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Bounded Value TheoremIf f is continuous on [a,b] then f is bounded on the interval. Suppose f is not bounded on [a,b]. (That means that given any large M, there exists a number c in [a,b] such that |f(c)| > M) For any c in [a,b], and given any positive real number, e, there
exists a positive real number, d, such that Since f is not bounded on [a,b], the interval can be cut in half, and f is not bounded on one of the halves. That is, either f is not bounded on [a,a+b/2] or f is not bounded on [a+b/2,b], because if it were bounded on both halves, then it would be bounded on the whole interval. The interval can be successively bisected, so that we obtain a sequence of successively smaller closed intervals of arbitrarily small length, converging on a single point, say, c in [a,b], on which f is unbounded. (We know c exists, because the set of real numbers are closed -- that is, the set of reals contains the limits of all convergent sequences of real numbers. This follows from the "completeness property" of the reals.) Since f is continuous at c, given
any positive real number, e, there
exists a positive real number, d, such that But at least one closed sub-interval of [a,b] -- I'll call it [a',b'] -- contains c and is contained within (c-d,c+d), such that f is unbounded on [a',b']. Let M = |f(c)| + e . Since f is unbounded on [a',b'] there exists a c' in [a',b'] such that
|f(c')| > M, By the triangle inequality, |f(c) - f(c')| ³ |f(c')| - |f(c)| > e . But by continuity, |f(c') - f(c)| < e, a contradiction, proving the bounded value theorem. How is this theorem used? The Bounded Value Theorem is used to prove the Extreme Value Theorem, that if f is continuous on [a,b] then f has a maximum and a minimum on the interval. Internet References
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