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

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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
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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
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is continuous at x = a m .
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