Functions of Several Variables (Rudin)

Functions of Several Variables (Rudin) Definition: Let X and Y be finite-dimensional real vector spaces. Then L(X, Y ) denotes the set of all linear transformations from X to Y and L(X) denotes the set of all linear transformations from X to X. Given A L(R n, R m ), we set A = sup x = Ax. Remar: Any x R n can be expressed as x = x x 0 where x 0 =. Hence Ax = A( x x 0 ) = Ax 0 x A x. Theorem: (a) If A L(R n, R m ) then A < and A is uniformly continuous. (b) For A, B L(R n, R m ), A + B A + B and A = A. (c) For A L(R n, R m ) and B L(R m, R ), BA B A. Proof: (a) For each i n write Ae i = a i. For x = we have hence Ax = x Ae + + x n Ae n x a + x n a n a + + a n, A a + + a n. Since A(x y) A x y, A is uniformly continuous. (b) For x =, (A + B)x = Ax + Bx Ax + Bx A + B, therefore A + B A + B. It is clear that A = A. (c) For x =, BAx B Ax BA, therefore BA B A. Theorem: The distance function d(a, B) = A B is a metric on L(R n, R m ). Proof: We have A A = 0, A B = B A, A B implies (A B)x 0 for some x with x = implies A B 0, and A C = A B + B C A B + B C. Theorem: (a) A sequence in R n converges if and only if it is Cauchy. (b) A sequence in L(R n, R m ) converges is and only if it is Cauchy.

Proof: Convergent sequences are Cauchy. Conversely, suppose (x ) is Cauchy in R n. Let π i : R n R be projection onto the i th coordinate. For > j we have π i (x j ) π i (x ) x j x 0 as j, therefore (π i (x )) is Cauchy for each i and converges to some a i R. Therefore a x. Now suppose (A ) is Cauchy in L(R n, R m ). For any x R n, the sequence (A x) is Cauchy: for n > m we have a n. A m x A n x = (A m A n )x A m A n x 0 as m. Hence (A x) is Cauchy. Define A : R n R n via Ax = lim Ax. Since each A is linear, A is linear. Since and A A (A A )e + + (A A )e n (A A )e i = Ae i A e i 0 as, A A 0 as, hence A A. Theorem: Let Ω = {A L(R n ) : A exists }. (a) A L(R n ) and I A < implies A Ω. (b) Ω is open in L(R n ) and A A is continuous on Ω. Proof: Suppose I A <. We will prove that A exists and that A. Set B A I = i=0 (I A)i. For > j we have B B j = (I A) i (I A) i 0. i=j+ i=j+ Since i=0 I A i converges, (B ) is Cauchy, hence converges to some B. We have AB = (I (I A))B = I (I A) + I

and AB AB A A B 0 as 0, therefore AB I and AB AB as 0. Hence AB = I and A = B. Moreover B (I A) i i=0 I A. The triangle inequality satisfied by implies that B = lim B I A. Now let A Ω be given. Then I B = I A + A B I A + A B, so when B A < I A we have I B < and B Ω. Therefore Ω is open. Given (A ) in Ω with A I we have I A for all sufficiently large, which implies A Thereore A I = A (I A ) I A I A I A. I. Now suppose (A ) Ω and A A Ω. Then A A I = A (A A) A A A 0, hence A A I, hence A A I, hence A hence A A = (A A I)A A A I A 0, A. Therefore the mapping A A is continuous. Differentiation: Let E R n be an open set, let x E and let f : E R m be given. Given A L(R n, R m ) we say that f = A if and only if there exists V : R n R m such that f(x + h) = f + Ah + h V (h) 3

where V (h) 0 as h 0. The operator A is unique: If where as h 0 then f(x + h) = f + Bh + h W (h) W (h) 0 (A B)h h V (h) + h W (h). Setting h n = t n e i where t n > 0 and t n 0 as n, we obtain t n (A B)e i t n V (t n e i ) + t n W (t n e i ), (A B)e i V (t n e i ) + W (t n e i ) 0 as n, hence A B annihilates e,..., e n, hence A = B. Theorem: Let E be open in R a and let F be open in R b. If f : E R b satisfies f = A and g : F R c satisfies g (f) = B and f(e) F, then (g f) = g (f)f. Proof: Write and Then We have f(x + h) = f + Ah + h V (h) g(f + ) = g(f) + B + W (). g(f(x + h)) = g(f + Ah + h V (h)) = g(f) + BAh + B h V (h) + Ah + h V (h) W (Ah + h V (h)). so we can write Ah + h V (h) h ( A + V (h) ), Ah + h V (h) = h θ(h) where 0 θ(h) A + V (h). This yields g(f(x + h)) = g(f) + BAh + h (BV (h) + θ(h)w (Ah + h V (h))). 4

When h 0 we have V (h) 0, Ah + h V (h) 0, W (Ah + h V (h)) 0, BV (h) 0, and θ(h) bounded. Therefore BV (h)+θ(h)w (Ah+ h V (h)) 0. This implies g (f) = BA = g (ff. Matrix representation of f : Let E be open in R n and assume f : E R m satisfy f = A for some x E. Then f(x + te j ) = f + Ate j + t V (te j ), t (f(x + te j) f) = Ae j + t t V (te j). Letting t 0 we obtain f Ae j =.. f m Hence the matrix ( representation ) of A with respect to the standard bases for R n and R m f is i. Theorem: Let E R n be open and let f : E R n be given. Write f(x,..., x n ) = (f (x,..., x n ),..., f m (x,..., x m )). Then f exists and is continuous on E if and only if each function f i exists and is continuous on E. Proof: Assume f exists and is continuous on E. Then f f.. = f e f m f m j f (y)e j f f (y), (u) hence for each i we have f i f i (y) f f (y). Since f f (y) as x y, f i f i as x y. : E R Conversely, assume each f i is continuous on E. Fix i. We will prove that f i exists and is continuous on E. Fix x E and choose any y B ɛ E. 5

Then there is a polygonal path y 0, y,..., y n in B ɛ from x to y, where y 0 = x, y n = y, and for j n, y j = y j + h j e j. By the Mean Value Theorem, where 0 c ij h j. Hence f i (y j ) f i (y j ) = f i (y j + c ij e j )h j f i (y) f i = n j= f i (y j + c ij e j )h j. Therefore f(y) f = By continuity of each f i ( ) fi (y j + c ij e j ) (y x). we can write f i (y j + c ij e j ) = f i + m ij (y x) where m ij (y x) 0 as y x. Setting h = h e + + h n e n and M(h) = (m ij (h)) we obtain ( ) fi f(x + h) = f + h + M(h)h. Writing h = h u h where u h = we have M(h)h = h M(h)u h with M(h)u h M(h) M(h)e + + M(h)e n 0 as ( h 0. ) This implies that f exists and has standard basis representation f i. The mapping x f is continuous on E because the partial derivatives are continuous on E. Contraction Principle: Let C be a compact subset of R n and let f : E E satisfy f f(y) c x y for some fixed c <. Then there is a unique x E such that f = x. Proof: Fix x 0 C. We have f (n+) (x 0 ) f (n) (x 0 ) c f (n) (x 0 ) f (n ) (x 0 ) c n f(x 0 ) x 0, 6

hence for n > m we have f (n) (x 0 ) f (m) (x 0 ) n =m f (+) (x 0 ) f () (x 0 ) n =m c f(x 0 ) x 0. Since =0 c converges, the sequence (f (n) (x 0 )) is Cauchy, hence converges to some x C. By continuity of f, the sequence (f (n+) (x 0 )) converges to f. But the limit of this sequence is also x, therefore f = x. The number x is unique: suppose f = x and f(y) = y. Then x y = f f(y) c x y and since c <, x y = 0. Inverse Function Theorem: Suppose f : E R m is continuously differentiable on E and that f (a) is invertible. Then a U E for some open set U, f is injective on U, f(u) is open, and f : f(u) U is continuously differentiable. Proof: Assume first that f(0) = 0, f (0) = I, and that f is continuously differentiable on E. Then we now that I f < implies f is invertible. Choose δ > 0 so that x B δ (0) implies I f <. A function with derivative I f on B δ is φ c = x f+c. This implies that φ c (y) φ c y x for all x, y B δ: set γ(t) = φ c (( t)x+ty), 0 t. Then for 0 < t <, γ (t) = φ c(γ(t))γ (t) = (I f (γ(t))(y x) I f (γ(t) y x y x. Hence φ c (y) φ c = (φ c (y) φ c ) γ() (φ c (y) φ c ) γ(0) = (φ c (y) φ c ) γ (t ) φ c (y) φ c y x for some t (0, ). So φ c is a potential contraction. We will prove that f(b δ (0)) is open. Let f(x 0 ) f(b δ (0)) be given. Suppose B r [x 0 ] B δ (0). Let c B r (f(x 0)) be given. We will show that φ c is a contraction of B r [x 0 ]. This yields x B r [x 0 ] B δ (0) such that φ c (x ) = x, which implies c = f(x ) f(b δ (0)). Hence B r (f(x 0)) f(b δ (0)). For x B r [x 0 ] we have φ c x 0 φ c φ c (x 0 ) + φ c (x 0 ) x 0 x x 0 + c f(x 0 ) r +r = r, 7

hence φ c B r [x 0 ]. Now suppose that f(x ) = f(x ) for x, x B δ (0). Let the common value be c. Then φ c (x ) = x and φ c (x ) = x, hence x = x. Therefore f maps B δ (0) bijectively onto f(b δ (0)). Let g = f on f(b δ (0)). We first prove that g is continuous. Write y = f and y + = f(x + h). Then φ y = x and φ y (x + h) = x + h, therefore h = φ y (x + h) φ y h, = h (h ) h h h, g(y + ) g(y) = h, hence g is continuous. Now write f(x + h) = f + Ah + h V (h) where V (h) 0 as h 0 and g(y + ) = g(y) + R(). Applying f, y + = f(g(y)) + AR() + R() V (R()), Hence we can write = AR() + R() V (R()), R() = A A R() V (R()). where g(y + ) = g(y) + A + W () W () = R() A V (R()). We showed above that R(), hence R(). By continuity of g, R() 0 as 0. Hence W () 0 as 0. We have shown that g (y) = f (g(y)), which is continuous by continuity of g, f, and A A. 8

More generally, suppose f(a) = b and f (a) is invertible. Set F = p f q where p = f (a) (f b) and q = x + a. Then F satisfies the hypotheses above. We have f = p F q, f = q F p. The open set on which f is invertible is B δ (a), where x a < δ implies I f (a) f <. 9