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Date: 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?


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 

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 

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   
Associated Topics:
College Analysis
High School Analysis

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