I'm going to prove that if f and g are continuous functions, then so is fg.(adsbygoogle = window.adsbygoogle || []).push({});

We define [itex]fg \; :\; E \rightarrow \bb{F}[/itex] by

[itex](fg)(t) \;=\; f(t)g(t)[/itex] for all [itex]t \in E[/itex]

(F is an ordered field and E is a subset of F).

Proof

[tex]|(fg)(x) - (fg)(a)| = |f(x)g(x) - f(a)g(a)| =[/tex]

Add and subtract f(x)g(a)

[tex]|f(x)g(x) - f(x)g(a) \;+\; f(x)g(a) - f(a)g(a)| \leq [/tex]

We use the triangle inequality

[tex]\leq |f(x)g(x) - f(x)g(a)| \;+\; |f(x)g(a) - f(a)g(a)| = [/tex]

We factor out f(x) and g(a)

[tex]|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)|[/tex]

From the definition pf continuity we have

[itex]\delta_1[/itex] such that [itex]|f(x) - f(a)| < \frac{\epsilon}{2}[/itex] for [itex]|x-a|[/itex]

[itex]\delta_2[/itex] such that [itex]|g(x) - g(a)| < \frac{\epsilon}{2}[/itex] for [itex]|x-a|[/itex]

But we still have |f(x)| and |g(a)|.

From the definition, we can deduce[*]- This part I'm not really sure about.

[tex]|f(x) - f(a)| < \frac{\epsilon}{2} \;\Rightarrow\; |f(x)| < \frac{\epsilon}{2} + |f(a)| [/tex]

[tex]|g(x) - g(a)| < \frac{\epsilon}{2} \;\Rightarrow\; |g(a)| < \frac{\epsilon}{2} + |g(x)| [/tex]

And we can choose new delta-values:

[tex]\delta_3[/tex] such that [tex]|f(x) - f(a)| \;<\; \frac{\epsilon}{2(\frac{\epsilon}{2}+ |f(a)|)}[/tex] for [tex]|x-a|[/tex]

[tex]\delta_4[/tex] such that [tex]|g(x) - g(a)| \;<\; \frac{\epsilon}{2(\frac{\epsilon}{2}+ |g(x)|)}[/tex] for [tex]|x-a|[/tex]

In conclusion, we take [itex]\delta_5 = \min(\delta_3, \delta_4)[/itex]

And get:

[tex]|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)| <[/tex]

[tex](\frac{\epsilon}{2} + |f(a)|\cdot \frac{\epsilon}{2(\frac{\epsilon}{2}\cdot |f(a)|)}) + (\frac{\epsilon}{2} + |g(x)|\cdot \frac{\epsilon}{2(\frac{\epsilon}{2}\cdot |g(x)|)}) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon[/tex]

And the proof is concluded. Quad Erad Demonstrandum... if it is right, of course.

Hope it wasn't too long!

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# Homework Help: Epsilon-Delta proof. Have I done it right?

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