Multivariable Advanced Calculus

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86 CONTINUOUS FUNCTIONS≤ |a| |f (x) − f (y)| + |b| |g (x) − g (y)|() ()εε< |a|+ |b|2 (|a| + |b| + 1) 2 (|a| + |b| + 1)Now consider 2.) There exists δ 1 > 0 such that if |y − x| < δ 1 , then |f (x) − f (y)| 0 be given. There exists δ 2 such that if |x − y| < δ 2 , then|g (x) − g (y)|
872≤2[|f (x) g (y) − f (y) g (y) + f (y) g (y) − f (y) g (x)|]|g (x)|2≤2[|g (y)| |f (x) − f (y)| + |f (y)| |g (y) − g (x)|]|g (x)|[ ]2 3≤|g (x)| 2 2 |g (x)| |f (x) − f (y)| + (1 + |f (x)|) |g (y) − g (x)|2≤2(1 + 2 |f (x)| + 2 |g (x)|) [|f (x) − f (y)| + |g (y) − g (x)|]|g (x)|≡ M [|f (x) − f (y)| + |g (y) − g (x)|]where M is defined by2M ≡2(1 + 2 |f (x)| + 2 |g (x)|)|g (x)|Now let δ 2 be such that if |x − y| < δ 2 , then|f (x) − f (y)| < ε 2 M −1and let δ 3 be such that if |x − y| < δ 3 , then|g (y) − g (x)| < ε 2 M −1 .Then if 0 < δ ≤ min (δ 0 , δ 1 , δ 2 , δ 3 ) , and |x − y| < δ, everything holds andf (x)∣g (x) − f (y)g (y) ∣ ≤ M [|f (x) − f (y)| + |g (y) − g (x)|][ ε< M2 M −1 + ε 2 M −1] = ε.This completes the proof of the second part of 2.)Note that in these proofs no effort is made to find some sort of “best” δ. The problemis one which has a yes or a no answer. Either it is or it is not continuous.Now consider 3.). If f is continuous at x, f (x) ∈ D (g) , and g is continuous atf (x) ,then g ◦ f is continuous at x. Let ε > 0 be given. Then there exists η > 0 suchthat if |y − f (x)| < η and y ∈ D (g) , it follows that |g (y) − g (f (x))| < ε. Fromcontinuity of f at x, there exists δ > 0 such that if |x − z| < δ and z∈ D (f) , then|f (z) − f (x)| < η. Then if |x − z| < δ and z ∈ D (g ◦ f) ⊆ D (f) , all the above hold andso|g (f (z)) − g (f (x))| < ε.This proves part 3.)To verify part 4.), let ε > 0 be given and let δ = ε. Then if ||x − y|| < δ, the triangleinequality implies|f (x) − f (y)| = |||x|| − ||y|||≤ ||x − y|| < δ = ε.This proves part 4.)Next consider 5.) Suppose first f is continuous. Let U be open and let x ∈ f −1 (U) .This means f (x) ∈ U. Since U is open, there exists ε > 0 such that B (f (x) , ε) ⊆ U. Bycontinuity, there exists δ > 0 such that if y ∈ B (x, δ) , then f (y) ∈ B (f (x) , ε) and sothis shows B (x,δ) ⊆ f −1 (U) which implies f −1 (U) is open since x is an arbitrary point
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4 CONTENTS5 Continuous Functions 85
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6 CONTENTS11.5 Integration Of Diffe
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8 CONTENTS
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10 INTRODUCTION
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12 SOME FUNDAMENTAL CONCEPTSthere i
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14 SOME FUNDAMENTAL CONCEPTSthat is
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16 SOME FUNDAMENTAL CONCEPTSFollow
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18 SOME FUNDAMENTAL CONCEPTSDefinit
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20 SOME FUNDAMENTAL CONCEPTSTheorem
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22 SOME FUNDAMENTAL CONCEPTSNext ob
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24 BASIC LINEAR ALGEBRA6 · (2, 6)(
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26 BASIC LINEAR ALGEBRA3.2 Subspace
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28 BASIC LINEAR ALGEBRATheorem 3.2.
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30 BASIC LINEAR ALGEBRAProof: This
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32 BASIC LINEAR ALGEBRAwhich showsL
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34 BASIC LINEAR ALGEBRAsuch thatl i
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Page 74 and 75: 74 SEQUENCESTheorem 4.1.8 Let {x n
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Page 88 and 89: 88 CONTINUOUS FUNCTIONSof f −1 (U
Page 90 and 91: 90 CONTINUOUS FUNCTIONSProof: First
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Page 102 and 103: 102 CONTINUOUS FUNCTIONSTherefore,
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Page 112 and 113: 112 CONTINUOUS FUNCTIONS5.9 Ascoli
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Page 118 and 119: 118 CONTINUOUS FUNCTIONS28. Let X b
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Page 122 and 123: 122 THE DERIVATIVEor equivalently,|
Page 124 and 125: 124 THE DERIVATIVEIt follows≡limt
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Page 132 and 133: 132 THE DERIVATIVE6.7.1 Some Standa
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136 THE DERIVATIVEAs implied above,
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138 THE DERIVATIVEwhich requires ea
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140 THE DERIVATIVETheorem 6.10.3 (i
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142 THE DERIVATIVEThen there exist
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144 THE DERIVATIVENow the idea is t
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146 THE DERIVATIVETheorem 6.11.5 If
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148 THE DERIVATIVEare linearly inde
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150 THE DERIVATIVEShow {x 1 , · ·
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152 THE DERIVATIVE25. Let (x, y) be
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154 MEASURES AND MEASURABLE FUNCTIO
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182 THE ABSTRACT LEBESGUE INTEGRALL
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184 THE ABSTRACT LEBESGUE INTEGRALa
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186 THE ABSTRACT LEBESGUE INTEGRALD
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192 THE ABSTRACT LEBESGUE INTEGRALP
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194 THE ABSTRACT LEBESGUE INTEGRALP
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196 THE ABSTRACT LEBESGUE INTEGRALF
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198 THE ABSTRACT LEBESGUE INTEGRALT
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202 THE ABSTRACT LEBESGUE INTEGRALC
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204 THE ABSTRACT LEBESGUE INTEGRALN
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206 THE ABSTRACT LEBESGUE INTEGRALc
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208 THE LEBESGUE INTEGRAL FOR FUNCT
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258 BROUWER DEGREEDefinition 10.1.1
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260 BROUWER DEGREEIt is obvious tha
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262 BROUWER DEGREETherefore, for an
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264 BROUWER DEGREE= −A.Therefore,
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266 BROUWER DEGREEProof: From Lemma
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268 BROUWER DEGREEfor all t ∈ [0,
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270 BROUWER DEGREELemma 10.3.2 Let
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272 BROUWER DEGREELemma 10.4.2 Let
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274 BROUWER DEGREEwhich is clearly
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276 BROUWER DEGREEThe product formu
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278 BROUWER DEGREEProof: Let B (y,3
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280 BROUWER DEGREEand y /∈ C. Sin
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282 BROUWER DEGREEwhich is the same
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284 BROUWER DEGREEFor y ∈ S, let
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286 BROUWER DEGREELetdist ( )y, WjC
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288 BROUWER DEGREE∞∑∫∞∑
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290 BROUWER DEGREE11. Establish the
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292 BROUWER DEGREE19. Using Problem
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294 INTEGRATION OF DIFFERENTIAL FOR
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326 THE LAPLACE AND POISSON EQUATIO
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350 THE JORDAN CURVE THEOREMNow the
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352 THE JORDAN CURVE THEOREMA solid
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354 THE JORDAN CURVE THEOREMLemma 1
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356 THE JORDAN CURVE THEOREMwhere R
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358 THE JORDAN CURVE THEOREMa QHKbN
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360 THE JORDAN CURVE THEOREM
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362 LINE INTEGRALSLet x t ≡ tx 1
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364 LINE INTEGRALSActually, people
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366 LINE INTEGRALSand if t is in th
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368 LINE INTEGRALSProof: The functi
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370 LINE INTEGRALSProof of the clai
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372 LINE INTEGRALS(γ s − 2h + t
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374 LINE INTEGRALS≤≤n∑∫ tj|
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376 LINE INTEGRALSand so|F (γ (t))
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378 LINE INTEGRALSLet m be the last
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380 LINE INTEGRALSLemma 14.3.5 In t
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382 LINE INTEGRALSLemma 14.3.6 Let
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384 LINE INTEGRALSwheref (x, y) ≡
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386 LINE INTEGRALSwhere B δ is the
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388 LINE INTEGRALSto Green’s theo
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390 LINE INTEGRALSDefinition 14.4.2
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392 LINE INTEGRALS14.5 Interpretati
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394 LINE INTEGRALS✻a × b✣θc
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396 LINE INTEGRALSSuppose then that
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398 LINE INTEGRALSProposition 14.6.
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400 LINE INTEGRALSCombining these l
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402 LINE INTEGRALSIn this case the
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404 LINE INTEGRALSThen as in the pr
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406 LINE INTEGRALSConsider only h
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408 LINE INTEGRALSwhere |g (w)| <
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410 LINE INTEGRALS6. In the situati
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412 LINE INTEGRALS18. Using Problem
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414 LINE INTEGRALS24. Let f, g be a
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416 LINE INTEGRALSShow this equals
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418 LINE INTEGRALS
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420 HAUSDORFF MEASURES AND AREA FOR
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460 BIBLIOGRAPHY[21] Gurtin M. An i
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462 INDEXderivative, 307exact, 322i
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464 INDEXvectors, 25Vitaliconvergen
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Multivariable Advanced Calculus

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