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Math Help - Uniform Continuity Proof

  1. #1
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    Uniform Continuity Proof

    Let f, g be uniformly continuous functions on (a,b). Show that the function f+g is uniformly continuous on (a,b) by using the epsilon - delta definition.

    I'm pretty sure it will involve something along the lines of |x-y| => |f(x)+g(x)-f(y)-g(y)|< epsilon and then some triangle inequality of something.

    Thanks
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  2. #2
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    Quote Originally Posted by tbyou87 View Post
    Let f, g be uniformly continuous functions on (a,b). Show that the function f+g is uniformly continuous on (a,b) by using the epsilon - delta definition.

    I'm pretty sure it will involve something along the lines of |x-y| => |f(x)+g(x)-f(y)-g(y)|< epsilon and then some triangle inequality of something.

    Thanks
    |f(x)+g(x) - f(y) - g(y)| \leq |f(x) - f(y)| + |g(x) - g(y)|.
    And each one can be made arbitrary small.
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  3. #3
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    Complete Proof:
    Since f is uniformly continuous on (a,b),
    for every eps1>0, there exists delt1>0 s.t. x,y ele (a,b) and |x-y|< delt1 => |f(x)-f(y)|< eps1
    Since g is uniformly continuous on (a,b),
    for every eps2>0, there exists delt2>0 s.t. x,y ele (a,b) and |x-y|< delt2 => |g(x)-f(y)|< eps2
    Let eps = eps1 + eps2 >0 and let delt = delt1 + delt2.
    Then |x-y|<delt1+delt2<delt => |f(x)-f(y)+g(x)-g(y)|<= |f(x)-f(y)|+|g(x)-g(y)|< eps1+eps2=eps
    Thus f+g is uniformly continuous.

    Is this ok?
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  4. #4
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    Quote Originally Posted by tbyou87 View Post
    Complete Proof:
    Since f is uniformly continuous on (a,b),
    for every eps1>0, there exists delt1>0 s.t. x,y ele (a,b) and |x-y|< delt1 => |f(x)-f(y)|< eps1
    Since g is uniformly continuous on (a,b),
    for every eps2>0, there exists delt2>0 s.t. x,y ele (a,b) and |x-y|< delt2 => |g(x)-f(y)|< eps2
    Let eps = eps1 + eps2 >0 and let delt = delt1 + delt2.
    Then |x-y|<delt1+delt2<delt => |f(x)-f(y)+g(x)-g(y)|<= |f(x)-f(y)|+|g(x)-g(y)|< eps1+eps2=eps
    Thus f+g is uniformly continuous.

    Is this ok?
    Sorry for being a little late.

    No let \delta = \min \{ \delta_1 , \delta_2 \}.
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