Calculus on Normed Vector Spaces (Universitext)

Calculus on Normed Vector Spaces (Universitext)

Rodney Coleman

Language: English

Pages: 249

ISBN: 1461438934

Format: PDF / Kindle (mobi) / ePub

This book serves as an introduction to calculus on normed vector spaces at a higher undergraduate or beginning graduate level. The prerequisites include basic calculus and linear algebra, as well as a certain mathematical maturity. All the important topology and functional analysis topics are introduced where necessary.

In its attempt to show how calculus on normed vector spaces extends the basic calculus of functions of several variables, this book is one of the few textbooks to bridge the gap between the available elementary texts and high level texts.  The inclusion of many non-trivial applications of the theory and interesting exercises provides motivation for the reader.

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.x/.y x/ C j j because x is a local minimum. Letting ky xk . .y x// 0; go to 0, we obtain the result. Remark. The above inequality is called Euler’s inequality . If x is a maximum, then x is a minimum of the function f restricted to X and so we obtain the inequality in the opposite direction. Corollary 2.4. Let O be an open subset of a normed vector space E and X an affine subspace of E: X D a C V , where V is a vector subspace of E and a 2 X . We suppose that O \ X ¤ ;. If f is a

centred on x and lying in int X . Given that the norm N is k k1 , B.x; r/ is a closed cube and so a polyhedron. From the lemma f has an N N upper bound on B.x; r/: there exists k such that f .x/ Ä k for all x 2 B.x; r/. Let N h ¤ 0 be such that khk Ä r. Then, for ˛ 2 Œ0; 1, x C ˛h 2 B.x; r/ and f .x C ˛h/ D f ..1 ˛/x C ˛.x C h// Ä .1 ˛/f .x/ C ˛f .x C h/: (7.1) Also,  f .x/ D f à h/ Ä ˛ 1 .x C ˛h/ C .x 1C˛ 1C˛ 1 .f .x C˛h/C˛f .x h//; 1C˛ from which we obtain .1 C ˛/f .x/ Ä f .x C ˛h/ C

dK2 is convex. Proof. We have ..dK2 /0 .x C h/ .dK2 /0 .x//h D 2hx C h D 2hh PK .x C h/; hi 2hx PK .x/; hi PK .x C h/ C PK .x/; hi 2 2hPK .x C h/ PK .x/; hi 2khk2 2kPK .x C h/ PK .x/kkhk D 2khk because PK is 1-Lipschitz. It follows that dK2 is convex. 0; t u If a function is 2-differentiable, then we may use the second differential to determine whether the function is convex or not. This is often easier to use than the (first) differential. Let O be an open subset of a normed vector

/ D .x1 ; : : : ; xr ; 0; : : : ; 0/ for all x 2 U 0 . Proof. For the case where E D Rm and F D Rn we have already proved the result (Propositions 8.2, 8.3 and 8.4). For the general case it is sufficient to notice that E and Rm are isomorphic as are F and Rn . t u Exercise 8.5. Let E and F be normed vector spaces of respective dimensions m and n, O an open subset of E and f a C 1 -mapping from O into F . Suppose that a 2 O with rk f 0 .a/ D m (resp. rk f 0 .a/ D n). Show that there is a

d: The function f is strictly convex on the convex set S D fx 2 Rn W C x D d g. If is the smallest eigenvalue of the matrix A, then f .x/ D 1 t x Ax 2 bt x 1 kxk22 2 kbk2 kxk2 184 8 The Inverse and Implicit Mapping Theorems and so f .x/ approaches 1 when kxk2 approaches 1. As S is closed, f has a minimum xN on S , which is unique, because f is strictly convex. We are minimizing f on Rn , which is open, under the constraints i .x/ n X D cij xj di j D1 for i D 1; : : : ; m. Now, r i

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