1. ## 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 $\displaystyle (0,\infty)$

2. Originally Posted by Math2010
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 $\displaystyle f$.

3. 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?

4. $\displaystyle |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| < |g(a)||f(x)-f(a)| < M \frac { \epsilon } { M +1 }$

5. what happens to $\displaystyle |f(x)||g(x)-g(a)|$?

and
for $\displaystyle M \frac { \epsilon } { M +1 }$ should this be $\displaystyle M\frac{\epsilon}{M}$?

$\displaystyle |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| < |g(a)||f(x)-f(a)| < M \frac { \epsilon } { M +1 }$
Is it not |f(x)||g(x)-g(a)| + |g(a)||f(x)-f(a)| > |g(a)||f(x)-f(a)|

Since $\displaystyle |f(x)||g(x)-g(a)|\geq 0$

7. Originally Posted by Plato
$\displaystyle \left( {\exists B > 0} \right)\left( {\forall x} \right)\left[ {\left| {g(x)} \right| \leqslant B} \right]$. $\displaystyle g$ 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??

8. Originally Posted by xalk
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.

9. Originally Posted by Plato
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 $\displaystyle (0,\infty)$???

10. Originally Posted by xalk
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.