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Thread: Uniform Again

  1. #1
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    Uniform Again

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

    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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    $\displaystyle |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 $\displaystyle |f(x)||g(x)-g(a)|$?

    and
    for $\displaystyle M \frac { \epsilon } { M +1 }$ should this be $\displaystyle M\frac{\epsilon}{M}$?
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    Quote Originally Posted by tttcomrader View Post
    $\displaystyle |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 $\displaystyle |f(x)||g(x)-g(a)|\geq 0$
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  7. #7
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    Quote Originally Posted by Plato View Post
    $\displaystyle \left( {\exists B > 0} \right)\left( {\forall x} \right)\left[ {\left| {g(x)} \right| \leqslant B} \right]$. $\displaystyle g$ is bounded.
    If $\displaystyle \varepsilon > 0$ then use $\displaystyle \frac{\varepsilon}{B+1}$ in the uniformity of $\displaystyle f$.
    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 $\displaystyle g$ was continuous.
    Without that it clearly not true.
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    Quote Originally Posted by Plato View Post
    No I don't. I missread it. I assumed that $\displaystyle g$ was continuous.
    Without that it clearly not true.
    You mean that if we have :

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

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

    Then fg is uniformly cintinuous in $\displaystyle (0,\infty)$???
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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 $\displaystyle (0,\infty)$

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

    Then fg is uniformly cintinuous in $\displaystyle (0,\infty)$???
    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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