MAT 257, Handou 6: Ocober 7-2, 20. I. Assignmen. Finish reading Chaper 2 of Spiva, rereading earlier secions as necessary. handou and fill in some missing deails! II. Higher derivaives. Also, read his a. (muli)-linear algebra preinaries. Le L(R n ; R m ) denoe he space of linear maps from R n o R m, and more generally, le L (R n ; R m ) denoe he space of -linear maps from (R n ) R m. Thus, an elemen of L (R n ; R m ) is a funcion Λ(R n ) R m such ha Λ(v,..., v i, av i + bw i,..., v i+,..., v ) = aλ(v,..., v i, v i,..., v i+,..., v ) + bλ(v,..., v i, w i,..., v i+,..., v ) for all vecors v,..., v, w i R n, scalars a, b R, and i {,..., }. We have seen ha here is a naural isomorphism beween he space of linear maps R n L(R n ; R m ) and he space L 2 (R n ; R m ) of bilinear maps R n R n R m. This is an insance of a more general fac: Lemma. For 2, is a naural isomorphism beween he space of linear maps R n L (R n ; R m ) and L (R n ; R m ). Proof. Given a linear map Λ : R n L (R n ; R m ), define Λ L (R n ; R m ) by () Λ(v,..., v ) = Λ(v )(v 2,...,, v ). The righ-hand side denoes he elemen of R m obained as before in he case = 2, ha is: Λ(v ) is by definiion an elemen of L (R n ; R m ) and so we can le i ac on he vecors v 2,..., v o ge an elemen of R m. One can hen verify ha he funcion Λ : (R n ) R m defined in his way is a -linear map R n R m. Conversely, given Λ L (R n ; R m ), we can use () o define Λ : R n L (R n ; R m ) (where now he lef-hand side is given and we are defining he righ-hand side). I is hen rouine o verify ha Λ is indeed a linear map R n L (R n ; R m ). In wha follows, we will abuse noaion somewha by idenifying Λ L (R n ; R m ) wih he space of linear maps R n L (R n ; R m ). So for example, if Λ L (R n ; R n ) and v is a vecor, we will wrie Λ(v ) (insead of Λ(v ), as above, which would be more correc) o denoe he elemen of L (R n ; R m ) defined by (). Some norms on L (R n ; R m ) include he Euclidean norm Λ = Λ(e i,..., e i ) 2 and he operaor norm i,...,i = Λ = sup{λ(v,..., v ) : v i for all i {,..., }}. (You can chec ha hese are norms. exercise!) General consideraions ell us ha here exis consans 0 < c C (depending on ) such ha (2) c Λ Λ C Λ for all Λ L (R n ; R m ). b. definiion of he h derivaive. We now define he h derivaive inducively.
2 Definiion. A funcion f : R n R m is said o be imes differeniable a a poin a R n if here exiss an open neighbourhood U of a such ha he ( )s derivaive D f(x) L (R n ; R m ) exiss a every x U, and if here exiss Λ L (R n ; R m ) such ha D f(a + h) D f(a) Λ(v) = 0. h 0 h When his holds, Λ is said o be he h derivaive of f a a, and is wrien D We have used he Euclidean norm in he definiion of he derivaive, bu as in he case = 2, in view of (2) we could jus as well have used he operaor norm. (Indeed any oher norm on L (R n ; R m ) would be fine as well and would lead o a compleely equivalen definiion.) c. facs abou higher derivaives. The proofs of some of hese facs are very smiler o he = 2 cases, which we have already seen. Proofs of ohers appear a he end of his handou. An informal summary is:. when a funcion f is imes differeniable a a poin a, all h order parial derivaives exis, and we can essenially idenify D f wih he collecion of all h order parial derivaives (made ino a -linear operaor in a paricular way, see formula (6) below.) 2. I can however happen ha all h-order parial derivaives of a funcion f exis a a poin a, bu ha he funcion is no imes differeniable a ha poin. In his siuaion, all bes are off. However, we will almos never encouner his siuaion. And if all h order parial derivaives exis and are coninuous in an open se U, hen f is imes differeniable everywhere in U. Theorem. Assume ha f; R n R m is imes differeniable a a. Then for every i,..., i {,..., n} (3) D f(a)(e i,..., e i ) = D i (D i2 (D i f) )) = D i,...,i In paricular, he derivaive on he righ-hand side (called a h order parial derivaive) exiss. More generally, (4) D f(v,..., v ) = D v (D v2 ( (D v f) )), where he righ-hand side denoes he resul of ieraed direcion differeniaion of f in he direcions v,..., v. Moreover, parial derivaives are independen of he order of differeniaion, so ha if σ is any permuaion of {,..., }, hen (5) D i,,i f(a) = D iσ(),,i σ() Finally, for any vecors v,..., v, where v j = (v i,..., vn j ) (6) D f(a)(v,..., v ) = i,...,i = v i D i,...,i Noe ha boh sides of equaions (3) - (6) are vecors wih m componens. The second equaliy in (3) is rue by our noaion, so he poin is he firs equaliy. Equaion (5) is a special case of he more general fac ha D f(a)(v,..., v ) = D f(a)(v π(),..., v π() ) for all v,..., v. In oher words, D f is a symmeric -linear map (R n ) R m, If h order parial and/or direcional derivaives exis bu f is no imes differeniable a a, hen various pahologies can occur, e.g. parial derivaives depending on he order of differenaion. However, we will rarely encouner such siuaions in his class. And none of hese unpleasan siuaions occur if he h order parial derivaives are coninuous, as follows from he nex heorem.
Theorem 2. Assume ha f : R n R m and ha every parial derivaive of order of every componen of f exiss and is coninuous in an open se U R n. Then f is -imes differeniable a every poin of U. I is worh poining ou he following resul, which we will need laer. (The proof follows easily from (4).) Lemma 2. Suppose ha f : R n R is imes differeniable a every poin in an open se U R n, and coninuously differeniable a a poin a U. For any v R n if we define g() = f(a + v), hen he h derivaive of g exiss a = 0, and (7) ( d d ) g(0) = D f(a)(v,..., v) 3 The lef-hand side of (7) denoes he h derivaive of g, evaluaed a = 0. (I am no sure wha noaion you are used o. Below I will also use he noaion ( d d ) g() =0 for he same hing.) The same resul holds for funcions R n R m, where (7) is hen undersood o hold for each componen. Finally, we also sae he following resul, which expresses D f(v,..., v ) as a 0 i of a cerain linear combinaion of values of f a he verices of a parallelepiped generaed by he vecors v,..., v. This is a generalizaion of problem 3 on homewor 3. Before reading he saemen, hin abou wha you expec i o be, based on he = 2 case. (And afer reading he saemen, which is a lile opaque, hin abou wha i says...) Lemma 3. If f : R n R m is imes differeniable a a, for every v,..., v R n (8) D f(a)(v,..., v ) = 0 ( ) P = σi f(a + σ {0,} σ i v i ) i= d. Some proofs. Proof of (pars of) Theorem. Sep. Firs, we prove (4). We will argue by inducion on. The case = is Exercise 2-29(c) in Spiva. We may assume ha v,..., v are all nonzero, as oherwise boh sides of (4) vanish and i obviously holds. The assumpion ha f is -imes differeniable, and he equivalence of he euclidean norm and he operaor norm, implies ha for any vecor v, D f(a + v ) D f(a) D f(a)(v ) = 0. 0 Noe also ha for any T L (R n, R), and any nonzero vecors v 2,..., v, T (v 2,..., v ) = v 2 v T ( v 2 v 2,..., v v ) v 2 v T using he definiion of he operaor norm T. Applying his o T = D f(a + v ) D f(a) D f(a)(v ), which is an elemen of L (R n ; R m ), we find ha D f(a + v ) D f(a) D f(a)(v ) 0 = v v 0 v D f(a + v )(v 2,..., v ) D f(a)(v 2,... v ) D f(a)(v, v 2,..., v ) = 0, 0
4 We rewrie he las erm, using he lineariy of D f(a), o obain D f(a + v )(v 2,..., v ) D f(a)(v 2,... v ) D f(a)(v, v 2,..., v ) = 0. 0 This says ha he direcional derivaive D v h(a) exiss, for h(x) = D f(x)(v 2,..., v ). Ideniy (4) now follows by invoing he inducion hypohesis. Sep 2. Conclusion (3) is a special case of (4), as noed above. Sep 3. Nex, (5) can be proved by using he = 2 case o swich he order of differeniaion, a pair of (adjacen) derivaives a a ime. We omi he deails. As an exercise, hin abou how o wrie up his argumen correcly. Sep 4. Finally, o prove (6), we can wrie each v j as v j = n i = vi i j e i j, where e ij denoes he sandard basis vecor in he i j direcion. Then muli-lineariy of D f, D f(a)(v,..., v ) = D f(a)( i = i e ii,..., i = v i e i ) = i,...,i = v i D f(a)(e i,..., e i ). Thus he conclusion follows from (3). Nex we give he Proof of Theorem 2. We will give he proof for m =, As in he case of firs derivaives, his implies he general case. The idea is as follows: We have in (6) a formula for D f(a) in erms of parial derivaives of f, a formula ha mus hold if, as we hope o prove, f is imes differeniable. So we will use his formula and he exisence of parial derivaives o define a candidae for D f, and hen verify (using he = case of he resul we are rying o prove, i.e. Theorem 2-8 in Spiva, which we already now) ha our candidae for D f in fac has he properies i is supposed o have. The main difficuly is noaion. Fix a poin a U. By assumpion, he following definiion maes sense, since all h order parial derivaives exis everywhere in U: Λ(v,..., v ) = i,...,i = v i D i,...,i (Compare (6).) We will show ha Λ = D In view of he definiion of he Euclidean norm on L (R n ; R), i suffices o prove ha for every i,..., i {,..., n}, (9) Df(a + h)(e i,...,, e i ) Df(a)(e i,...,, e i ) Λ(h, e i,...,, e i ) = 0. h 0 h (Proving his in deail is an exercise.) So we fix some such i,..., i for he duraion of he proof. By Theorem, we now ha D f(e i,..., e i ) = D i,...,i Le us simplify noaion by wriing D α f(a) o denoe D i,...,i f for mos of he res of his proof. Now since f is assumed o be -imes differeniable in a neighbourhood of a, we can use Lemma o rewrie lef-hand side of (9) = h 0 h D αf(a + h) D α f(a) Λ(h, e,..., e ) Also, our hypoheses implies ha D α f is coninuously differeniable in U, hence Theorem 2-8 in Spiva implies ha D α f is a differeniable funcion, and ha he Jacobian marix (D α f) (a)
(here a row vecor) is given by (D α f) (a) = (D D α f(a),..., D n D α f(a)). In paricular, D(D α f)(a)(h) = (D α f) (a) = D j D α f(a)h j = j= h j D i i j If we scruinize he definiion of Λ, we see ha he righ-hand side is exacly Λ(h, e i,..., e i ). Thus lef-hand side of (9) = h 0 h D αf(a + h) D α f(a) D(D α f)(a)(h) And he above i clearly equals 0, by definiion of wha i means for D α f o be differeniable. Thus we have proved (9), and his complees he proof of he heorem (up o deails lef as exercises for he reader.) Proof of Lemma 2. According o (4), if f is imes differeniable a a, hen j= D f(v,..., v) = D v (D v ( D v f) ), and an easy inducion argumen shows ha he righ-hand side can be rewrien as ( d d ) g(0) for g() = f(a + v). We omi he proof of Lemma 3. The idea is o use inducion on, and o argue in a way similar o he proof of Problem 3 in Homewor 3. The noaion is more complicaed, bu oherwise he proof is raher similar. 5