ContinuityDate: 11/24/2001 at 14:45:44 From: Cathy Subject: Real Analysis - Continuity In my math class, we were asked to prove the following: Suppose f and g are continuous on I = [a,b] and f(x) < g(x) for all x in I. Prove there exists a k > 0 such that f(x) + k is less than or equal to g(x) for all x in I. I have trouble starting this proof. Can you help me? Thanks. Date: 11/24/2001 at 18:34:59 From: Doctor Paul Subject: Re: Real Analysis - Continuity f and g are continuous on I, so g-f is certainly continuous on I as well. let h(x) = (g-f)(x) Certainly, h takes a minimum value at some point x_0 in I (this follows from that fact that I is closed and bounded as long as we assume that a and b are real numbers - and I think that is a safe assumption here). Note that the point x_0 doesn't have to be unique; there could be many points in I at which h takes on its minimum value; x_0 is just one of them. Choose k = h(x_0) = (g-f)(x_0) = g(x_0) - f(x_0) Then f(x) + k <= g(x) for all x in I. In particular, at x_0 and at all other points that give h its minumum value [ ie, {x in I | h(x) = h(x_0) } ], we have f(x) + k = g(x) and at all other points [ ie, {x in I | h(x) != h(x_0)} ] we have: f(x) + k < g(x) I guess another way to do this problem is to choose k = 1/2 * h(x_0) Then for all x in I we have f(x) + k < g(x) and that's acceptable as well. I hope this helps. Please write back if you'd like to talk about this some more or if you have questions about what I've written above. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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