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Unformatted text preview: (This exercise appeared in the last analysis qualifier. You have the advantage over the people who took the exam of knowing that it has to do with something we covered recently.) Solution. Claim that the sequence { f n } is equicontinuous and bounded. In fact, given x,y ∈ [0 , 1] we have that  f n ( x ) f n ( y )  ≤  x y  by the Mean Value Theorem, thus given ² > 0 we can take δ = ² to conclude that  f n ( x ) f n ( y )  < ² for all x,y ∈ [0 , 1] with  x y  < δ , n = 1 , 2 , 3 ,... . This establishes equicontinuity. For boundedness, we can use that f n ( x ) = f n (0) + Z x f n ( t ) dt, 3 hence  f n ( x )  ≤  f n (0)  + Z x  f n ( t )  dt ≤ 1 + Z xdt = 1 + x ≤ 2 for all n ∈ N , x ∈ [0 , 1]. The claim is established. By the Theorem of ArzelaAscoli, it follows that { f n } has a uniformly convergent subsequence { f n k } . Since this sequence also converges pointwise, the limit must be f . As the uniform limit of a sequence of continuous functions, f must be continuous. The rest of the exercise is a bit ambiguous. I think I copied it from a non final version of the qualifier without checking too carefully. “Must the sequence converge?” can be interpreted in at least two ways: (a) Given all the other hypothesis, is the hypothesis lim n →∞ f n ( x ) = f ( x ) for all x ∈ [0 , 1] redundant; is it needed? Interpreted in this way the answer is no; the sequence does not have to converge; counterexamples provided upon request. (b) Must the full sequence converge uniformly ? The answer in this case is yes, and I made this an exercise Homework 5. In either case, the answer to the last question is yes. I will only grade your proof of the continuity of f (or lack thereof). 4. Let [ a,b ] be a closed and bounded interval and let E ⊂ C ([ a,b ]). Assume that E is equicontinuous and bounded. Define f : [ a,b ] → R by: if a ≤ x ≤ b , then f ( x ) = sup { g ( x ) : g ∈ E} . Prove: f is continuous. Solution. Because E is bounded, sup { g ( x ) : g ∈ E} < ∞ for all x ∈ [ a,b ]; f is a well defined, real valued, function on [ a,b ]. By equicontinuity, given ² > 0 there is δ > 0 such that if x,y ∈ [ a,b ] satisfy  x y  < δ , then  g ( x ) g ( y )  < ²/ 2 for all for all g ∈ E . Now let x,y ∈ [ a,b ] and assume  x y  < δ . By definition of f ( x ) as the sup of the set of all values g ( x ), g ∈ E , there is g ∈ E such that f ( x ) ² 2 < g ( x ) , thus f ( x ) < g ( x ) + ² / 2 . By the choice of δ and because g ∈ E ,  x y  < δ ,  g ( x ) g ( y )  < ² 2 , thus g ( x ) < g ( y ) + ² 2 . By the definition of f ( y ) as a sup, g ( y ) ≤ f ( y ) . Putting it all together, we get f ( x ) < g ( x ) + ² 2 < g ( y ) + ² 2 + ² 2 ≤ f ( y ) + ² 2 + ² 2 = f ( y ) + ² ; that is f ( x ) f ( y ) < ² . Changing the roles of x and y , we similarly get that f ( y ) f ( x ) < ² , thus proving that  f ( x ) f ( y )  < ² if  x y  < δ . Continuity follows. 4...
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 Spring '11
 Speinklo
 Metric space, Limit of a sequence, Hilbert space, Cauchy sequence, Banach space

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