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Thread: Multiplication of uniformly continuous functions

  1. #1
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    Multiplication of uniformly continuous functions

    I'm trying to prove:

    If $\displaystyle f$:$\displaystyle A --> R$ and $\displaystyle g$:$\displaystyle A --> R$ are uniformly continuous functions, then $\displaystyle f * g$ is uniformly continuous.

    I've gotten to the fact that
    |f(x)g(x) - f(y)g(y)| <= |f(x)| * |g(x)-g(y)| + |g(y)| * |f(x) - f(y)|

    But I'm not sure where to go from here...
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  2. #2
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    Quote Originally Posted by moses View Post
    I'm trying to prove:

    If $\displaystyle f$:$\displaystyle A --> R$ and $\displaystyle g$:$\displaystyle A --> R$ are uniformly continuous functions, then $\displaystyle f * g$ is uniformly continuous.

    I've gotten to the fact that
    |f(x)g(x) - f(y)g(y)| <= |f(x)| * |g(x)-g(y)| + |g(y)| * |f(x) - f(y)|

    But I'm not sure where to go from here...
    We really need some more information to solve this problem. We need some information about the set $\displaystyle A$.

    For example let $\displaystyle f(x)=g(x)=x$ both of these are uniformly continuous on the set $\displaystyle [0,\infty)$ let $\displaystyle \delta =\epsilon$

    Now consider their prouduct $\displaystyle h:[0,\infty) \to \mathbb{R},h(x)=f(x)g(x)=x^2$

    on this unbounded set this is not uniformly continuous.

    So if A is compact. The proof is easy! because continuous functions on compact sets are bounded.

    If A is an open interval $\displaystyle (a,b)$ the function is uniformly continuous if and only if it can be continuously extended to the closed compact set $\displaystyle [a,b]$
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