Continuous Function 1 - StudentOfFortune.com

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$3.00 Continuous Function 1

From Mathematics: Calculus
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Due on Apr. 30, 2008
Asked on Apr 28, 2008 at 10:19:12PM

Q:
a)If a function f is continuous on (a,b) and f(a^+) and
f(b^-) are both finite, then prove that f is bounded on (a,b). Explain why the converse is not true.

b)Prove that a continuous function f:(a,b)--->R is uniformly continuous on (a,b) iff f can be extended continuously to [a,b]

c)use the result of part (a) to prove that f(x)=sin 1/x is not uniformly continuous on (0,1) but g(x)=x sin 1/x is uniformly continuou on (0,1)

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cxxliu
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$2.50 missing part b), "<=" direction. need bolzano-weierstrass theorem or heine-borel theorem

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Posted on Apr 30, 2008 at 7:20:58PM

A:
Preview: ... ) >= M. Since f is unbounded positively, some x in (a,b) satisfies f(x) >= M, hence I_M is nonempty. We claim that I_{M+1} is a proper subset of I_M. Proof: It is trivial that I_{M+1} is a subset of I_M. To show that the containment is proper, let a and a' be the left boundaries of I_M and I_{M+1} respectively. Clearly, a <= a', f(a) < M + 0.5 (otherwise I_M would have included some points less than a), and f(a') > M+0.5 (otherwise I_{M+1} wouldn't be minimal, since f is continuous). Hence a does not equal a', and so a < a', which shows that I_{M+1} is a proper subset of I_M. Now, I_M, I_{M+1}, I_{M+2}, ... ...

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