Solutions Of Some Exercises - Springer
Solutions of Some ExercisesIn this section the formulas are numbered (S1), (S2), etc, in order to avoid anyconfusion with formulas from the previous sections.1.11. The equality f, x x 2 implies that x f . Corollary 1.3 implies thatF (x) is nonempty. It is clear from the second form of F (x) that F (x) is closedand convex.2. In a strictly convex normed space any nonempty convex set that is contained ina sphere is reduced to a single point.3. Note that11f, y f y f 2 y 2.22Conversely, assume that f satisfies(S1)1y22 1x22 f, y x y E.First choose y λx with λ R in (S1); by varying λ one sees that f, x x 2 .Next choose y in (S1) such that y δ 0; it follows thatf, y 1 2 1δ x 2.22Therefore we obtainδ f sup f, y y Ey δ1 2 1δ x 2.22The conclusion follows by choosing δ x .4. If f F (x) one hasH. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations,DOI 10.1007/978-0-387-70914-7, Springer Science Business Media, LLC 2011371
372Solutions of Some Exercises1y22 1x22 f, y x1x22 1y22 g, x y .and if g F (y) one hasAdding these inequalities leads to f g, x y 0. On the other hand, notethatf g, x y x 2 y 2 f, y g, x x2 y2 2 xy .5. By question 4 we already know that x y . On the other hand, we haveF (x) F (y), x y [ x2 F (x), y ] [ y2 F (y), x ],and both terms in brackets are 0. It follows that x 2 y 2 F (x), y F (y), x , which implies that F (x) F (y) and thus F (x) F (y) by question 2.1.21(a).fE max fi .1 i n1(b). f F (x) iff for every 1 i n one hasfi (sign xi ) x 1anything in the interval [ x 1 , x 1 ]2(a).fE n if xi 0,if xi 0. fi .i 12(b). Given x E consider the setI {1 i n; xi x }.Then f F (x) iff one has(i) fi 0 i / I, (ii) fi xi 0 i I and i I fi x3.fE n . 1/2 fi 2i 1and f F (x) iff one has fi xi i 1, 2, . . . , n. More generally,
Solutions of Some Exercises373fE n 1/p p fi ,i 1where 1/p 1/p 1, and f F (x) iff one has fi xi p 2 xi / x i 1, 2, . . . , n.p 2p1.31. f E 1 (note that f (t α ) 1/(1 α) α 0).12. If there exists such a u we would have 0 (1 u)dt 0 and thus u 1; absurd.1.51. Let P denote the family of all linearly independent subsets of E. It is easy to seethat P (ordered by the usual inclusion) is inductive. Zorn’s lemma implies that Phas a maximal element, denoted by (ei )i I , which is clearly an algebraic basis.Since ei 0 i I , one may assume, by normalization, that ei 1 i I .2. Since E is infinite-dimensional one may assume that N I . There exists a(unique) linear functional on E such that f (ei ) i if i N and f (ei ) 0 ifi I \N.3. Assume that I is countable, i.e., I N. Consider the vector space Fn spanned by(ei )i i n . Fn is closed (see Section 11.1) and, moreover, n 1 Fn E. It followsfrom the Baire category theorem that there exists some n0 such that Int(Fn0 ) .Thus E Fn0 ; absurd.1.71. Let x, y C, so that x lim xn and y lim yn with xn , yn C. Thus tx (1 t)y lim[txn (1 t)yn ] and therefore tx (1 t)y C t [0, 1].Assume x, y Int C, so that there exists some r 0 such that B(x, r) C andB(y, r) C. It follows thattB(x, r) (1 t)B(y, r) C t [0, 1].But tB(x, r) (1 t)B(y, r) B(tx (1 t)y, r) (why?).2. Let r 0 be such that B(y, r) C. One hastx (1 t)B(y, r) C t [0, 1],and therefore B(tx (1 t)y, (1 t)r) C. It follows that tx (1 t)y Int C t [0, 1).3. Fix any y0 Int C. Given x C one has x limn [(1 n1 )x n1 y0 ]. But(1 n1 )x n1 y0 Int C and therefore x Int C. This proves that C Int C andhence C Int C.1.81. We already know that
374Solutions of Some Exercisesp(λx) λp(x) λ 0, x Ep(x y) p(x) p(y) x, y E.andIt remains to check that(i) p( x) p(x) x E, which follows from the symmetry of C.(ii) p(x) 0 x 0, which follows from the fact that C is bounded. Moreprecisely, let L 0 be such that x L x C. It is easy to see thatp(x) 1x x E.L 2. C is not bounded. Consider for example the sequence un (t) n/(1 nt) and/01/21 check that un C, while un n. Here p(u) 0 u(t) 2 dtis a normthat is not equivalent to u .1.91. LetP λ (λ1 , λ2 , . . . , λn ) R ; λi 0 i andnn λi 1 ,i 1so that P is a compact subset of Rn and Cn is the image of P under the continuousmap λ ni 1 λi xi .2. Apply Hahn–Banach, second geometric form, to Cn and {0}. Normalize the linearfunctional associated to the hyperplane that separates Cn and {0}.4. Apply the above construction to C A B.1.10(A) (B) is obvious.(B) (A). Let G be the vector space spanned by the xi ’s (i I ). Given x Gwrite x i J βi xi and set g(x) i J βi αi . Assumption (B) implies thatthis definition makes sense and that g(x) M x x G. Next, extend g toall of E using Corollary 1.2.1.11(A) (B) is again obvious.(B) (A). Assume first that the fi ’s are linearly independent (1 i n). Setα (α1 , α2 , . . . , αn ) Rn . Consider the map ϕ : E Rn defined byϕ(x) ( f1 , x , . . . , fn , x ) .Let C {x E; x M ε}. One has to show that α ϕ(C). Suppose,by contradiction, that α / ϕ(C) and separate ϕ(C) and {α} (see Exercise 1.9).Hence, there exists some β (β1 , β2 , . . . , βn ) Rn , β 0, such that) * β · ϕ(x) β · α x C, i.e.,βi fi , x βi αi x C.
Solutions of Some Exercises375 Itβi fi βi αi . Using asumption (B) one finds that follows that (M ε)βi fi 0. Since the fi ’s are linearly independent one concludes that β 0;absurd.In the general case, apply the above result to a maximal linearly independentsubset of (fi )1 i n .1.151. It is clear that C C and that C is closed. Conversely, assume that x0 C and x0 / C. One may strictly separate {x0 } and C, so that there exist some f0 E and some α0 R such thatf0 , x α0 f0 , x0 x C.Since 0 C it follows that α0 0; letting f (1/α0 )f0 , one hasf, x 1 f, x0 x C.Thus f C and we are led to a contradiction, since x0 C .2. If C is a linear subspace thenC {f E ; f, x 0 x C} C .1.18(a)(b) bif f a, if f a. f log f f if f 0,ϕ (f ) 0if f 0, if f 0.ϕ (f ) (d)ϕ (f ) f .ϕ (f ) 0.(e)ϕ (f ) (f)ϕ (f ) (1 f 2 )1/2 .(g)ϕ (f ) (h)1 ϕ (f ) f pp(i)ϕ (f ) (c) 1 log f 1 22f 0 if f 0,if f 0.if f 1,if f 1.with11 1.p pif 0 f 1,otherwise.
376Solutions of Some Exercises1 p p fif f 0,0if f 0.(j)ϕ (f ) (k)ϕ (f ) (l)ϕ (f ) f p1 f pif f 0,if f 0.1 f p .p 1.20 The conjugate functions are defined on p with1p 1p 1 by 12if k 1 k fk , otherwise. k/(k 1)k/(k 1) ,if k 2 ak fk k 2 ak fk ϕ (f ) otherwise,ϕ (f ) (a)(b)1412k 1 k fk (k 1).k k/(k 1)0if f 1,ϕ (f ) otherwise.with ak (c)1.212. ϕ IA , where A {[f1 , f2 ]; f1 0, f2 0, and 4f1 f2 1}.3. One hasinf {ϕ(x) ψ(x)} 0x Eandϕ ID ,where D {[f1 , f2 ]; f2 0}.It follows thatϕ ( f ) ψ (f ) f E ,and thussup { ϕ ( f ) ψ (f )} .f E 4. The assumptions of Theorem 1.12 are not satisfied: there is no element x0 Esuch that ϕ(x0 ) , ψ(x0 ) , and ϕ is continuous at x0 .1.221. Write thatx a x y y a .Taking inf a A leads to ϕ(x) x y ϕ(y). Then exchange x and y.
Solutions of Some Exercises3772. Let x, y E and t [0, 1] be fixed. Given ε 0 there exist some a A andsome b A such thatx a ϕ(x) εandy b ϕ(y) ε.Thereforetx (1 t)y [ta (1 t)b] tϕ(x) (1 t)ϕ(y) ε.But ta (1 t)b A, so thatϕ(tx (1 t)y) tϕ(x) (1 t)ϕ(y) ε ε 0.3. Since A is closed, one has A {x E; ϕ(x) 0}, and therefore A is convex ifϕ is convex.4. One hasϕ (f ) sup { f, x inf x a }x Ea A sup sup{ f, x x a }x E a A sup sup { f, x x a }a A x E (IA ) (f ) IBE (f ).1.231. Let f D(ϕ ) D(ψ ). For every x, y E one hasf, x y ϕ(x y) ϕ (f ),f, y ψ(y) ψ (f ).Adding these inequalities leads to(ϕ ψ)(x) f, x ϕ (f ) ψ(f ).In particular, (ϕ ψ)(x) . Also, we have(ϕ ψ) (f ) sup { f, x inf [ϕ(x y) ψ(y)]}x Ey E sup sup { f, x ϕ(x y) ψ(y)}x E y E sup sup { f, x ϕ(x y) ψ(y)}y E x E ϕ (f ) ψ (f ).2. One has to check that f, g E and x E,f, x ϕ(x) ψ(x) ϕ (f g) ψ (g).
378Solutions of Some ExercisesThis becomes obvious by writingf, x f g, x g, x .3. Given f E , one has to prove that(S1)sup { f, x ϕ(x) ψ(x)} inf {ϕ (f g) ψ (g)}.g Ex ENote thatsup { f, x ϕ(x) ψ(x)} inf {ϕ̃(x) ψ(x)}x Ex Ewith ϕ̃(x) ϕ(x) f, x . Applying Theorem 1.12 to the functions ϕ̃ and ψleads toinf {ϕ̃(x) ψ(x)} sup { ϕ̃ ( g) ψ (g)},x Eg E which corresponds precisely to (S1).4. Clearly one has(ϕ ψ ) (x) sup { f, x inf [ϕ (f g) ψ (g)]}f E g E sup sup { f, x ϕ (f g) ψ (g)}f E g E sup sup { f, x ϕ (f g) ψ (g)}g E f E ϕ (x) ψ (x).1.241. One knows (Proposition 1.10) that there exist some f E and a constant Csuch that ϕ(y) f, y C y E. Choosing n f , one has ϕn (x) n x C .2. The function ϕn is the inf-convolution of two convex functions; thus ϕn is convex(see question 7 in Exercise 1.23). In order to prove that ϕn (x1 ) ϕn (x2 ) n x1 x2 , use the same argument as in question 1 of Exercise 1.22.3. (ϕn ) InBE ϕ (by question 1 of Exercise 1.23).5. By question 1 we have ϕ(y) f y C y E, which leads ton x yn fyn C ϕ(x) 1/n.It follows that yn remains bounded as n , and therefore limn x yn 0. On the other hand, we have ϕn (x) ϕ(yn ) 1/n, and sinceϕ is l.s.c. we conclude that lim inf n ϕn (x) ϕ(x).6. Suppose, by contradiction, that there exists a constant C such that ϕn (x) Calong a subsequence still denoted by ϕn (x). Choosing yn as in question 5 we see
Solutions of Some Exercises379that yn x. Moreover, ϕ(yn ) C 1/n and thus ϕ(x) lim inf n ϕ(yn ) C; absurd.1.254. For each fixed t 0 the function(1 'x ty 2 x 22tis convex. Thus the function y [x, y] is convex as a limit of convex functions.On the other hand, G(x, y) supt 0 { 2t1 [ x ty 2 x 2 ]} is l.s.c. as asupremum of continuous functions.5. One already knows (see question 3 of Exercise 1.1) thaty 1x ty22 1x22 f, tyand therefore[x, y] f, y x, y E, f F (x).On the other hand, one hasϕ (f ) 1f22 f, x 1x22andψ (f ) 0 if f, y α 0,if f, y α 0.It is easy to check that infz E {ϕ(z) ψ(z)} 0. It follows from Theorem 1.12that there exists some f0 E such that ϕ (f0 ) ψ ( f0 ) 0, i.e., f0 , y αand 21 f0 2 f0 , x 21 x 2 0. Consequently, we have f0 x andf0 , x x 2 , i.e., f0 F (x). xi p 2 xi yi6. (a) 1 p , [x, y] .p 2' x p( (b) p 1, [x, y] x 1(signx)y y .iiixi 0xi 0(c) p , [x, y] maxi I {xi yi }, where I {1 i n; xi x }.1.27 Let T : E F be a continuous linear extension of T . It is easy to check thatE N (T ) G and N (T ) G {0}, so that N (T ) is a complement of G; absurd.2.1 Without loss of generality we may assume that x0 0.1. Let X {x E; x ρ} with ρ 0 small enough that X D(ϕ). The setsFn are closed and n 1 Fn X. By the Baire category theorem there is somen0 such that Int(Fn0 ) . Let x1 E and ρ1 0 be such that B(x1 , ρ1 ) Fn0 .Given any x E with x ρ1 /2 write x 21 (x1 2x) 21 ( x1 ) to concludethat ϕ(x) 21 n0 21 ϕ( x1 ).
380Solutions of Some Exercises2. There exist some ξ E and some constant t [0, 1] such that ξx2 tx1 (1 t)ξ . It follows that R andϕ(x2 ) tϕ(x1 ) (1 t)Mand consequently ϕ(x2 ) ϕ(x1 ) (1 t)[M ϕ(x1 )]. But x2 x1 (1 t)(ξ x1 ) and thus x2 x1 (1 t)(R r). Hence we haveϕ(x2 ) ϕ(x1 ) x2 x1[M ϕ(x1 )].R rOn the other hand, if x2 0 one obtains t x1 (1 t)R and therefore(1 t) 1x1 .x1 R2It follows that ϕ(0) ϕ(x1 ) 21 [M ϕ(x1 )], so that M ϕ(x1 ) 2[M ϕ(0)].2.2 We have p(0) p(xn ) p( xn ) 0, so that p(0) 0. On the other handp(0) 2p(0) by (i). Thus p(0) 0.Next we prove that p(αn xn ) 0. Argue by contradiction and assume that p(αn xn ) 2ε along a subsequence, for some ε 0. Passing to a further subsequence we may assume that αn α for some α R. For simplicity we stilldenote (xn ) and (αn ) the correspondingsequences. The sets Fn are closed and n 1 Fn R. Applying the Baire category theorem,we find some n0 such that Int Fn0 . Hence, there exist some λ0 R and someδ 0 such that p((λ0 t)xk ) ε k n0 , t with t δ. On the other hand,note thatp(αk xk ) p((λ0 αk α)xk ) p((α λ0 )xk ), p(αk xk ) p((λ0 αk α)xk ) p((λ0 α)xk ).Hence we obtain p(αk xk ) 2ε for k large enough. A contradiction.Finally, writep(αn xn ) p(αx) p(αn (xn x)) p(αn x) p(αx) 0andp(αn x) p(αn (x xn )) p(αn xn ),so thatp(αn xn ) p(αx) p(αn (xn x)) p(αn x) p(αx) 0.2.4 By (i) there exists a linear operator T : E F such that a(x, y) T x, y F ,F x, y. The aim is to show that T is a bounded operator, i.e., T (BE )is bounded in F . In view of Corollary 2.5 it suffices to fix y F and to check thatT (BE ), y is bounded. This follows from (ii).
Solutions of Some Exercises3812.61. One has Axn A(x0 x), xn x0 x 0 and thus Axn , x εn Axn C(x)with εn xn x0 and C(x) A(x0 x) (1 x ) (assuming εn 1 n).It follows from Exercise 2.5 that (Axn ) is bounded; absurd.2. Assume that there is a sequence (xn ) in D(A) such that xn x0 and Axn .Choose r 0 such that B(x0 , r) conv D(A). For every x E with x rwritex0 x m m ti yi with ti 0 i,i 1ti 1,andyi D(A) ii 1(of course ti , yi , and m depend on x). We haveAxn Ayi , xn yi 0and thus ti Axn , xn yi ti Ayi , xn yi . It follows thatAxn , xn x0 x m ti Ayi , xn yi ,i 1which leads toAxn , x εn Axn C(x) with εn xn x0 and C(x) mi 1 ti Ayi (1 x0 yi ).3. Let x0 Int D(A). Following the same argument as in question 1, one shows thatthere exist two constants R 0 and C such thatf C x D(A) with x x0 R and f Ax. 2.7 For every x p set Tn x ni 1 αi xi , so that Tn x converges to a limit forevery x p . It follows from Corollary 2.3 that there exists a constant C such that Tn x C x p x p , Choosing x appropriately, one sees that α p and α n.p C.2.8 Method (ii). Let us check that the graph of T is closed. Let (xn ) be a sequence in E such that xn x and T xn f . Passing to the limit in the inequalityT xn T y, xn y 0 leads tof T y, x y 0 y E.Choosing y x tz with t R and z E, one sees that f T x.2.101. If T (M) is closed then M N (T ) T 1 (T (M)) is also closed. Conversely,assume that M N (T ) is closed. Since T is surjective, one has T ((M N (T ))c )
382Solutions of Some Exercises(T (M))c . The open mapping theorem implies that T ((M N (T ))c ) is open andthus T (M) is closed.2. If M is any closed subspace and N is any finite-dimensional space then M Nis closed (see Section 11.1).2.11 By the open mapping theorem there is a constant c 0 such that T (BE ) cBF . Let (en ) denote the canonical basis of 1 , i.e.,en (0, 0, . . . , 0, 1 , 0, . . . ).(n)There exists some un E such that un 1/c and T (un ) en . Given y (y1 , y2 , . . . , yn , . . . ) 1 , set Sy i 1 yi ei . Clearly the series converges and Shas all the required properties.2.12 Without loss of generality we may assume that T is surjective (otherwise,replace E by R(T )). Assume by contradiction that there is a sequence (xn ) in E suchthatxn E 1 and T xn F xn 1/n.By the open mapping theorem there is a constant c 0 such that T (BE ) cBF .Since T xn F 1/n, there exists some yn E such thatT xn T ynandynE 1/nc.Write xn yn zn with zn N (T ), yn E 0 and zn E 1. On the otherhand, xn 1/n; hence zn (1/n) yn (1/n) M yn E , and consequently zn 0. This is impossible, since the normsE and are equivalent on thefinite-dimensional space N (T ).2.13 First, let T O so that T 1 L(F, E) (by Corollary 2.7). Then T U Ofor every U L(E, F ) with U small enough. Indeed, the equation T x U x fmay be written as x T 1 (f U x); it has a unique solution (for every f F )provided T 1 U 1 (by Banach’s fixed-point theorem; see Theorem 5.7).Next, let T . In view of Theorem 2.13, R(T ) is closed and has a complementin F . Let P : F R(T ) be a continuous projection. The operator P T is bijectivefrom E onto R(T ) and hence the above analysis applies. Let U L(E, F ) be suchthat U δ; the operator (P T P U ) : E R(T ) is bijective if δ is smallenough and thus (P T P U ) 1 is well-defined as an element of L(R(T ), E). SetS (P T P U ) 1 P . Clearly S L(F, E) and S(T U ) IE .2.14 E/N(T ) and the canonical surjection π : E E,1. Consider the quotient space Eso that π x E dist(x, N (T )) x E. T induces an injective operator T onMore precisely, write T T π with T L(Ẽ, F ), so that R(T ) R(T ).E.On the other hand, Corollary 2.7 shows that R(T ) is closed iff there is a constantC such that
Solutions of Some Exercises383yE C T yπxE C T πxor equivalently y E, x E.The last inequality readsdist(x, N (T )) C T x x E.2.15 The operator T : E1 E2 F is linear, bounded, and surjective. Moreover,N(T ) N(T1 ) N (T2 ) (since R(T1 ) R(T2 ) {0}). Applying Exercise 2.10 withM E1 {0}, one sees that T (M) R(T1 ) is closed provided M N (T ) is closed.But M N(T ) E1 N (T2 ) is indeed closed.2.16 Let π denote the canonical surjection from E onto E/L (see Section 11.2).Consider the operator T : G E/L defined by T x π x for x G. We havedist(x, N (T )) dist(x, G L) C dist(x, L) C T x x G.It follows (see Exercise 2.14) that R(T ) π(G) is closed. Therefore π 1 [π(G)] G L is closed.2.19 Recall that N (A ) R(A) .1. Let u N (A) and v D(A); we haveA(u tv), u tv C A(u tv)2 t R,which implies that Av, u 0. Thus N (A) R(A) .2. D(A) equipped with the graph norm is a Banach space. R(A) equipped with thenorm of E is a Banach space. The operator A : D(A) R(A) satisfies theassumptions of the open mapping theorem. Hence there is a constant C such that f R(A), v D(A) with Av f and vD(A) C f .In particular, v C f . Given u D(A), the above result applied to f Aushows that there is some v D(A) such that Au Av and v C Au . Sinceu v N (A) R(A) , we haveAu, u Av, u Av, v Avv C Au 2 .2.211. Distinguish two cases:Case (i): f (a) 1. Then N (A) Ra and R(A) N (f ).Case (ii): f (a) 1. Then N (A) {0} and R(A) E.2. A is not closed. Otherwise the closed graph theorem would imply that A isbounded and consequently that f is continuous.3. D(A ) {u E ; u, a 0} and A u u u D(A ).
384Solutions of Some Exercises4. N(A ) {0} and R(A ) {u E ; u, a 0}.5. R(A) {0} and R(A ) Ra (note that N (f ) is dense in E; see Exercise 1.6).It follows that N (A ) R(A) and N (A) R(A ) .Observe that in Case (ii), N (A) R(A ) .6. If A is not closed it may happen that N (A) R(A ) .2.221. Clearly D(A) is dense in E. In order to check that A is closed let (uj ) be asequence in D(A) such that uj u in E and Auj f in E. It follows thatjun un nj andjnun fnj n.Thus nun fn n, so that u D(A) and Au f.2.D(A ) {v (vn ) ; (nvn ) },A v (nvn ) and D(A ) c0 .2.241. We have D(B ) {v G ; T v D(A )} and B A T .2. If D(A) E and T 0, then B is not closed. Indeed, let (un ) be a sequence in/ D(A). Then Bun 0 but u / D(B).D(A) such that un u with u 2.252. By Corollary 2.7, T 1 L(F, E). Since T 1 T IE and T T 1 IF , it followsthat T (T 1 ) IE and (T 1 ) T
Solutions of Some Exercises In this section the formulas are numbered (S1), (S2), etc, in order to avoid any confusion with formulas from the previous sections. 1.1 1. The equality f,x x 2 implies that x f. Corollary 1.3 implies that F(x)is nonempty. It is clear from the second form of F(x)that F(x)is closed and convex. 2.
piano exercises czerny, czerny piano exercises imslp, carl czerny 101 exercises piano pdf, carl czerny 101 exercises piano, czerny hanon piano exercises, czerny piano exercises youtube May 4, 2020 — I always teach Hanon, since it exercises all five fingers equally, and I
Preface This manual contains solutions/answers to all exercises in the text Precalculus: Functions and Graphs, Thirteenth Edition, by Earl W. Swokowski and Jeffery A. Cole.A Student's Solutions Manualis also available; it contains solutions for the odd-numbered exercises in each section and for the Discussion Exercises, as well as solutions for all the exercises in the
Springer-Verlags unter Springer. de. Dem Springer-Verlag, insbesondere Frau Martina Siedler und Frau Rose Marie Doyon danken wir an dies er Stelle fUr die gute Zusammenarbeit und die Unterstiitzung unserer Autoren. Wir wiinschen allen Lesern der 2. Auflage viel Erfolg bei der Examensvorbe reitung und einen guten Start in die berufliche Zukunft.
Free (OCLC) MARC records COUNTER-compliant usage statistics MyCopy – a unique service from Springer: low cost, soft cover editions of English language eBooks eBooks from Springer are an unparalleled resource for scientific research. Springer’s eBook
Available on Springer’s Leading Research Platforms All Adis journal and newsletter content is available on SpringerLink , Springer for R&D, and Springer for Hospitals & Health. These robust, mobile-optimized platforms enable your end users to access over 8 million research documents anytime, anywhere, including thousands
Eerderes Ommen 1929 1929 nee nee Springer, L.A. landsch onbekend Eese, de Steenwijkerland 1402 1600 1800 1900 Springer, L.A. class, landsch Coniferenlaan, lanen Egheria/Ten Cate Losser 1909 1911 nee nee Springer, L.A. gemengd onbekend Eshorst Rijssen en Holten 1914 1915 nee nee Springer, L.A. gemengd vijvers
This book is now out of print. The authors are grateful to Springer-Verlag for the permission to make this postscript file accessible. ISBN 3-540-19468-1 Springer-Verlag Berlin Heidelberg New York ISBN -387-19468-1 Springer-Verlag New York Berlin Heidelberg "c Springer-Verlag Berlin Heidelberg 1988
Printed on acid-free paper Springer is part of Springer Science Business Media (www.springer.com) Springer New York Dordrecht Heidelberg London M. Tamer Özsu David R. Cheriton School of Computer Science University of Waterloo Waterloo Ontario Canada N2L 3G1 ISBN 978-1-4419-8833-1 e-ISBN 978-1-4419-8834-8 DOI 10.1007/978-1-4419-8834-8
