1. Sep 5, 2011

tylerc1991

1. The problem statement, all variables and given/known data

Given two functions $f$ and $g$, if $f$ and $g$ are continuous at a point $x$, then the function $h = fg$ is continuous at $x$.

2. Relevant equations

Lemma 1
If a function $f$ is continuous at a point $x$, then f is bounded on some interval centered at $x$. That is, there exists an $M \geq 0$ and a $\delta > 0$ such that for all $y$,
$|x - y| < \delta \implies |f(y)| \leq M.$

3. The attempt at a solution

Let $f$ and $g$ be functions that are continuous at a point $x$.
Define a new function $h$ as $h = fg$.
Let $\varepsilon > 0$ and $\delta_1 > 0$.
If $f(x) = 0$, then it is trivially true that for all $y$,
$\displaystyle |x - y| < \delta_1 \implies |f(x)||g(x) - g(y)| < \varepsilon,$
so assume $f(x) \neq 0$.
Now, suppose $\displaystyle \mu = \frac{\varepsilon}{|f(x)|}$.
Clearly $\mu$ is a positive real number.
Since $g$ is continuous at $x$, for all $\varepsilon$, there exists a $\delta_1$ such that
$|x - y| < \delta_1 \implies |g(x) - g(y)| < \mu. \quad \quad (1)$
Let $M = 0$. Lemma 1 states that, since $g$ is continuous, there exists a $\delta_2 > 0$ such that $|x - y| < \delta_2 \implies |g(y)| \leq M = 0.$ So,
$|x - y| < \delta_2 \implies |g(y)| = 0. \quad \quad (2)$
Let $\delta = \text{min}( \delta_1, \delta_2 )$. Then $|x - y| < \delta$ implies
$|h(x) - h(y)| = |f(x)g(x) - f(y)g(y)|$
$= |f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)|$
$\leq |f(x)g(x) - f(x)g(y)| + |f(x)g(y) - f(y)g(y)|$
$= |f(x)||g(x) - g(y)| + |g(y)||f(x) - f(y)|$
$< |f(x)| \mu + 0 = \varepsilon.$ QED

Pointing out any mistakes / giving constructive criticism is welcome. Thank you very much!

2. Sep 5, 2011

micromass

Staff Emeritus
You can't just set M=0. All lemma 1 says is that there exists an M, but that M is potentially very large. You have no control over what value M has. So saying that M=0 is not ok.

3. Sep 5, 2011

tylerc1991

Could I say that $M = \frac{\varepsilon}{2|f(x) - f(y)|}$?
Then $M \geq 0$

Then I would have to reset $\mu = \frac{\varepsilon}{2|f(x)|}$ in order for everything to cancel and add up to $\varepsilon$

4. Sep 5, 2011

micromass

Staff Emeritus
No, you cannot choose M. You can choose epsilon though.

5. Sep 5, 2011

tylerc1991

So the lemma is pretty pointless in this case? Can I try something like:

I know that since $f$ is continuous, I can choose a $\delta > 0$ such that

$|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{2|g(y)|}$?

6. Sep 5, 2011

micromass

Staff Emeritus
No since your "epsilon" is $\frac{\varepsilon}{2|g(y)|}$ and is dependent on y. It shouldn't be dependent on y. It can depend on x, though.

7. Sep 5, 2011

tylerc1991

OK, since $g$ is continuous, we may say that there exists a $\delta_1$

$|x - y| < \delta_1 \implies |g(y)| < |g(x)| + 1$.

Also, since $f$ is continuous, there exists a $\delta_2$ such that

$|x - y| < \delta_2 \implies |f(x) - f(y)| < \frac{\varepsilon}{2(|g(x)| + 1)}$.

This would cancel fine and would add to $\varepsilon$ in the end, but this way seems so roundabout. Is there no nicer way?

8. Sep 5, 2011

micromass

Staff Emeritus
Thay seems ok!