Continuity of Functions Proof | f and g Continuous at x | h = fg Continuity

[tylerc1991]

Homework Statement

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.

Homework 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.

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!

 

[micromass]

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,

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.

 

[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

 

[micromass]

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 to \varepsilon

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

 

[tylerc1991]

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

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)|}?

 

[micromass]

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)|}?

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.

 

[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?

 

[micromass]

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?

Thay seems ok!