NOTES WEEK 11 DAY 2 SCOT ADAMS

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1 NOTES WEEK 11 DAY 2 SCOT ADAMS In Proposition 0.1 below, Bp0, δq is just the open interval p δ, δq and Bp0, x q is just the closed interval r x, x s. PROPOSITION 0.1. Let g : R R and let δ ą 0. Assume that Bp0, δq Ď domrg 1 s. Let x P Bp0, δq. Then there exists t P Bp0, x q such that rgpxqs rgp0qs rg 1 ptqsx. Proof. Unassigned exercise. (Hint: Use the Mean Value Theorem.) NOTE: We need to use B instead of B because, in Proposition 0.1, we allow the possibilty that x 0. NOTE: If we replace there exists t P Bp0, x q by the stronger there exists t P rx0, xys, then Proposition 0.1 would remain true and would be a slightly better result. However, we choose the formulation above, because it s easily stated and generalizes to higher dimensions, see Theorem 0.5 below. DEFINITION 0.2. For all m P N and for all i P t0,..., mu, we define λ m i : R m Ñ R m by λ m i pxq px 1,..., x i, 0,..., 0q, and we define ρ m i : R m Ñ R m by ρ m i pxq p0,..., 0, x m i`1,..., x m q. P R m, [ λ m mpxq ρ m mpxq x s and r λ m 0 pxq ρ m 0 pxq 0 m s. DEFINITION 0.3. For all m P N, for all i P t0,..., mu, for any set S, for all f : R m S, we define L i f : f λ m i : R m S and R i f : f ρ m i : R m S. Note: For all x P R m, both pl m fqpxq pr m fqpxq fpxq and pl 0 fqpxq pr 0 fqpxq fp0 m q. It follows both that L m f R m f f and that L 0 f R 0 f C fp0mq R m C f 0 m P CpR m, R n q. DEFINITION 0.4. For all m P N, for all i P t1,..., mu, for any vector space S, for all f : R m S, we define L i f : pl ifq pl i 1 fq and R i f : pr ifq pr i 1 fq. Date: April 6, 2017 Printout date: April 7,

2 2 SCOT ADAMS Note: For all x P R n, we have rfpxqs rfp0 m qs rpl mfqpxqs ` ` rpl 1 fqpxqs. Throughout this lecture, we will use to denote the L 8 norm on all Euclidean spaces. So, for all m P N, for all x P R m, we have x maxt x 1,..., x m u; consequently, for all i P t1,..., mu, we have x i ď x. For all m P N, for all x P R m, for all r ą 0, we define the open L 8 ball Bpx, rq : B px, rq ty P R m s.t. y x ă ru and the closed L 8 ball Bpx, rq : B px, rq ty P R m s.t. y x ď ru. We now get our first Multi-Variable Mean Value Theorem, which we ll call MVMVT1: THEOREM 0.5. Let m P N, f : R m R, i P t1,..., mu, δ ą 0. Assume that Bp0 m, δq Ď domrb i fs. Let x P Bp0 m, δq. Then there exists c P Bp0 m, x q such that pl i fqpxq rpb ifqpcqsx i. Proof. Define α : R Ñ R m by αptq px 1,..., x i 1, t, 0,..., 0q. Since x P Bp0 m, δq, it follows that x ă δ. Then, for all j P t1,..., mu, we have x i ď x ă δ. P Bp0, δq, αptq maxt x 1,..., x i 1, t, 0u ă δ. That is, α pbp0, δqq Ď Bp0 m, δq. Define g : f α : R R. Then, for all t P R, it follows from the definitions of g 1 ptq and pb i fqpαptqq, that g 1 ptq pb i fqpαptqq. Then g 1 pb i fq α. So, since α pbp0, δqq Ď Bp0 m, δq Ď domrb i fs, we get Bp0, δq Ď domrg 1 s. As x i ă δ, we get x i P p δ, δq Bp0, δq. By Proposition 0.1 (with x replaced by x i ), choose t P Bp0, x i q such that rgpx i qs rgp0qs rg 1 ptqsx i. Since t P Bp0, x i q, we get t ď x i. Then t ď x i ď x, so t P Bp0, x q. Then αptq P α pbp0, x qq Ď Bp0 m, x q. Let c : αptq. Then c P Bp0 m, x q, and we wish to prove: pl i fqpxq rpb ifqpcqsx i. We have gpx i q fpαpx i qq fpx 1,..., x i 1, x i, 0,..., 0q pl i fqpxq. Also gp0q fpαpx i qq fpx 1,..., x i 1, 0, 0,..., 0q pl i 1 fqpxq. Then pl i fqpxq rpl ifqpxqs rpl i 1 fqpxqs rgpx i qs rgp0qs rg 1 ptqsx i rpb i fqpαptqqsx i,

3 NOTES WEEK 11 DAY 2 3 DEFINITION 0.6. For all m, n P N, for all k P N 0, we define Om,npkq : OR m,r npkq and O m,npkq : O R m,r npkq. We sometimes omit the commas, and write O mn and Omn. We now get our second Multi-Variable Mean Value Theorem, which we ll call MVMVT2: THEOREM 0.7. Let m P N, f : R m R, i P t1,..., mu, k P N 0. Assume that B i f P Om1pkq. Then L i f P O m1pk ` 1q. Proof. Since B i f P Om1pkq, choose δ ą 0 such that Bp0 m, δq P domrb i fs. Combining Theorem 0.5 with the Axiom of Choice, we choose a function c : Bp0 m, δq Ñ R m such that, for all x P Bp0 m, δq, we have both cpxq P Bp0 m, x q and pl i fqpxq rpb ifqpcpxqqsx i. For all x P Bp0 m, δq, since cpxq P Bp0 m, x q, we see that cpxq ď x. Then c P O mm p1q. Then pb i fq c P rom1pkqs ro mm p1qs Ď Om1pkq. Let P SBFpR, Rq be ordinary multiplication of real numbers, defined by a b ab. Let P P LpR m, Rq be the ith standard projection, defined by P pxq x i. Then P P LpR m, Rq Ď O m1 p1q. For all x P Bp0 m, δq, we have pl i fqpxq rpb ifqpcpxqqsx i rpb i fqpcpxqqsrp pxqs rpb i fqpcpxqqs rp pxqs rppb i fq cqpxqs rp pxqs prpb i fq cs P qpxq. Then L i f rpb ifq cs P. Recall: pb i fq c P Om1pkq. Then L i f P ro m1pkqs ro m1 p1qs Ď Om1pk ` 1q, We now get our third Multi-Variable Mean Value Theorem, which we ll call MVMVT3: THEOREM 0.8. Let m P N, f : R m R, i P t1,..., mu, k P N 0. Assume that B i f P O m1 pkq. Then L i f P O m1pk ` 1q. To prove this, capitalize all the Os appearing in the preceding proof:

4 4 SCOT ADAMS Proof. Since B i f P O m1 pkq, choose δ ą 0 such that Bp0 m, δq P domrb i fs. Combining Theorem 0.5 with the Axiom of Choice, we choose a function c : Bp0 m, δq Ñ R m such that, for all x P Bp0 m, δq, we have both cpxq P Bp0 m, x q and pl i fqpxq rpb ifqpcpxqqsx i. For all x P Bp0 m, δq, since cpxq P Bp0 m, x q, we see that cpxq ď x. Then c P O mm p1q. Then pb i fq c P ro m1 pkqs ro mm p1qs Ď O m1 pkq. Let P SBFpR, Rq be ordinary multiplication of real numbers, defined by a b ab. Let P P LpR m, Rq be the ith standard projection, defined by P pxq x i. Then P P LpR m, Rq Ď O m1 p1q. For all x P Bp0 m, δq, we have pl i fqpxq rpb ifqpcpxqqsx i rpb i fqpcpxqqsrp pxqs rpb i fqpcpxqqs rp pxqs rppb i fq cqpxqs rp pxqs prpb i fq cs P qpxq. Then L i f rpb ifq cs P. Recall: pb i fq c P O m1 pkq. Then L i f P ro m1pkqs ro m1 p1qs Ď O m1 pk ` 1q, REMARK 0.9. Let m, n P N and let g : R m R n and let k P N 0. Define J : t1,..., nu. Then (i) r g P Omnpkq s ô P J, g j P Om1pkq s. (ii) r g P O mn pkq s ô P J, g j P O m1 pkq s, and Proof. Unassigned HW. We can improve Theorem 0.7 and Theorem 0.8, by upgrading the target from R to R n. We call the results MVMVT2 and MVMVT3. WARNING: Because of the spiral path t ÞÑ p cos t, sin t, t q : R Ñ R 3, Theorem 0.5 does not upgrade, even when m 1 and i 1. Let s upgrade MVMVT2 to MVMVT2 : THEOREM Let m, n P N, f : R m R n, i P t1,..., mu, k P N 0. Assume that B i f P Omnpkq. Then L i f P O mnpk ` 1q.

5 NOTES WEEK 11 DAY 2 5 Proof. Let J : t1,..., nu. By (i) of Remark 0.9, we wish to show, for all j P J, that pl i fq j P Om1pk ` 1q. Let j P J be given. We wish to show that pl i fq j P Om1pk ` 1q. Unassigned HW: Show pl i fq j L i pf jq and pb i fq j B i pf j q. Since B i f P Omnpkq, by (i) of Remark 0.9, we have pb i fq j P Om1pkq. Then B i pf j q pb i fq j P Om1pkq. Then, by Theorem 0.7, it follows that L i pf jq P Om1pk`1q. Then pl i fq j L i pf j q P Om1pk`1q, Now, let s upgrade MVMVT3 to MVMVT3 : THEOREM Let m, n P N, f : R m R n, i P t1,..., mu, k P N 0. Assume that B i f P O mn pkq. Then L i f P O mnpk ` 1q. We capitalize Os and change (i) to (ii): Proof. Let J : t1,..., nu. By (ii) of Remark 0.9, we wish to show, for all j P J, that pl i fq j P O m1 pk ` 1q. Let j P J be given. We wish to show that pl i fq j P O m1 pk ` 1q. Unassigned HW: Show pl i fq j L i pf jq and pb i fq j B i pf j q. Since B i f P O mn pkq, by (ii) of Remark 0.9, we have pb i fq j P O m1 pkq. Then B i pf j q pb i fq j P O m1 pkq. Then, by Theorem 0.7, it follows that L i pf jq P O m1 pk ` 1q. Then pl i fq j L i pf j q P O m1 pk ` 1q. DEFINITION For all m, n P N, we define 0 m,n : 0 R m,r n. We sometimes omit the commas, and write 0 mn. Note: 0 mn C 0n R m. Thus, for all x P R m, we have 0 mn pxq 0 n. For functions f and g, we will use the notation g Ě f to mean that g is an extension of f, i.e., that both ( domrgs Ě domrfs ) and Next, MVMVT4: P domrfs, gpxq fpxq q. THEOREM Let m, n P N, f : R m R n, k P N 0. Let I : t1,..., mu Assume, for all i P I, that B i f P Omnpkq. Assume that fp0 m q 0 n. Then f P Omnpk ` 1q. Proof. For all x P R n, rfpxqs rfp0 m qs rpl mfqpxqs` `rpl 1 fqpxqs. So, since fp0 m q 0 n, we conclude that f Ě pl mfq ` ` pl 1 fq. By Theorem 0.10, we see, for all i P I, that L i f P O mnpk`1q. Therefore f Ě pl mfq ` ` pl 1 fq P Omnpk ` 1q, and so f P Omnpk ` 1q.

6 6 SCOT ADAMS LEMMA Let V and W be finite dimensional vector spaces, let f, g : V W, let p P V and let q P W. Assume, for all x P V, that we have gpxq rfpp ` xqs q. Then all of the following hold: P V, pdgqpxq pdfqpp ` xq, (ii) p f is defined near p q ñ p g is defined near 0 V q (iii) p f is continuous at p q ñ p g is continuous 0 V q. and Proof. We leave (ii) and (iii) as unassigned HW and prove only (i). Let x P V be given. We wish to prove that pdgqpxq pdfqpp ` xq, i.e., that D x g D p`x f. It suffices to show: LINS x rgs LINS p`x rfs. For any φ : V W, we have LINS x rφs LINS 0V rφ T x s. Then both LINS x rgs LINS 0V rgx T s and LINS p`x rfs LINS 0V rfp`xs. T We therefore wish to show: LINS 0V rgx T s LINS 0V rfp`xs. T It suffices to show: gx T fp`x. T We wish to show, for all h P V, that gx T phq fp`xphq. T Let h P V be given. We wish to prove: gx T phq fp`xphq. T We have g T x phq rgpx ` hqs rgpxqs rpfpp ` x ` hqq qs rpfpp ` xqq qs rfpp ` x ` hqs rfpp ` xqs f T p`xphq, LEMMA Let m, n P N, let f, g : R m R n, let p P R m and let q P R n. P R m, gpxq rfpp ` xqs q. P R P t1,..., mu, pb i gqpxq pb i fqpp ` xq. Proof. Unassigned HW. THEOREM Let m, n P N, f : R m R n, k P N 0 and p P R m. Let I : t1,..., mu. Assume, for all i P I, that B i f is defined near p and continuous at p. Then p P dlinrfs. Proof. Let g : f T p. For all x P R n, we have gpxq rfpp ` xqs rfppqs. Then gp0 m q 0 n. Also, by (i) of Lemma 0.14, we see, for all x P R n, that pdgqpxq pdfqpp ` xq. Then pdgqp0 m q pdfqppq. Also, we have: [ ( 0 m P domrdgs ) iff ( p P domrdfs ) ]. Equivalently, we have: [ ( 0 m P dlinrgs ) iff ( p P dlinrfs ) ]. We wish to prove: 0 m P dlinrgs. Recall: For all x P R n, we have gpxq rfpp ` xqs rfppqs. Then, by Lemma 0.15, for all x P R n, for all i P I, pb i gqpxq pb i fqpp ` xq. By asumption, for all i P I, B i f is defined near p and continuous at p.

7 NOTES WEEK 11 DAY 2 7 Then, for all i P I, by (ii) and (iii) of Lemma 0.14 (with V replaced by R m, W by R n, f by B i f, g by B i g and q by 0 m ), we see that B i g is defined near 0 m and continuous at 0 m. For all i P I, let w i : pb i gqp0 m q. Define L P LpR m, R n q by Lpxq x 1 w 1 ` ` x m w m. Then Lp0 m q 0 n. Let h : g L. Then hp0 m q 0 n 0 n 0 n. Because L P LpR m, R n q, it follows that dlinrls R m. Then 0 m P R m dlinrls. We therefore have: [ ( 0 m P dlinrgs ) iff ( 0 m P dlinrhs ) ]. It therefore suffices to show that 0 m P dlinrhs. That is, we wish to prove that 0 m P domrdhs. It suffices to show that pdhqp0 m q 0 mn, i.e., that h T 0 m P Omnp1q. Since hp0 m q 0 n, it follows that h T 0 m h. So we wish to prove: h P Omnp1q. Then, by Theorem 0.13, it suffices to P I, B i h P Omn. Let i P I be given. We wish to prove: B i h P Omn. From the definition of L, we see, for all x P R m, that pb i Lqpxq w i. Then, for all x P R m, we have pb i hqpxq rpb i gqpxqs w i. Therefore pb i hqp0 m q w i w i 0 n. It remains to show that B i h is defined near 0 m and continuous at 0 m. Recall that B i g is defined near 0 m and continuous at 0 m, Therefore, by (i) and (ii) of Lemma 0.14 (with V replaced by R m, W by R n, f by B i f, g by B i g, p by 0 m and q by w i ), we conclude that B i h is defined near 0 m and continuous at 0 m,