:y~&.uW~

~~~ o~ ~ JtJJ.ll :I!JWI s:..,F.J I -~9ihH ~I t.;J~

cJ Jlf.ll JW .J ,~_,1.JI ~~ .J ,_)J_,:ill ~ .J :~~

Directional derivatives for functions of two variables

Partitions of intervals in IRlhnd in lffi.3

~W\ ~ .}-\

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9

(iv) xxx=O 'v'xElR3

(v) (x x y,x) = (x x y,y) = 0 'v'x,y E JR 3

(vi) llx X Yll = llxiiiiYII sine 'v' X, y E lR 3

~) '(vi) a...,at.;L\ l) y) X ~\ lP.:i ~)\)\

. v2 ~ ~.:> r ~ .b:-)i (iii) .{vl'v2,v3,v4} ~~\if 0:!..1..,4~ ~ c~l (iv)

. v2) vl LP.:! d.:) I)\ vtJ ~I (v) "' ~£ ~t 1~1~L:k>-~ v4,v3 'v2 'v1 o~l~ (vi)

~yG if ~ ~\) ~ lli:....J.I ~ _f.-1 ol;. ~I Y ~ C~ij .__,..,o~

10

.~jlp Gts"' 1~1 ~) b1 ~ 0~ / v2 ) v1 ~I 0i c.;i .2

~ ]Rn l) o..i...4ldl (d.:~l ~) o~l if~~~~ 0i ,~ . ~~ .3

.01..1...4k. ~ ~ y 0i o..i...4ldl ~~~ ~ .:>~IJ .~ J ... ( ~ ..,...o~ --l>-i 0i c.;t ~ ~ __,... o~l if ~ ~ L-its"' b1 .4

'1- $ 1-.w\ . L.-=-- ..__s ~ ~~(" 01 ~ . if ~ . ~ .r! u--

0ts"' b1 ~) b1 01..1...4~ v1, v2 E JRn 0i c.;i .5

llvl +v2ll2

= llv1ll2

+llv2ll2

. JR 2 l) v J.i-~ d.:_;k t../' oiJLll o~ 0i ~ :i

v 1 = ( xll ' xl2 ' ... ' xl n )

V2 =(x2Px22,···,x2n)

v n = ( xnl' xn2' . .. 'xnn) "'

o~l 0i c.;i .6

o.:>~l c.-its"' 1~1 k) 1~1 L:k>- ~ / ]Rn >-L,a.AJ\ l)

1 1

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27

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35

2.1 lf._)!l

:

lim [limf ( x, y )] = 1, lim [limf (x, y )] = -1 x-0 y- O y-O x-0

. lim f (x,y) ) r J i.J.>-~ 2 J-_} i~1 (:c,y) - (0, 0)

f(x,y) = (xy)2 +(x-y)2

0

(x,y) =/= (0,0)

(x,y)=(O,O)

lim [lim f ( x, y )] = lim[limf ( x, y )] = 0 x-o v-o y-o x-o

.o) y.- _y yj- lim f(x,y) ~ J (x ,y )-(0,0)

( ) {xsin(1/y) y =/= 0

f x,y = 0 y=O

~ (x,y)--+ (0,0) ~..1.:.>- f (x,y)--+ 0 0i ~t

~~[~~f(x,y)] =/= ~~[~~f(x,y)] ~(2) J-_} ~..lb ~~ Ja>

JL..aJ~\ 2.2

36

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40

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41

(i)

sin(x2 +y2)

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1 (x,y) =(O,O)

(ii) f ( x, Y) ={Yo tan~ x + o x=O

{1 0 < y < x 2

(iii) f ( x, y) = 0 otherwise

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Function8 on Euclidean Space l3

The oscillation o(f,a) of f at a is defined by o(f,a) = lim[M(a,!,o) - m(a,J,~)]. This limit always exists, since HO

M(a,!,o) - m(a,J,o) decreases as o decreases. There are two important facts about o(f,a).

1-10 Theorem. The boundedfunctionfis continuous at a if and only if o(f,a) = 0.

Proof. Let f be continuous at a. For every number € > 0 we can choose a number o > 0 so that if(x) - f(a)i < € for all x E A with lx - al < o; thus M(a,!,o) - m(a,J,o) ~ 2€. Since this is true for every €, we have o(f,a) = 0. The con-verse is similar and is left to the reader. I

I-ll Theorem. Let A C R" be closed. lf f: A --+ R is any bounded function, and € > 0, then I x E A: o(f,x) ;;:::: €) is closed.

Proof. Let B= lxEA: o(f,x)~e} . We wish to show that R" - B is open. If x E R" - B, then either x fl. A or else x E A and o(f,x) < €. In the first case, since A is closed, there is an open rectangle C containing x such that C C R" - A C R" - B. In the second case there is a o > 0 such that M(x,f,o) - m(x,f,o) < €. Let C be an open rectangle containing x such that lx - Yi < o for all y E C. Then if y E C there is a 01 such that lx - z l < ~ for all z satisfying lz - yj < h Thus M(y,J,01) - m(y,f,ol) < €, and consequently o(y,f) < €. Therefore C C R" - B. I

Problem•. 1-23. Iff: A-+ R"' and a E A, show that lim /(:r.) • b if and only if lim f(x) • b.- for i -= 1, ... ,m. :t-+a

:o-o 1-24. Prove that f: A ~ R m is continuous at a if and only if each f' is. 1-25. Prove that a linear transformation T : R" _, Rm is continuous.

H i nt: Use Problem 1-10. 1-26. Let A- f(x,y) E R 2: x >DandO< y < x 2}.

(a) Show that every straight line through (0,0) contains an interval around (0,0) which is in R2 - A.

(b) Define f: R 2 -+ R by f(x) - 0 if X e A and /(x) - 1 if :r. E A. For h E R1 define gA: R-+ R by g11(t) • f(lh) . Show that each gA is continuous at 0, but f is not continuous at (0,0).

3

In Exercises 4.29 through 4.33, we assume that/: S --+ Tis a function from one metric space (S,ds) to another (T,dr).

4.29 Prove thatfis continuous on S if, and only if, f 1 (intB) c int(f-1 (B)) for every subset B ofT.

Proof:(=>) Suppose thatfis continuous on S, and let B be a subset ofT. Since int(B) c B, we have f 1 (intB) c f 1 (B). Note thatf1 (intB) is open since a pull back of an open set under a continuous function is open. Hence, we have

intlj- 1 (intB)] = f 1 (intB) c int(j- 1 (B)). That is,f1 (intB) c int(j-1 (B)) for every subset B ofT.

( ) Suppose that f is continuous on S, and let A be a subset of S. Since flA) c cl(f(A) ), then (A c)f1 (f(A)) c f 1 (cl(f(A)) ). Note thatf1 (cl(f(A))) is closed since a pull back of a closed set under a continuous function is closed. Hence, we have

cl(A) c cl[f-1(cl(f(A)))] =f1(cl(f(A)))

which implies that

flcl(A)) cj[f-1(cl(f(A)))] c cl(f(A)).

() Suppose that f is continuous on S, then it is clear that f is continuous on every compact subset of S.

(

ifx E Q, andg(x) = Oifx E Qc.

Remark: It is the same rusult for min(f1, .. .fm) since max( a, b) + min( a, b) = a + b. ~· 4.21 Let f: S --+ R be continuous on an open set in Rn, assume that p E S, and assume

that flp) > 0. Prove that there is an n -ball B(p; r) such that j(x) > 0 for every x in the ball.

Proof: Since (p E )S is an open set in Rn, there exists a 81 > 0 such that B(p, 8 1) c S. Sinceflp) > 0, givens= ~) > 0, then there exists ann -ball B(p;82 ) such that as X E B(p;82) n S, we have

~) = flp)- & < j(x) < flp) + & = 3.tr:). Let 8 = min(8t,82), then as x E B(p;8), we have

j(x) > ~) > 0.

Remark: The exercise tells us that under the assumption of continuity at p, we roughly havethesamesigninaneighborhoodofp, ifflp) > 0 (orf(p) < o.)

4.22 Letfbe defined and continuous on a closed setS in R. Let

A = {X : X E S and j(x) = 0} . Prove that A is a closed subset of R.

Proof: Since A = J-1 ( {0} ), andfis continous on S, we have A is closed inS. And since Sis closed in R, we finally have A is closed in R.

Remark: 1. Roughly speaking, the property ofbeing closed has Transitivity. That is, in (M, d) let S c T c M, if S is closed in T, and Tis closed in M, then S is closed in M

Proof: Let X be an adherent point of sin M, then B M(X, r) n s * ¢ for every r > 0. Hence, B M(x, r) n T * ¢ for every r > 0. It means that x is also an adherent point ofT in M. Since Tis closed in M, we fmd that x E T. Note that since B M(x, r) n S * ¢ for every r > 0, we have (S c 1)

Br(x,r) n s = (BM(x,r) n T) n s = BM(x,r) n (S n T) = BM(x,r) n s * ¢. So, we have x is an adherent point of S in T. And since S is closed in T, we have x E S. Hence, we have proved that if x is an adherent point of S in M, then x E S. That is, S is closed inM.

Note: ( 1) Another proof of remark 1, since S is closed in T, there exists a closed subset U in Msuch that S = U n T, and since Tis closed in M, we have S is closed in M.

(2) There is a similar result, in (M, d) letS c T c M, if S is open in T, and Tis open in M, then Sis open in M. (Leave to the reader.)

2. Here is another statement like the exercise, but we should be cautioned. We write it as follows. Letfandgbe continuous on (S,d1) into (T,d2). LetA= {x: j(x) = g(x)}, show that A is closed in S.

Proof: Let x be an accumulation point of A, then there exists a sequence {xn} c A such that Xn --+ x. So, we have f(xn) = g(xn) for all n. Hence, by continuity off and g, we have

j(x) =!(limxn) = limfrxn) = limg(xn) = g(limxn) = g(x). n~oo n-+OCrl \ n-+oo n-+oo Hence, x E A. That is, A contains its all adherent point. So, A is closed.

1 = limg(ln.. ) n-+CX> n

= g ( Mm ~ ) by continuity of gat 0 = g(O)

= 0 which is absurb. Hence, g is not continuous on R by the exercise. To find such/, it suffices to consider fl..x) = eg(x).

Note: Such g (or!J is not measurable by Lusin Theorem.

4.19 Letfbe continuous on [a,b] and defineg as follows: g(a) = j(a) and, for a < x :S b, let g(x) be the maximum value ofjin the subinterval [a,x]. Show that g is continuous on [a, b].

Proof: Define g(x) = max {f(t) : t E [a, x]}, and choose any point c E [a, b], we want to show that g is continuous at c. Given c > 0, we want to fmd a 8 > 0 such that as X E (c- O,C + 8) n [a,b], we have

lg(x)- g(c)l < c. Since/is continuous at x = c, then there exists a 8' > 0 such that as x E (c-8',c+o')n[a,b], wehave

fl..c)- c/2 < fl..x) < fl..c) + c/2. Consider two cases as follows.

(1) max{f(t) : t E [a, c + 8'] n [a, b]} = j(pJ), where PI :s c- 8'. Asx E (c-8',c+8')n[a,b], wehaveg(x) =j(p 1)andg(c) = j(pJ). Hence, lg(x)- g(c) l = 0. (2) max{f(t) : t E [a, c + 8'] n [a, b]} = j(pl ), where PI > c- 8'. As X E (c- 8' ,c + 8') n [a,b], we have by (*)fl.. c)- c/2 :s g(x) :s fl.. c)+ c/2. Hence, lg(x)- g(c)l < c. So, ifwe choose 8 = 8', then for X E (c- 8,c + 8) n [a,b],

lg(x)- g(c )I < c by (1) and (2). Hence, g(x) is continuous at c. And since c is arbitrary, we have g(x) is continuous on [a, b].

8 Remark: It is the same result for min{f(t) : t E [a,x]} by the preceding method. ·~ .:E-- 4.20 Let/1, • • • ,Jm be m real-valued functions defined on Rn. Assume that each/k is

( Q.. 2-) continuous at the point a of S. Define a new function/ as follows : For each x inS, fl..x) is the largest of them numbers/1 (x), ... ,Jm(x) . Discuss the continuity off at a.

Proof: Assume that eachfk is continuous at the point a of S, then we have (fi + jj) and if;- jjl are continuous at a, where 1 :S i,j :S m. Since max( a, b) = (a+b);la-bl, then maxCfi,/2) is continuous at a since both (f1 + /2) and !fi-/ 2 1 are continuous at a. Define fl..x) = maxCfi, .. .fm), use Mathematical Induction to show thatj(x) is continuous at x = a as follows. As m = 2, we have proved it. Suppose m = k holds, i.e., max(f1, •• .fk) is continuous at x = a. Then as m = k + 1, we have

maxCfJ , .. .fk+I) = max[maxCfJ, .. .fk),fk+I] is continuous at x = a by induction hypothesis. Hence, by Mathematical Induction, we have prove thatfis continuos at x = a.

It is possible thatfand g is not continuous on R whihc implies that max((, g) is continuous on R. For example, let fl..x) = 0 if x E Q, and fl..x) = 1 if x E Qc and g(x) = 1

*

ifx E Q, and g(x) = 0 ifx E Qc.

Remark: It is the same rusult for min(/i, .. .fm) since max( a, b) + min( a, b) = a + b.

4.21 Letf: S--). R be continuous on an open set in R 11 , assume thatp E S, and assume that.f{p) > 0. Prove that there is ann -ball B(p;r) such thatj(x) > 0 for every x in the ball.

Proof: Since (p E )S is an open set in Rn, there exists a o 1 > 0 such that B(p, o 1) c S. Since.f{p) > 0, given & = ~) > 0, then there exists ann -ball B(p;o2 ) such that as X E B(p;oz) n S, we have

~) = j{p)- c < j(x) < j{p) + c = 3fl:!) . Leto = min(o 1,o2 ), thenasx E B(p;o), we have

j(x) > ~) > 0.

Remark: The exercise tells us that under the assumption of continuity at p, we roughly have the same sign in a neighborhood ofp, if.f{p) > 0 ( orf(p) < 0. )

4.22 Letfbe defined and continuous on a closed setS in R. Let

A= {x : X E Sandj(x) = o}. Prove that A is a closed subset of R.

Proof: Since A = f 1 ( {0} ), andfis continous on S, we have A is closed inS. And since Sis closed in R, we finally have A is closed in R.

Remark: 1. Roughly speaking, the property of being closed has Transitivity. That is, in (M, d) let S c T c M, if S is closed in T, and Tis closed in M, then S is closed in M.

Proof: Let x be an adherent point of S in M, then B M(x, r) n S * ¢ for every r > 0. Hence, B M(x, r) n T * ¢ for every r > 0. It means that x is also an adherent point ofT in M. Since Tis closed in M, we fmd that x E T. Note that since B M(x, r) n S * ¢ for every r > 0, we have (S c T)

Br(x,r) n s = (BM(x,r) n T) n s = BM(x,r) n (S n T) = BM(x,r) n s * ¢. So, we have x is an adherent point of Sin T. And since S is closed in T, we have x E S. Hence, we have proved that if x is an adherent point of S in M, then x E S. That is, S is closed in M.

Note: ( 1) Another proof of remark 1, since Sis closed in T, there exists a closed subset U in Msuch that S = U n T, and since Tis closed in M, we have S is closed in M.

(2) There is a similar result, in (M, d) letS c T c M, if S is open in T, and Tis open in M, then S is open in M. (Leave to the reader.)

2. Here is another statement like the exercise, but we should be cautioned. We write it as follows. Letfand g be continuous on (S,dJ) into (T,d2 ). LetA = {x : j(x) = g(x)} , show that A is closed in S.

Proof: Let x be an accumulation point of A, then there exists a sequence {xn} c A such that Xn --). x. So, we havef(xn) = g(xn) for all n. Hence, by continuity off and g, we have

j(x) =f(limxn) = limfTxn) = limg(x11 ) = g(limxn) = g(x). n-+oo n-+OC;l \ n-+oo n-+oo Hence, x E A. That is, A contains its all adherent point. So, A is closed.

I I

49

2.3 ~Jil

~ f ~\..ill 0t5" l~l . ~ f: I---t~ ~IJJIJ d...ol_.? o ~ I ~ .1

~..LY ~ ~ C ~ :> .)>-" J ~l! X2 J x1 if j5" ..LY ~ LS o,\?c

.~0~~ f

~4-JI :>yo:-J ~i . f: I---t~) d...ol_.? o_;j I ~ .2 w1 (x)- lim sup{f(z)- f(y): z,y E B£ (x)} £-+0+

.x ..LY ~ j{::}w1 (x)=0 0i ~I { 'xEI js:J

a.) l:J I J:! J lA:J I j l5:.; ~ i . 3

~I_? Ac~n (i)

A J 0 0.r" A ~ ~ J ~ ~ E c A ..G:- .J:! 'i (ii)

.~G ~b j ~ ~ f:A---t{O,l} (iii)

0i ~l! o:> J.b! ~ .F A J JL,a;'il 441..· "A f: A---t ~m ~t)' l~l .4

.o:> J.b! ~i f(A)

~l! A ~ t.? §' a.)b (xn) J JL,a;'J\ ~ f: A---t ~m c...jt)' 1~1 .5

·t.? §' a.)b 0_;>-~~ ·~ (f(xrJ) 0i ~ ~ f ~..\1 ~~l ~l! JL,a;'J\ ~ f: A---t ~m ~t)' 1~1 .6

. A ~ JL,a;'il ~ ~IJJ ~ J

JG~ dA : ~n ---t ~ j~ J a.)G:. _!)- ~ _F A C ~n ~ .7

dA (x) = inf {llx-all: a E A}

. A= {x E ~n : dA (x) = 0} 0iJ ~n ~ JL,a;'il a4k··A dA 0i ~i

Note: In remark 2, we CANNOT use the relation

j(x)- g(x)

since the difference "-" are not necessarily defined on the metric space (T, d2 ).

4.23 Given a function/: R --+ R, define two sets A and B in R2 as follows: A = {(x,y) : y < .f(x)},

B = {(x,y) : y > .f(x)}.

and Prove thatfis continuous on R if, and only if, both A and Bare open subsets of R2.

Proof: c~) Suppose thatfis continuous on R. Let (a, b) E A, then .I( a) > b. Sincefis continuous at a, then given G = J(atb > 0, there exists a (G > )8 > 0 such that as lx- al < 8, we have

}(a)+ b 2

=}(a)- G y. That is, B((a, b); 8) c A. So, A is open since every point of A is interior. Similarly for B.

( )8 > 0 such that

B(p 1,8) c A and B(p2 ,8) c B.

Hence, if(x,y) E B(p 1,8), then

(x- a) 2 + (y- (j{a)- G/2)) 2 < 82 andy< j(x).

So, it implies that

lx- al < 8, [y-}(a)+ G/21 < 8, andy< j(x).

Hence, as lx- al < 8, we have - 8 < y -./(a) + G/2

~}(a)- 8- G/2 < y < j(x)

~ ./(a) - G < y < j(x) ~}(a)- G < j(x).

And if (x,y) E B(p2, 8), then

(x- a) 2 + (y- (f{a) + G/2)) 2 < 82 andy> j(x). So, it implies that

lx- al < 8, [y -}(a)- G/21 < 8, andy > j(x).

Hence, as lx- al < 8, we have

j(x)

number

D.j{T) = sup{f(x)-fly) : x,y E 1}

is called the oscillation (or span) off on T. If x E S, the oscillation off at x is defined to be the number

cvj{x) = limD.j{B(x;h) n S). h-+O+

Prove that this limit always exists and that cvfx) = 0 if, and only if,Jis continuous at x.

Proof: 1. Note that sincefis bounded, say 1/(x)l :::; Mfor all x, we have 1/(x)-fl.y)l :S 2M for all x,y E S. So, D.j{T), the oscillation off on any subset Tof S, exists. In addition, we define g(h) = D.j{B(x;h) n S). Note that if T1 c T2(c S), we have D.j{T1) :S D.j{T2). Hence, the oscillation off at x, cvj{x) = limh-+O+ g(h) = g(O +)since g is an increasing function. That is, the limit of D.j{B(x; h) n S) always exists as h ~ O+.

2. ( =>) Suppose that cvfx) = 0, then given G > 0, there exists a 8 > 0 such that as h E (0, 8), we have

lg(h)l = g(h) = D.j{B(x;h) n S) < G/2. That is, as h E (0, 8), we have

sup{f(t) -fis) : t,s E B(x;h) n S} < d2 which implies that

- d2 0 such that as t E (x- 8,x + 8) n S, we have

1/(t)-fix)l < G/3. So, as t,s E (x- 8,x + 8) n S, we have

1/(t)-fis)l :S 1/(t)-fix)l+l/(x)-fis)l < d3+d3 = 2d3 which implies that

sup{(t)- fis) : t,s E (x- 8,x + 8) n S} :::; 2d3 < G. So, as h E (0, 8), we have

D.j{B(x;h) n S) = sup{(t)- fis) : t,s E (x- 8,x + 8) n S}

[a, b] = (U~1 J11n) U (Q n [a, b ]) is of the firse category which is absurb since every complete metric space is of the second category. So, this f cannot exist.

Note: 1 The Boundedness off can be removed since we we can accept the concept 00 > 0.

2. (J11n is closed in [a, b]) Given an accumulation point x of J 11m if x ~ J 11m we have wj{x) < 1/n. So, there exists a 1 -ball B(x) such that O.j{B(x) n [a, b]) < lin. Thus, no points of B(x) can belong to J 11n, contradicting that x is an accumulation point of J 11n. Hence, x E J 1tn and J11n is closed.

3. (Definition of a nowhere dense set) In a metric space (M, d), let A be a subset of M, we say A is nowhere dense inMif, and only if A contains no balls of M, (~ int(A) = ¢).

4. (Definition of a set of the first category and of the second category) A set A in a metric space M is of the first category if, and only if, A is the union of a countable number of nowhere dense sets. A set B is of the second category if, and only if, B is not of the first category.

5. (Theorem) A complete metric space is of the second category. We write another important theorem about a set of the second category below. (Baire Category Theorem) A nonempty open set in a complete metric space is of the

second category.

6. In the notes 3,4 and 5, the reader can see the reference, A First Course in Real Analysis written by M. H. Protter and C. B. Morrey, in pages 375-377.

[') ,) \ _L., ~ 4.25 Letfbe continuous on a compact interval [a, b]. Suppose thatfhas a local '\:: / maximum at x 1 and a local maximum at x2. Show that there must be a third point between

x1 and x2 wherefhas a local minimum. Note. To say thatfhas a local maximum at x1 means that there is an 1 -ball B(x1) such

that fix) :S fix!) for all X in B(xl) n [a, b]. Local minimum is similarly defined. Proof: Let x2 > x1. Suppose NOT, i.e., no points on (x1,x2) can be a local minimum

off Sincefis continuous on [x 1,x2], then inf{f(x) : x E [x1,x2]} =fix!) orfix2) by hypothesis. We consider two cases as follows:

(1) Ifinf{f(x) : x E [x1,x2]} =fix!), then

{

(i)fix) has a local maximum atx1 and

(ii)fix) ~ fix 1) for all x E [x1,x2].

By (i), there exists a 8 > 0 such that x E [x1,x1 + 8) c [x1,x2), we have (iii) fix) :S fix I) .

So, by (ii) and (iii), as x E [x 1,x1 + 8) , we have

fix) =fix!)

which contradicts the hypothesis that no points on (x 1,x2) can be a local minimum off (2) Ifinf{f(x) : x E [x 1,x2]} =fix!), it is similar, we omit it. Hence, from (1) and (2), we have there has a third point between x 1 and x2 where /has

a local minimum.

4.26 Letfbe a real-valued function, continuous on [0, 1 ], with the following property: For every real y, either there is no x in [0, 1] for which fix) = y or there is exactly one such x. Prove that f is strictly monotonic on [0, 1].

function. Since./{[0, 1]) = [0, 1] x [0, 1], we have the domain ofj-1 is [0, 1] x [0, 1] which is connected. Choose a special pointy E [0, 1] x [0, 1] so thatf1 (y) := x E (0, 1 ). Consider a continous function g = f 11 [O,IJx[O,IJ-{y}, then g : [0, 1] x [0, 1]- {y} --+ [O,x) U (x, 1] which is continous. However, it is impossible since [0, 1] x [0, 1]- {y} is connected but [O,x) U (x , 1] is not connected. So, suchfcannot exist.

Connectedness

4.36 Prove that a metric spaceS is disconnected if, and only if there is a nonempty subset A of S, A * S, which is both open and closed in S.

Proof: ( =>) Suppose that S is disconnected, then there exist two subset A, B in S such that

1. A, Bare open inS, 2. A * ¢and B * ¢, 3. An B = ¢, and 4. AU B = S. Note that since A, Bare open inS, we have A = S- B, B = S-A are closed inS. So, if S is disconnected, then there is a nonempty subset A of S, A * S, which is both open and closed inS.

(

[ (1 + __)_) 1/r _ 1 J lim 2x (_Q_) x-+oo (x1r)-llr ' 0

= l!ffi 7r ~lr x+-1 ( 1 + ix ) +-I by L-Hospital Rule. = 0.

Hence Xn --+ 0 as n --+ oo.

6. Here is a useful criterion for a function which is NOT uniformly continuous defined a subset A in a metric space. We say a function f is not uniformly continuous on a subset A in a metric space if, and only if, there exists Eo > 0, and two sequences {xn} and {yn} such that as

limxn- Yn = 0 n-+oo

which implies that

1/(xn) - Nn)\ ~ Eo for n is large enough.

The criterion is directly from the definition on uniform continuity. So, we omit the proof.

~ ~ 4.52 Assume thatfis uniformly continuous on a bounded setS in Rn. Prove that/ v '->) must be bounded on S.

\__,.._... Proof: Sincefis uniformly continuous on a bounded setS in Rn, given E = 1, then there exists a 8 > 0 such that as \\x- y 1\ < 8, x,y E S, we have

d(f(x ),fly)) < 1. Consider the closure of S, cl(S) is closed and bounded. Hence cl(S) is compact. Then for any open covering of cl(S), there is a finite subcover. That is,

cl(S) C UxEci(S) B(x; 812 ),

~ cl(S) c Ut7 B(xk;8/2), wherexk E cl(S),

~ S c Ut7 B(xk; 8/2 ), where xk E cl(S) . Note that if B(xk; 8/2) n S = ¢ for some k, then we remove this ball. So, we choose Yk E B(xk;8/2) n S, 1 ~ k ~nand thus we have

B(xk; 8/2) c B(yk; 8)for 1 ~ k ~ n,

since let z E B(xk;812),

\\z- Yk \\ ~ \\z- Xk \\ + \\xk- Yk \\ < 812 + 8/2 = 8. Hence, we have

S c Ut7 B(yk; 8), where Yk E S. Given x E S, then there exists B(yk; 8) for some k such that x E B(yk; 8). So,

d(j(x),j(xk)) < 1 ~JCx) E B(j(yk);1) Note that Ut7 B(j(yk); 1) is bounded since every B(j(yk); 1) is bounded. So, let B be a bounded ball so that Ut7 B(j(yk); 1) c B. Hence, we have every x E S, j(x) E B. That is, f is bounded.

Remark: If we know that the codomain is complete, then we can reduce the above proof. See Exercise 4.55.

4. 53 Let fbe a function defined on a set S in Rn and assume that j(S) c Rm. Let g be defined onj(S) with value in Rk, and let h denote the composite function defined by

-------- - - - - -

= (A ) r12r - 11 ~ oo, as n ~ oo since r < 0, which is absurb with (*). Hence, we know that xr is not uniformly continuous on (0, 1 ), for r < 0.

Ps: The reader should try to realize why xr is not uniformly continuous on (0, 1 ), for r < 0. The ruin of non-uniform continuity comes from that x is small enough.

6 A . ~· 4. 54 Assume f : S ~ Tis uniformly continuous on S, where S and Tare metric J ·~ ·) spaces. If {xn} is any Cauchy sequence inS, prove that {f(xn)} is a Cauchy sequence in T. (_Z .. J (Compare with Exercise 4.33.)

Proof: Given E > 0, we want to find a positive integer N such that as n, m ~ N, we have

d(f(xn),f{xm)) 0 such that as d(x,y) < 8, x,y E S, we have

d(f(x),fly)) 0, there is a postive integer N such that as n, m ~ N, we have

d(f(xn),f{xm)) < E. That is, {f(xn)} is a Cauchy sequence in T.

Remark: The reader should compare with Exercise 4.33 and Exercise 4.55 .

~- 4. 55 Let f : S ~ T be a function from a metric space S to another metric space T. Assume that f is uniformly continuous on a subset A of S and let Tis complete. Prove that there is a unique extension ofjto cl(A) which is uniformly continuous on cl(A).

Proof: Since cl(A) =AU A', it suffices to consider the case x E A'- A. Since x E A' -A, then there is a sequence {xn} ~ A with Xn ~ x. Note that this sequence is a Cauchy sequence, so we have by Exercise 4. 54, {f(x n)} is a Cauchy sequence in T since f is uniformly on A. In addition, since Tis complete, we know that {f(xn)} is a convergent sequence, say its limit L. Note that if there is another sequence {.Xn} c A with Xn ~ x, then {f(.Xn)} is also a convergent sequence, say its limit L' . Note that {xn} U {.Xn} is still a Cauchy sequence. So, we have

d(L,L') ~ d(L,flxn)) + d(f{xn),f{.Xn)) + d(f{.Xn) ,L' ) -+ 0 as n ~ oo. So, L = L'. That is, it is well-defined for g : cl(A) ~ Tby the following

(x) = { fix) ifx E A, g limn .... ooflxn) ifx E A' -A, wherexn ~X.

So, the function g is a extension ofjto cl(A). Claim that this g is uniformly continuous on cl(A). That is, given E > 0, we want to

find a 8 > 0 such that as d(x,y) < 8, x,y E cl(A), we have

d(g(x),g(y)) < E. Sincefis uniformly continuous on A, for E 1 = c/3, there is a 8' > 0 such that as

d(x,y) < 8' , x,y E A, we have

d(j(x),f(y)) < & 1•

Letx,y E cl(A), and thus we have {xn} c A withxn ~ x, and {yn} c A withyn ~ y. Choose 8 = 8' /3, then we have

d(xn,x) < 8'13 and d(yn,y) < 8'13 as n ~ N 1 So, as d(x,y) < 8 = 8'13, we have (n ~ N 1)

d(xn,Yn) ~ d(xn,X) + d(x,y) + d(y,yn) < 8'/3 + 8'13 + 8'/3 = 8'. Hence, we have as d(x,y) < 8, (n ~ N 1)

· d(g(x),g(y)) ~ d(g(x),f(xn)) + d(f(xn),fiyn)) + d(j(yn),f(y)) < d(g(x),f(xn)) + £ 1 + d(j(yn),g(y))

And since limn-+oofCXn) = g(x ), and limn-+oo/(yn) = g(y ), we can choose N ~ N 1 such that

d(g(x),f(xn)) < £ 1 and

d(j(yn ),g(y)) < £ 1• So, as d(x,y) < 8, (n ~ N) we have

d(g(x) ,g(y)) < 3&' = & by(*). That is, g is uniformly on cl(A).

It remains to show that g is a unique extension off to cl(A) which is uniformly continuous on cl(A). If there is another extension h ofjto cl(A) which is uniformly continuous on cl(A) , then given x E A'- A, we have, by continuity, (Say Xn ~ x)

h(x) = h(limxn) = limh(xn) = lim fl'xn) = limg(xn) = g(limxn) = g(x) n~oo n~oo n-+OCil ' n-+oo n-+oo

which implies that h(x ) = g(x) for all x E A' -A. Hence, we have h(x) = g(x) for all x E cl(A) . That is, g is a unique extension ofjto cl(A) which is uniformly continuous on cl(A).

Remark: 1. We do not require that A is bounded, in fact, A is any non-empty set in a metric space.

2. The exercise is a criterion for us to check that a given function is NOT uniformly continuous. For example, letf: (0, 1) ~ R by j(x) = llx. Since j(O +) does not exist, we know that f is not uniformly continuous. The reader should feel that a uniformly continuous is sometimes regarded as a smooth function. So, it is not surprising for us to know the exercise. Similarly to checkj(x) = x2 ,x E R, and so on.

3. Here is an exercise to make us know that a uniformly continuous is a smooth function. Let f : R ~ R be uniformly continuous, then there exist a, f3 > 0 such that

1/(x)l ~ alxl+ f3. Proof: Sincefis uniformly continuous on R, given & = 1, there is a 8 > 0 such that as

lx- Yl < 8, we have 1/(x)- f(y)l < 1.

Given any x E R, then there is the positive integer N such that N8 > lxl > (N- 1 )8. If x > 0, we consider

8 Yo = 0, YI = 812, Yz = 8, . .. ,y2N-I = N8- 2 'Y2N = x.

Then we have

*

*

N

Jf(x) -A o) I ~ z )N2k) - N2k-1 ) I + l!tY2k-1 ) - N2k-2) I k=l

~ 2Nby (*)

which implies that

Jf(x )I ~ 2N + j/(0)1

~ 2( 1 + ~) + jf(O)I since lxl > (N- 1)8

~ ~ lxl + (2 + l/(0)1). Similarly for x < 0. So, we have proved that jf(x) l ~ alxl + f3 for all x.

~k--"' 4. 56 In a metric space (S, d), let A be a nonempty subset of S. Define a function ~ ,'l ~) /A : S --)o R by the equation 1/ · } fA(x) = inf{d(x,y) : y E A}

for each x in S. The number /A (x) is called the distance from x to A. (a) Prove that/A is uniformly continuous on S. (b)Provethatcl(A) = {x : X E Sand/A(x) = o}. Proof: (a) Given c > 0, we want to find a 8 > 0 such that as d(x 1,x2) < 8, x 1,x2 E S,

we have

Consider (x 1,x2,y E S)

d(x1,y) ~ d(x1,x2) + d(x2,y), and d(x2,y) ~ d(x1,x2) + d(x1,y) So,

inf{d(x 1,y) : y E A} ~ d(x 1,x2) + inf{d(x2,y ) : y E A} and inf{d(x2,y) : y E A} ~ d(x 1,x2) + inf{d(x 1,y) : y E A}

which implies that

/A(xi)- jA(x2) ~ d(x1,x2) andfA(x2)- /A(xi) ~ d(x1 ,x2)

which implies that

lfA(xi)- jA(x2)l ~ d(x1 ,x2).

Hence, ifwe choose 8 = c, then we have as d(x 1,x2) < 8, x 1,x2 E S, we have

lfA(xl)- /A(X2)1 < c .

That is, fA is uniformly continuous on S. (b) Define K = {x : x E S andfA(x) = 0}, we want to show cl(A) = K. We prove it

by two steps. (c) Let X E cl(A), then B(x; r) n A * ¢for all r > 0. Choose Yk E B(x; Ilk) n A, then

we have

inf{d(x,y) : y E A} ~ d(x,Jk) --)o 0 ask --)o oo.

So, we havefA(x) = inf{d(x,y) : y E A} = 0. So, cl(A) c K. (:2) Let x E K, thenfA (x) = inf{d(x,y) : y E A} = 0. That is, given any c > 0, there

is an elementye E A such that d(x,ye) < c. That is,ye E B(x;c) nA . So, x is an adherent point of A. That is, x E cl(A). So, we have K c cl(A) .

From above saying, we know that cl(A) = { x : x E Sand fA (x) = 0}. Remark: 1. The function/A often appears in Analysis, so it is worth keeping it in mind.

In addition, part (b) comes from intuition. The reader may think it twice about distance 0.

2. Here is a good exercise to pratice. The statement is that suppose that K and Fare disjoint subsets in a metric space X, K is compact, F is closed. Prove that there exists a 8 > 0 such that d(p, q) > 8 if p E K, q E F. Show that the conclusion is may fail for two disjoint closed sets if neither is compact.

Proof: Suppose NOT, i.e., for any 8 > 0, there exist Po E K, and q8 E F such that d(pa,qo) :S 8. Let 8 = lin, then there exist two sequence {pn} c K, and {qn} c F such that d(pn, q n) :S lin. Note that {pn} c K, and K is compact, then there exists a subsequence {pnk} with limnr•ooPnk = p E K. Hence, we consider d(pnpqnk) :S ~k to get a contradiction. Since

d(pnk,p)+d(p,qnk) :S d(pnk,qnk) :S Jk, then let nk --+ oo, we have limnr•oo q nk = p. That is, p is an accumulation point ofF which implies that p E F. So, we get a contradiction since K n F = ¢. That is, there exists a 8 > 0 such that d(p,q) > 8 ifp E K, q E F.

We give an example to show that the conclusion does not hold. Let K= {(x,O): x E R}andF= {(x,l/x): x > 0}, thenKandFareclosd.Itisclearthat such 8 cannot be found.

Note: Two disjoint closed sets may has the distance 0, however; if one of closed sets is compact, then we have a distance 8 > 0. The reader can think ofthem in Rn, and note that a bounded and closed subsets in Rn is compact. It is why the example is given.

4. 57 In a metric space (S, d), let A and B be disjoint closed subsets of S. Prove that there exists disjoint open subsets U and V of S such that A c U and B c V. Hint. Let g(x) =/A (x) -fs(x ), in the notation of Exercise 4.56, and consider g-1 ( -oo, 0) and g-1 (0, +oo ).

Proof: Let g(x) =fA (x) - fs(x ), then by Exercixe 4.56, we have g(x) is uniformly continuous on S. So, g(x) is continuous on S. Consider g-1 ( -oo, 0) and g-1 (0, +oo ), and note that A, Bare disjoint and closed, then we have by part (b) in Exercise 4.56,

g(x) < 0 if x E A and

g(x) > 0 if x E B. So, wehaveA cg-1(-oo,O) := U, andB c g-1(0,+oo) := V.

Discontinuities

4. 58 Locate and classify the discontinuities of the functions f defined on R 1 by the following equations:

(a)flx) = sinx/x ifx * O,flO) = 0. Solution:fis continuous on R- {0}, and since limx-+O si~x = 1, we know thatfhas a

removable discontinuity at 0.

(b)flx) = e11x ifx * O,flO) = 0. Solution:fis continuous onR- {0}, and since limx-+o+ellx = oo and limx-+o- e 11x = 0,

we know thatfhas an irremovable discontinuity at 0.

(c)flx) = e11x+sinllxifx * O,flO) = 0. Solution:fis continuous on R- {0}, and since the limitflx) does not exist as x --+ 0,

we know thatfhas a irremovable discontinuity at 0.

I .

65

3.4 JL=;.o

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f(x) = {~xli' sin(l/llxll) :1 ~ 0~ ..:U~ J '~~~ ~I..U I t../' Df(O) J 0 ~ JLA.:.,;,")U ~li f

lf(x)- j(O)I < ~ = llxll Vx =f= 0 llx- Oil - llxll

=> lim If ( x) - f ( 0 )I = 0 x-.o llx- Oil

c>_r:-i ~if)

DJi (x) = 2x1 sin ( 1/llxll) -~~II cos ( 1/llxll) V x 1 0.

. x1 -+ 0 ~~~~if ~ _?i 'i D1f 0i l>~ x = (x1,0,· .. ,0) ~ y.

. 0 ~ ~ e--) D1j 0i ~-· - · ·~

3.1 ~)~

J: ~ u..b:-) ~i f ~\..ill ~_,)-I u~l ~I .1 L.....!G ~ b ~ j(x) = (b,x) .(i)

f Cx) = llxll4 (ii) ~ ~b T ~ f (x) = (x, Tx) (iii)

11. 11.

j(x ) = LLaijxix.i (iv) i=l j=l

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(ii) f( x, y) = {0

xysin(1/(x2 +y

2)) (x,y) =f= (0,0)

(x,y) = (0,0)

(iii) j(x,y,z )= x2y4z3 + x2

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~\ ~ ~ J ~t u oli.l J a X>- ~1.?.1 ~ f ~\.ill c.-i\5' 1)1 .5

0i J 'k =f= 0 ~ 'ku oli.l y ~Li,'jll

DJ.uf (a)= kDuf (a).

~JL; J o~ _r-->.-y Du+vf(a) H 'Dvf(a) J Duf(a) 0~ J bl ~ Duf(a)+Dvf(a)

0i ~t 'j, g: A C ~n --t ~

Du (oJ + {3g) (a) = aDuf(a) + f3 Dug(a) Du (f · g)( a)= g(a)Duf(a)+ f(a)Dug(a)

u oli.l j5:J J B ~_,:.ill o }:JI l) & a p Duf (a)= 0 0L.S' 1)1 . 7 Duf(a) = 0 ~IS'} J~ 0i t-k" y; bL. . B ~ ~G" j 0i c.. ~t

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67

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68

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lim llf (x) -f (a)- T (x- a)ll = O. x-.a llx- all

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Solut ions by Erin P. J. Pearse

First , the derivatives are

x2 (x~- xi) Dd(x 1, x2) = ( 2 + 2 ) 2 , x1 x2

and so clearly exist wherever (x, y) =f (0, 0). To check the origin, consider that along the axes,

0 0 Dd(x1, 0) = 4 = 0, and Dd(O, x2) = 4 = 0. x1 x2

However , for f to be continuous at (0, 0), we must have f(O, 0) = lim f (x, y ),

(x ,y)---> (0,0)

no matter how (x , y) ---+ (0 , 0)! Define 'Y(t ) : lR---+ JR2 by

'Y( t ) = (x (t ), y(t)) = (t, t),

so that we approach the origin along the diagonal. Then

t2 1 1 E~6 f(x (t ), y(t )) = ~~ f(t , t ) = ~~ 2t2 = ~~ 2 = 2 =f 0.

3 ~ 7. Suppose that f is a JR-valued defined in an open set E ~ JRn, and that the v-f' partial derivatives Dd, . .. , Dnf are bounded in E . Prove that f is continuous

in E . Take m = 1 as in 9.21. P ick x E E and consider the ball of radius r

B (x, r ) ~E. Choose hE JRn such that llhl l < r. Define Vk E JRn by

Vk = (h1 , . .. , hk , 0, . .. , 0) for k = 1, .. . , n .

Then as in (42), n

f (x + h) - f (x ) = :L [f (x + vj) - f (x + Vj-1 )] . j =l

Since lvkl < r for 1 :S k :S n , and since B = B (x , r ) is convex, the segments with endpoints x + Vj_1 and x + Vj lie in B . Then Vj = Vj- 1 + hjej , where

hjej = (0 , .. . , 0, hj, 0, .. . , 0).

Then the Mean Value Theorem applies to the partials and shows that the yth summand is

hj(Dj f) (x + Vj-1 + tj hjej) for some t j E (0, 1). By hypothesis , we have M such that

I(Dj f) (x) l :S M j = 1, . . . , n , \:fx E E.

Applying this to the absolute value of the above difference , n

lf (x +h) - f (x)l :S :L lhj l · M :S nMmax{lhj l} :S nMIIhl l· j =1

15

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CHAPTER 7 489

4. By theorem 2.21, jD/(£)(u)j = (''V/(£))r.u ~ II 'Vf(£)11 whenever !l u ll= 1, and equality holds if and only if u = u0• Now ID/(£)(u0)l = sup ID/(s)(u)j

llul l - 1 = II D/(£)11- Use 3. 5. Let u2+v2 = 1. Since {f(tu, tv)-/(0, 0)}/ t = u2v, D(u,v>f(O, 0) exists. Also if(tu, tv)-f(O, O)j = ju 2vtj ~ It I and so/ is continuous at (0, O)T .

If/is differentiable at (O,O)T, then, sinceDtf(O, 0) = Da/(0,0) = 0, D/(0, 0) = 0. But

· ·th/() - 2 . if(tu,tv)-f(0, 0)-0(tu,tv)l/l t l =iu2vi+O as t -+ 0 . . ecreasmg wt a - , . ·--

' = 1 and /(a) f; d = 0 1/k c!:" 6. Since /(w, 0) = /(0, w) = 0 for all w, Dtf(O,_ 0), f?d\0, 0) exist and are 0. ' Ja g . • )JY ~ut, when w *