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Math Help - Uniform Again

  1. #1
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    Uniform Again

    if f is unif. cont. on (0,\infty) and g is positive and bounded on (0,\infty), then f \cdot g is unif. cont. on (0,\infty)
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  2. #2
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    Quote Originally Posted by Math2010 View Post
    if f is unif. cont. on (0,\infty) and g is positive and bounded on (0,\infty), then f \cdot g is unif. cont. on (0,\infty)
    \left( {\exists B > 0} \right)\left( {\forall x} \right)\left[ {\left| {g(x)} \right| \leqslant B} \right]. g is bounded.
    If \varepsilon  > 0 then use \frac{\varepsilon}{B+1} in the uniformity of f.
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    I understand that M>0. |g(x)| \le M, \forall x since g is bounded. We know f is unif. cont. so for any \epsilon > 0 there exists delta > 0 such that |f(x)-f(a)| < \frac{\epsilon}{M+1}, \forall x,a \in A with |x-a| < \delta. Here is where I think I go wrong and do not understand. Do I simplify |fg(x) - fg(a)| to be  \le |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)|?

    g not being uniformly continuous is really throwing me off. I believe I need to use something from g being continuous?
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  4. #4
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     |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| < |g(a)||f(x)-f(a)| < M \frac { \epsilon } { M +1 }
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  5. #5
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    what happens to |f(x)||g(x)-g(a)|?

    and
    for M \frac { \epsilon } { M +1 } should this be M\frac{\epsilon}{M}?
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  6. #6
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    Quote Originally Posted by tttcomrader View Post
     |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| < |g(a)||f(x)-f(a)| < M \frac { \epsilon } { M +1 }
    Is it not |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| > |g(a)||f(x)-f(a)|

    Since |f(x)||g(x)-g(a)|\geq 0
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  7. #7
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    Quote Originally Posted by Plato View Post
    \left( {\exists B > 0} \right)\left( {\forall x} \right)\left[ {\left| {g(x)} \right| \leqslant B} \right]. g is bounded.
    If \varepsilon > 0 then use \frac{\varepsilon}{B+1} in the uniformity of f.
    Plato, do you still think that this theorem is true??
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  8. #8
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    Quote Originally Posted by xalk View Post
    Plato, do you still think that this theorem is true??
    No I don't. I missread it. I assumed that g was continuous.
    Without that it clearly not true.
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  9. #9
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    Quote Originally Posted by Plato View Post
    No I don't. I missread it. I assumed that g was continuous.
    Without that it clearly not true.
    You mean that if we have :

    1) f uniformly continuous in (0,\infty)

    2) g continuous , bounded and g>0 in (0,\infty)

    Then fg is uniformly cintinuous in (0,\infty)???
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  10. #10
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    Quote Originally Posted by xalk View Post
    You mean that if we have :

    1) f uniformly continuous in (0,\infty)

    2) g continuous , bounded and g>0 in (0,\infty)

    Then fg is uniformly cintinuous in (0,\infty)???
    No again. I cannot remember, why but I thought at the time the question was something different.
    I cannot tell you why I posted what I did.
    I think I was working out two similiar problems.
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