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Uniform continuity Proof

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data
    1. Consider the function f(x) = x^3. Prove that (a) it is not uniformly continuous on R, but that (b) it is uniformly continuous on any interval of the type [-a, a]
    2. Suppose that f is uniformly continuous on a region S, and g is uniformly continuous on the region f(S). Show that the composite function g(f(x)) is uniformly continuous on S.

    2. Relevant equations
    A function is uniformly continuous on a set S if for every e>0 exists d>0 such that |x-y|<d implies |f(x)-f(y)|<e for any x,y in S

    3. The attempt at a solution

    1a) I'm strugginling how to prove that it is not uniformly continuous on R. I don't really know where to start...I mean intuitively its pretty obviously its not uniformly continuous since the slope increases without bound , so no delta you choose for |x-y|<d can imply |f(x)-f(y)|<e for every x,y (due to the fact that f(x)-f(y) tends to infinity as x, y go to inifnity under the constraint x = y+d)
    But I really have no idea how to "prove" this..The only thing i can think of to do, but I cant htink of a reason to do it (lol) is to show that
    If i take |x-y|<d with x>y, d>o then i have

    |f(x)-f(y)| is equal to or greater than (y+d)^3 - y^3 = 3y^2d + 3d^2y + d^3 and I don't really know what to do from here.

    1b) By mean favlue theorem, f(x)-f(y) = f'(c) (x-y)
    so |f(x)-f(y)| ≤ |f'(c)| |x-y| ≤ sup on [-a,a] of f' * |x-y| = 3a^2 |x-y|

    since Take d = e/3a^2

    Then , if |x-y| < e/3a^2
    We have

    |f(x)-f(y)| ≤ 3a^2 |x-y| ≤ 3a^2 * e/3a^2 = e

    So it has been proven

    2. This one I don't even know where to begin :S Help!
  2. jcsd
  3. Jul 20, 2009 #2
    Some hints:

    1a) Start with the negation of the definition of uniform continuity: A function is not uniformly continuous on A if there is an [tex]\varepsilon > 0[/tex] such that for any [tex]\delta > 0[/tex], there exist [tex]x,y \in A[/tex] satisfying [tex]\left|x-y\right| < \delta[/tex] and [tex]\left|f(x)-f(y)\right| \geq \varepsilon[/tex].

    1b) I don't think you need to apply the mean value theorem here. Once you factor out the (x-y) term I think you can easily bound the [tex]\left|f(x)-f(y)\right|[/tex] by choosing the appropriate delta.

    2. Standard composition of functions limiting argument. Start with the fact that g is uniformly continuous. Use the "delta" in the definition of continuity of g as the "epsilon" in the definition of the continuity of f.
  4. Jul 20, 2009 #3
    For 2, I was thinkin something along those lines, but would it be as simple as saying
    We know:
    |x-y| < d1 ---> |f(x)-f(y)|<e1

    |f(x)-f(y)|<d2 ----> |g(f(x)) - g(f(y))| < e2
    want to show |x-y|<d ----> |g(f(x)) - g(f(y))| < e
    set e1 = d2

    |x-y|<d1 ------> |f(x)-(f(y)|<d2 -------> |g(f(x)) - g(f(y))| < e2

    So it is proven?
  5. Jul 21, 2009 #4
    Yeah, that is the basic idea, but the actual write-up should be organized a bit better. For instance, you should understand why you want to estimate the proximity of g(a) and g(b) for a,b in f(S) first. Then the modulus of continuity of g (the "delta") tells us exactly how close we need f(x) and f(y) to be (x,y in A) for |g(f(x)) - g(f(y))| < e to be satisfied, so we can choose the "epsilon" accordingly in the definition of the continuity of f.

    As for generalizations of 1b), the simplest one can be found in the second thread by JG89 under the similar threads are at the bottom of this page.
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