 1 x is not uniformly continuous on (0, ?), however (though it is on [a, ?) for any a > 0 ). Thanks Arturo, all the information is starting to come together much more clearly. I've also noticed while poking around that people talk about a function being continuous at a point.
What is the statement of uniform continuity of a function on R?
For a related question, see Intuition for uniform continuity of a function on R. Uniform continuity implies continuity, but is strictly stronger, and in fact, the function at hand is continuous on (0, ?), but not uniformly continuous. The statement of continuity would be that for all ? and all x in the range, there exists a ? such that...
Is f uniformly continuous?
(b) Theorem: If f : D ? R is continuous and D is closed and bounded then f is uniformly continuous. Proof: Suppose {u n} and {v n} are sequences in D satisfying {u n?v n} ? 0. Assume by way of contradiction that {f(u n)?f(v n)} ? 0.
Continuity and Uniform Continuity
521May 12, 2010
1.ThroughoutSwill denote a subset of the real numbersRandf:S!R
will be a real valued function dened onS. The setSmay be bounded likeS= (0;5) =fx2R: 0< x <5g
or innite likeS= (0;1) =fx2R: 0< xg:
It may even be all ofR. The valuef(x) of the functionfat the pointx2S will be dened by a formula (or formulas). Denition 2.The functionfis said to becontinuous onSi8x02S8" >09 >08x2S
jxx0j< =) jf(x)f(x0)j< "Hencefis not continuous1onSi
9x02S9" >08 >09x2S
jxx0j< andjf(x)f(x0)j " Denition 3.The functionfis said to beuniformly continuous onSi8" >09 >08x02S8x2S
jxx0j< =) jf(x)f(x0)j< "Hencefis not uniformly continuous onSi
9" >08 >09x02S9x2S
jxx0j< andjf(x)f(x0)j " :1 For an example of a function which isnotcontinuous see Example 22 below. 14.The only dierence between the two denitions is the order of the quan-
tiers. When you provefis continuous your proof will have the formChoosex02S. Choose" >0. Let=(x0;"). Choosex2S.
Assumejxx0j< .Thereforejf(x)f(x0)j< ".
The expression for(x0;") can involve bothx0and"but must be independent ofx. The order of the quaniers in the denition signals this; in the proofx has not yet been chosen at the point whereis dened so the denition of must not involvex. (Therepresent the proof thatjf(x)f(x0)j< " follows from the earlier steps in the proof.) When you provefis uniformly continuous your proof will have the formChoose" >0. Let=("). Choosex02S. Choosex2S.
Assumejxx0j< .Thereforejf(x)f(x0)j< ".
so the expression forcan only involve"and must not involve eitherxor x 0. It is obvious that a uniformly continuous function is continuous: if we can nd awhich works for allx0, we can nd one (the same one) which works for any particularx0. We will see below that there are continuous functions which are not uniformly continuous. Example 5.LetS=Randf(x) = 3x+7. Thenfis uniformly continuous onS. Proof.Choose" >0. Let="=3. Choosex02R. Choosex2R. Assume jxx0j< . Then jf(x)f(x0)j=j(3x+ 7)(3x0+ 7)j= 3jxx0j<3=":Example 6.LetS=fx2R: 0< x <4gandf(x) =x2. Thenfis uniformly continuous onS. Proof.Choose" >0. Let="=8. Choosex02S. Choosex2S. Thus0< x0<4 and 0< x <4 so 0< x+x0<8. Assumejxx0j< . Then
jf(x)f(x0)j=jx2x20j= (x+x0)jxx0j<(4 + 4)=":27.In both of the preceeding proofs the functionfsatised an inequality of
form jf(x1)f(x2)j Mjx1x2j(1) forx1;x22S. In Example 5 we had j(3x1+ 7)(3x2+ 7)j 3jx1x2j and in Example 6 we had jx21x22j 8jx1x2j for 0< x1;x2<4. An inequality of form (1) is called aLipschitz inequality and the constantMis called the correspondingLipschitz constant. Theorem 8.Iffsatises (1) forx1;x22S, thenfis uniformly continuous onS. Proof.Choose" >0. Let="=M. Choosex02S. Choosex2S. Assume thatjxx0j< . Then jf(x)f(x0)j Mjxx0j< M=":9.The Lipschitz constant depend might depend on the interval. For example, jx21x22j= (x1+x2)jx1x2j 2ajx1x2j for 0< x1;x2< abut the functionf(x) =x2does not satisfy a Lipschitz inequality on the whole interval (0;1) since jx21x22j= (x1+x2)jx1x2j> Mjx1x2j ifx1=Mandx2=x1+ 1. In fact, Example 10.The functionf(x) =x2is continuous but not uniformly con- tinuous on the intervalS= (0;1).Proof.We showfis continuous onS, i.e.
8x02S8" >09 >08x2S
jxx0j< =) jx2x20j< " 3 Choosex0. Leta=x0+ 1 and= min(1;"=2a). (Note thatdepends on x0sinceadoes.) Choosex2S. Assumejxx0j< . Thenjxx0j<1 so
x < x0+ 1 =asox;x0< aso
jx2x20j= (x+x0)jxx0j 2ajxx0j<2a2a"2a=" as required. We show thatfis not uniformly continuous onS, i.e.9" >08 >09x02S9x2S
jxx0j< andjx2x20j " Let"= 1. Choose >0. Letx0= 1=andx=x0+=2. Thenjxx0j= =2< but jx2x20j= 1 +2 2 12= 1 +24
>1 ="as required. (Note thatx0is large whenis small.)11.According to the Mean Value Theorem from calculus for a dierentiable
functionfwe have f(x1)f(x2) =f0(c)(x2x1): forsomecbetweenx1andx2. (The slope (f(x1)f(x2))=(x1x2) of the secant line joining the two points (x1;f(x1)) and (x2;f(x2)) on the graph is the same as the slopef0(c) of the tangent point at the intermediate point (c;f(c)).) Ifx1andx2lie in some intervalSandjf0(c)j Mforallc2S we conclude that the Lipschitz inequality (1) holds onS. We don't want to use the Mean Value Theorem without rst proving it, but we certainly can use it to guess an appropriate value ofMand then prove the inequality by other means.12.For example, consider the functionf(x) =x1dened on the interval
S= (a;1) wherea >0. Forx1;x22Sthe Mean Value Theorem says that x11x12=c2(x1x2) wherecis betweenx1andx2. Ifx1;x22Sthen
c2S(ascis betweenx1andx2) and hencec > asoc2< a2. We can prove the inequality jx11x12j a2jx2x2j 4 forx1;x2aas follows. Firsta2x1x2sinceax1andax2. Then jx11x12j=jx1x2jx1x2jx1x2ja
2(2) where we have used the fact that1< 1if 0< < . It follows that that the functionf(x) is uniformly continuous on any interval (a;1) where a >0. Notice however that the Lipschitz constantM=a2depends on the interval. In fact, the functionf(x) =x1doesnotsatisfy a Lipshitz inequality on the interval (0;1).13.We can discover a Lipscitz inequality for the square root functionf(x) =pxin much the same way. Consider the functionf(x) =pxdened on the
intervalS= (a;1) wherea >0. Forx1;x22Sthe Mean Value Theorem says thatpx 1px2= (x1x2)=(2pc) wherecis betweenx1andx2. If
x1;x22Sthenc2S(ascis betweenx1andx2) and hencec > aso
(2pc)1<(2pa)1. We can prove the inequality j px 1px2j jx1x2j2
pa (3) forx1;x2aas follows: Divide the equation px 1px 2)(px 1+px2) = ((px
1)2(px
2)2) =x1x2
by ( px 1+px2), take absolute values, and use (px
1+px2)2pa. Again the
Lipschitz constantM= (2pa)1depends on the interval and the function doesnotsatisfy a Lipschitz inequality on the interval (0;1). Example 14.The functionf(x) =x1is continuous but not uniformly continuous on the intervalS= (0;1).Proof.We showfis continuous onS, i.e.
8x02S8" >09 >08x2S
jxx0j< =)1x 1x 0 Choosex0. Leta=x0=2 and= min(x0a;a2"). Choosex2S. Assume jxx0j< . Thenx0x jxx0j< x0asox0=2=2< but
1x 1x 0 =1x 0=21x 0 =1x 01 =" as required.Example 15.The functionf(x) =pxis uniformly continuous on the setS= (0;1).
Remark 16.This example shows that a function can be uniformly contin- uous on a set even though it does not satisfy a Lipschitz inequality on that set, i.e. the method of Theorem 8 is not the only method for proving a function uniformly continuous. The proof we give will use the following idea. After choosing" >0 we specify two numbersaandbwhich will depend on ". These numbers will satisfy 0< a < b. We will chooseso that (among other things) < ba. Then after we choosex;x02Sand assume that jxx0j< we will be able to conclude that either bothx0andxare less thanbor both are greater thana. We will choosebso small thatpxandpx0are within"of zero forx;x0< b. We will use a Lipschitz inequality to
handle the case wherex;x0> a. We give the details of this proof after some preliminary lemmas. The only properties of that square root function that we will use are thatpxis dened forx0 and satises px0;(px)2=x;px 2=x:Lemma 17.The square root function is increasing:
0a < b=)pa <
pb:Proof.Assume 0a < b. Ifpbpathenb= (pb)2(pa)2=a
contradictinga < b. Hencepa < pb.Lemma 18. pab=pa pbfora;b0. 6Proof.(pa
pb)2= (pa)2(pb)2=absopa pb=q( pa pb)2=pab.Lemma 19.Assumea < b. Then for any two numbersxandyat least one of the four alternatives (i)x < b&y < b, (ii)ax&ay, (iii)x < a&by, (iv)y < a&bx. Proof.Exactly one of the three alternativesx < a,ax < b,bxholds and exactly one of the three alternativesy < a,ay < b,byholds. There are thus nine cases which we can arrange in a table:x < a ax < b bxy < a(i) (i) (iv) ay < b(i) (i);(ii) (ii) by(iii) (ii) (ii) In each entry of the table we have indicated the alternative (iiv) whichholds in the corresponding case.Proof.Now we prove what is claimed in Example 15, viz. that the square
root function is uniformly continuous on the positive real numbers, i.e.8" >09 >08x;x0>0
jxx0j< =) jpxpx 0j< " Choose" >0. Let= min("2=2;p2"2). Choosex;x0>0. Assumejxx0j< . Reada="2=2 andb="2in Lemma 19: we need consider only four cases: (i)x < "2&x0< "2, (ii)"2=2x&"2=2x0, (iii)x < "2=2 &"2x0, (iv)x0< "2=2 &"2x. Cases (iii) and (iv) contradict the assumption thatjxx0j< "2=2 =ba so we need only consider cases (i) and (ii). In case (i) we havepx < "andpx0< "by Lemma 17 sojpxpx
0j max(px;
px0)< ". In case (ii) we
use the inequality (3) and get j pxpx0j jxx0j2
p"2=2=jxx0jp2" Remark 20.Of course, min("2=2;p2"2) ="2=2. The more complicated formula is given to help the reader understand how the proof was discovered. Remark 21.We can also prove that the square root function is uniformly continuous by taking="2and using the inequality pa+bpa+pb: (To prove the inequality square both sides.) Ifjxx0j< "2then either x 0x < x0+"2(which impliespx
0px < px 0+") or elsexx0< x+"2
(which impliespxpx 0 0j< ". This
proof is simpler than the one given but is harder to nd since it uses a clever inequality which the reader might not think of. Example 22.The functionfdened by
f(x) =1 ifx >0 0 ifx0
is not continuous. Proof.In other words,
9x02R9" >08 >09x2R
jxx0j< andjf(x)f(x0)j " Letx0= 0. (No other value ofx0will work here.) Let"= 1. Choose >0. Letx==2. Thenjxx0j==2< butjf(x)f(x0)j=j10j= 1".8
quotesdbs_dbs17.pdfusesText_23
0x < x0+"2(which impliespx
0px < px0+") or elsexx0< x+"2
(which impliespxpx0 0j< ". This
proof is simpler than the one given but is harder to nd since it uses a clever inequality which the reader might not think of. Example 22.The functionfdened by
f(x) =1 ifx >0 0 ifx0
is not continuous. Proof.In other words,
9x02R9" >08 >09x2R
jxx0j< andjf(x)f(x0)j " Letx0= 0. (No other value ofx0will work here.) Let"= 1. Choose >0. Letx==2. Thenjxx0j==2< butjf(x)f(x0)j=j10j= 1".8
quotesdbs_dbs17.pdfusesText_23
0j< ". This
proof is simpler than the one given but is harder to nd since it uses a clever inequality which the reader might not think of.Example 22.The functionfdened by
f(x) =1 ifx >00 ifx0
is not continuous.Proof.In other words,
9x02R9" >08 >09x2R
jxx0j< andjf(x)f(x0)j " Letx0= 0. (No other value ofx0will work here.) Let"= 1. Choose >0.Letx==2. Thenjxx0j==2< butjf(x)f(x0)j=j10j= 1".8
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