{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

royden5

# royden5 - Royden Real Analysis 3rd ed Chapter 5 Problem 5-3...

This preview shows pages 1–4. Sign up to view the full content.

Royden, Real Analysis 3rd ed. Chapter 5 Problem 5-3 a. Obvious. But it should be D + f ( c ) D + f ( c ) 0 D f ( c ) D f ( c ) . b. At the end point a we can only have D + f ( a ) and D + f ( a ) . If f has a local maximum at a, we have D + f ( a ) D + f ( a ) 0 . Problem 5-4 We may assume D + f ε > 0 . Otherwise apply the re- sult to f ( x ) + ε x and then let ε 0 . Suppose the contrary that there exist c < d such that f ( c ) > f ( d ) . Consider max [ a,d ] f , which must be attained at some point in [ a, d ) . By as- sumption there exist, for any x [ a, b ) , arbitrarily small δ x > 0 such that f ( x + δ x ) f ( x ) εδ x 2 which implies no point in [ a, d ) can be a local maximum. A contradiction. Problem 5-6 If g ( γ + h ) 6 = g ( γ ) , then f g ( γ + h ) f g ( γ ) h = f ( g ( γ + h )) f ( g ( γ )) g ( γ + h ) g ( γ ) · g ( γ + h ) g ( γ ) h (1) a. If g 0 ( γ ) > 0 , then g ( γ + h ) g ( γ ) h · g 0 ( γ ) > 0 for h small and positive. Since the second term on the right-hand side of (1) converges to g 0 ( γ ) , we get by taking lim sup that D + f g ( γ ) = D + f ( g ( x )) · g 0 ( γ ) . b. Similar to part a , except that g ( γ + h ) < g ( γ ) and the second term is negative. The fact that we need is the following lim a n · b n = lim a n · lim b n , if lim b n < 0 . c. If g ( γ + h ) = g ( γ ) then the left hand side of (1) is 0 . We need only to consider the case g ( γ + h ) 6 = g ( γ ) . Since g 0 ( γ ) = 0 , g ( γ + h ) g ( γ ) h = o (1) as h 0 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The fi rst term on the right-hand side is bounded by max( | D + f ( g ( γ )) | , | D + f ( g ( γ )) | , | D f ( g ( γ )) | , | D f ( g ( γ )) | ). When multiplying together, the limit is 0 . Problem 5-7 a. By Theorem 5 we may assume f is a monotone in- creasing function. Then it is clear ( why ? ) that lim x c f ( x ) = f ( c ) and lim x c f ( x ) = f ( c +) . Moreover f ( c ) f ( c ) f ( c +) and f ( x ) is continuous at x = c if and only if f ( c ) = f ( c +) . Hence the set of discontinuous points D is { x : f ( c ) < f ( c +) } and may be written as D = n =1 D n , when D n = ½ x [ a, b ] : f ( c +) f ( c ) > 1 n ¾ . Enough to show each D n is countable. Let a < x 1 < x 2 < x 3 < · · · < x m < b belong to D n . With x 0 = a and x m +1 = b, f ( b ) f ( a ) = ( f ( b ) f ( x m + ε )) + ( f ( x m + ε ) f ( x m ε )) + · · · + ( f ( x 1 + ε ) f ( x 1 ε )) + f ( x 1 ε ) + f ( a ) = [ f ( b ) f ( x m + ε )] + m X k =1 ( f ( x k + ε ) f ( x k ε )) + [ f ( x 1 ε ) + f ( a )] . By letting ε small enough and using the monotonicity of f, f ( b ) f ( a ) m X k =1 ( f ( x k + ε ) f ( x k ε )) m X k =1 ( f ( x k +) f ( x k )) m n . Hence m ( f ( b ) f ( a )) · n. That is | D n | ( f ( b ) f ( a )) · n + 2 and is fi nite. b. Let Q = { a 1 , a 2 , . . . } . For each a n de fi ne f n ( x ) = χ ( a n , 1] ( x ) so that f n is increasing and discontinuous at x = a n . ( When a n = 1 , we set f n ( x ) = χ { 1 } ( x ) for this purpose. ) The following function is what we want: f ( x ) = X n =1 2 n f n ( x ) . Since P n =1 2 n < , the Weierstrass M Theorem tells us fi rst that f is continuous at any irrational points and then X n 6 = m 2 n f n ( x ) = f ( x ) 2 m f m ( x ) 2
is continuous at x = a m .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}