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

[tylerc1991]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

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

 

[micromass]

Let [itex] M = 0 [/itex]. Lemma 1 states that, since [itex] g [/itex] is continuous, there exists a [itex] \delta_2 > 0 [/itex] such that [itex] |x - y| < \delta_2 \implies |g(y)| \leq M = 0. [/itex] 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 [itex]M = \frac{\varepsilon}{2|f(x) - f(y)|}[/itex]?
Then [itex]M \geq 0[/itex]

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

 

[micromass]

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

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

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 [itex]f[/itex] is continuous, I can choose a [itex]\delta > 0[/itex] such that

[itex]|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{2|g(y)|}[/itex]?

 

[micromass]

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

I know that since [itex]f[/itex] is continuous, I can choose a [itex]\delta > 0[/itex] such that

[itex]|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{2|g(y)|}[/itex]?

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

 

[tylerc1991]

OK, since [itex]g[/itex] is continuous, we may say that there exists a [itex]\delta_1[/itex]

[itex]|x - y| < \delta_1 \implies |g(y)| < |g(x)| + 1[/itex].

Also, since [itex]f[/itex] is continuous, there exists a [itex]\delta_2[/itex] such that

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

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

 

[micromass]

OK, since [itex]g[/itex] is continuous, we may say that there exists a [itex]\delta_1[/itex]

[itex]|x - y| < \delta_1 \implies |g(y)| < |g(x)| + 1[/itex].

Also, since [itex]f[/itex] is continuous, there exists a [itex]\delta_2[/itex] such that

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

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

Thay seems ok!