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a) Prove that there exists a unique function defined on R and satisfying the following conditions:f (1) = a(a > 0, a = 1),f (x1 ) · f (x2 ) = f (x1 + x2 ),f (x) → f (x0 )as x → x0 .b) Prove that there exists a unique function defined on R+ and satisfying thefollowing conditions:f (a) = 1 (a > 0, a = 1),f (x1 ) + f (x2 ) = f (x1 · x2 ),f (x) → f (x0 )for x0 ∈ R+ and R+ x → x0 .Hint: Look again at the construction of the exponential function and logarithm discussed in Example 10.2. a) Establish a one-to-one correspondence ϕ : R → R+ such that ϕ(x + y) =ϕ(x) · ϕ(y) for any x, y ∈ R, that is, so that the operation of multiplication in theimage (R+ ) corresponds to the operation of addition in the pre-image (R).
Theexistence of such a mapping means that the groups (R, +) and (R+ , ·) are identicalas algebraic objects, or, as we say, they are isomorphic.b) Prove that the groups (R, +) and (R\0, ·) are not isomorphic.3. Find the following limits.a)b)c)d)limx→+0 x x ;limx→+∞ x 1/x ;log (1+x);limx→0 a xxlimx→0 a x−1 .4. Show that1+11+ · · · + = ln n + c + o(1)2nas n → ∞,where c is a constant. (The number c = 0.57721 . .
. is called Euler’s constant.)Hint: One can use the relation n+1111lnas n → ∞.= ln 1 += +O 2nnnn5. Show that∞a) if two series ∞n=1 an andn=1 bn with positive terms are such that an ∼ bnas n → ∞, then the two series either both converge or both diverge;1b) the series ∞n=1 sin np converges only for p > 1.3.2 The Limit of a Function1476. Show thata) if an ≥ an+1 > 0 for all n ∈ N and the series ∞n=1 an converges, then an =o( n1 ) as n → ∞;b) if bn = o( n1 ), one can always construct a convergent series ∞n=1 an such thatbn = o(an ) as n →∞;∞with positive terms converges, then the series ∞c) if a seriesn=1 An , n=1 an ∞∞where An =k=n αk −k=n+1 ak also converges, and an = o(An ) as n → ∞;d) if a series ∞awithpositive terms diverges, then the series ∞nn=1n=2 An ,n−1nwhere An =k=1 ak −k=1 ak also diverges, and An = o(an ) as n → ∞.It follows from c) and d) that no convergent (resp. divergent) series can serve asa universal standard of comparison to establish the convergence (resp.
divergence)of other series.7. Show thata) the series ∞n=1 ln an , where an > 0, n ∈ N, converges if and only if the se· · · an } has a finite nonzero limit.quence {Πn = a1∞b) the series∞ n=1 ln(1 + αn ), where |αn | < 1, converges absolutely if and onlyif the series n=1 αn converges absolutely.Hint: See part a) of Exercise 5.(sequence of numbers Πn =8.
An infinite product ∞k=1 ek is said to converge if the ((n∞ehasafinitenonzerolimitΠ.WethensetΠ=k=1 kk=1 ek .Show that(en → 1 as n → ∞;a) if an infinite product ∞n=1 en converges, then(∞>0),thentheinfiniteproductb) if ∀n∈N(enn=1 en converges if and only ifthe series ∞lneconverges;nn=1(∞c) if en = 1 + αn and the αn are all of the same∞sign, then the infinite product(1+α)convergesifandonlyiftheseriesnn=1n=1 αn converges.(∞9. a) Find the product n=1 (1 + x 2n−1 ).(x14b) Find ∞n=1 cos 2n and prove the following theorem of Vièteπ=2&12·1 12+21'12·&1 12+21 12+2.12···c) Find the function f (x) iff (0) = 1,14 F.Viète (1540–1603) – French mathematician, one of the creators of modern symbolic algebra.1483Limitsf (2x) = cos2 x · f (x),f (x) → f (0)as x → 0.Hint: x = 2 · x2 .10.
Show thatbna) if bn+1= 1 + βn , n = 1, 2, . . . , and the series ∞n=1 βn converges absolutely,then the limit limn→∞ bn = b ∈ R exists;n= 1 + pn + αn , n = 1, 2, . . . , and the series ∞b) if αan+1n=1 αn converges absolutely, then an ∼ ncp as n → ∞;∞panc) if the series ∞n=1 an is such that an+1 = 1 + n + αn and the seriesn=1 αn∞converges absolutely, then n=1 an converges absolutely for p > 1 and divergesfor p ≤ 1 (Gauss’ test for absolute convergence of a series).11. Show thatlimn→∞1 + an+1ann≥efor any sequence {an } with positive terms, and that this estimate cannot be improved.Chapter 4Continuous Functions4.1 Basic Definitions and Examples4.1.1 Continuity of a Function at a PointLet f be a real-valued function defined in a neighborhood of a point a ∈ R.
Inintuitive terms the function f is continuous at a if its value f (x) approaches thevalue f (a) that it assumes at the point a itself as x gets nearer to a.We shall now make this description of the concept of continuity of a function ata point precise.Definition 0 A function f is continuous at the point a if for any neighborhoodV (f (a)) of its value f (a) at a there is a neighborhood U (a) of a whose imageunder the mapping f is contained in V (f (a)).We now give the expression of this concept in logical symbolism, along with twoother versions of it that are frequently used in analysis. (f is continuous at a) := ∀V f (a) ∃U (a) f U (a) ⊂ V f (a) ,∀ε > 0 ∃U (a) ∀x ∈ U (a) f (x) − f (a) < ε ,∀ε > 0 ∃δ > 0 ∀x ∈ R |x − a| < δ ⇒ f (x) − f (a) < ε .The equivalence of these statements for real-valued functions follows from thefact (already noted several times) that any neighborhood of a point contains a symmetric neighborhood of the point.For example, if for any ε-neighborhood V ε (f (a)) of f (a) one can choosea neighborhood U (a) of a such that ∀x ∈ U (a) (|f (x) − f (a)| < ε), that is,f (U (a)) ⊂ V ε (f (a)), then for any neighborhood V (f (a)) one can also choose acorresponding neighborhood of a.
Indeed, it suffices first to take an ε-neighborhoodof f (a) with V ε (f (a)) ⊂ V (f (a)), and then find U (a) corresponding to V ε (f (a)).Then f (U (a)) ⊆ V ε (f (a)) ⊂ V (f (a)).© Springer-Verlag Berlin Heidelberg 2015V.A. Zorich, Mathematical Analysis I, Universitext,DOI 10.1007/978-3-662-48792-1_41491504Continuous FunctionsThus, if a function is continuous at a in the sense of the second of these definitions, it is also continuous at a in the sense of the original definition. The converseis obvious, so that the equivalence of the two statements is established.We leave the rest of the verification to the reader.To avoid being distracted from the basic concept being defined, that of continuityat a point, we assumed for simplicity to begin with that the function f was definedin a whole neighborhood of a.
We now consider the general case.Let f : E → R be a real-valued function defined on some set E ⊂ R and a apoint of the domain of definition of the function.Definition 1 A function f : E → R is continuous at the point a ∈ E if for everyneighborhood V (f (a)) of the value f (a) that the function assumes at a there existsa neighborhood UE (a) of a in E 1 whose image f (UE (a)) is contained in V (f (a)).Thus(f : E → R is continuous at a ∈ E) := = ∀V f (a) ∃UE (a) f UE (a) ⊂ V f (a) .Of course, Definition 1 can also be written in the ε–δ-form discussed above.
Wherenumerical estimates are needed, this will be useful, and even necessary.We now write these versions of Definition 1.(f : E → R is continuous at a ∈ E) :== ∀ε > 0 ∃UE (a) ∀x ∈ UE (a) f (x) − f (a) < ε ,or(f : E → R is continuous at a ∈ E) :== ∀ε > 0 ∃δ > 0 ∀x ∈ E |x − a| < δ ⇒ f (x) − f (a) < ε .We now discuss in detail the concept of continuity of a function at a point.10 If a is an isolated point, that is, not a limit point of E, there is a neighborhood U (a) of a containing no points of E except a itself. In this case UE (a) = a,and therefore f (UE (a)) = f (a) ⊂ V (f (a)) for any neighborhood V (f (a)).
Thusa function is obviously continuous at any isolated point of its domain of definition.This, however, is a degenerate case.20 The substantive part of the concept of continuity thus involves the case whena ∈ E and a is a limit point of E. It is clear from Definition 1 that(f : E → R is continuous at a ∈ E, where a is a limit point of E) ⇔⇔ lim f (x) = f (a) .Ex→a1 Werecall that UE (a) = E ∩ U (a).4.1 Basic Definitions and Examples151Proof In fact, if a is a limit point of E, then the base E x → a of deleted neighborhoods ŮE (a) = UE (a)\a of a is defined.If f is continuous at a, then, by finding a neighborhood UE (a) for the neighborhood V (f (a)) such that f (UE (a)) ⊂ V (f (a)), we will simultaneously havef (ŮE (a)) ⊂ V (f (a)).
By definition of limit, therefore, limEx→a f (x) = f (a).Conversely, if we know that limEx→a f (x) = f (a), then, given a neighborhoodV (f (a)), we find a deleted neighborhood ŮE (a) such that f (ŮE (a)) ⊂ V (f (a)).But since f (a) ∈ V (f (a)), we then have also f (UE (a)) ⊂ V (f (a)). By Definition 1 this means that f is continuous at a ∈ E.30 Since the relation limEx→a f (x) = f (a) can be rewritten aslim f (x) = fEx→alim x ,Ex→awe now arrive at the useful conclusion that the continuous functions (operations)and only the continuous ones commute with the operation of passing to the limit ata point.
This means that the number f (a) obtained by carrying out the operation fon the number a can be approximated as closely as desired by the values obtainedby carrying out the operation f on values of x that approximate a with suitableaccuracy.40 If we remark that for a ∈ E the neighborhoods UE (a) of a form a base Ba(whether a is a limit point or an isolated point of E), we see that Definition 1 ofcontinuity of a function at the point a is the same as the definition of the statementthat the number f (a) – the value of the function at a – is the limit of the functionover this base, that is(f : E → R is continuous at a ∈ E) ⇔ lim f (x) = f (a) .Ba50 We remark, however, that if limBa f (x) exists, since a ∈ UE (a) for everyneighborhood UE (a), it follows that this limit must necessarily be f (a).Thus, continuity of a function f : E → R at a point a ∈ E is equivalent to theexistence of the limit of this function over the base Ba of neighborhoods (not deletedneighborhoods) UE (a) of a ∈ E.Thus(f : E → R is continuous at a ∈ E) ⇔ ∃ lim f (x) .Ba60 By the Cauchy criterion for the existence of a limit, we can now say that afunction is continuous at a point a ∈ E if and only if for every ε > 0 there exists aneighborhood UE (a) of a in E on which the oscillation ω(f ; UE (a)) of the functionis less than ε.Definition 2 The quantity ω(f ; a) = limδ→+0 ω(f ; UEδ (a)) (where UEδ (a) is theδ-neighborhood of a in E) is called the oscillation of f : E → R at a.1524Continuous FunctionsFormally the symbol ω(f ; X) has already been taken; it denotes the oscillationof the function on the set X.
However, we shall never consider the oscillation of afunction on a set consisting of a single point (it would obviously be zero); thereforethe symbol ω(f ; a), where a is a point, will always denote the concept of oscillationat a point just defined in Definition 2.The oscillation of a function on a subset of a set does not exceed its oscillationon the set itself, so that ω(f ; UEδ (a)) is a nondecreasing function of δ. Since it isnonnegative, either it has a finite limit as δ → +0, or else ω(f ; UEδ (a)) = +∞ forevery δ > 0.
In the latter case we naturally set ω(f ; a) = +∞.70 Using Definition 2 we can summarize what was said in 60 as follows: a function is continuous at a point if and only if its oscillation at that point is zero. Let usmake this explicit:(f : E → R is continuous at a ∈ E) ⇔ ω(f ; a) = 0 .Definition 3 A function f : E → R is continuous on the set E if it is continuous ateach point of E.The set of all continuous real-valued functions defined on a set E will be denotedC(E; R) or, more, C(E).We have now discussed the concept of continuity of a function. Let us considersome examples.Example 1 If f : E → R is a constant function, then f ∈ C(E).
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