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!