(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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].

2. Relevant 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]

3. 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!

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# Homework Help: Continuity proof check please

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