Epsilon-Delta proof. Have I done it right?

[MrGandalf]

I'm going to prove that if f and g are continuous functions, then so is fg.

We define $fg \; :\; E \rightarrow \bb{F}$ by
$(fg)(t) \;=\; f(t)g(t)$ for all $t \in E$
(F is an ordered field and E is a subset of F).

Proof
$$|(fg)(x) - (fg)(a)| = |f(x)g(x) - f(a)g(a)| =$$

Add and subtract f(x)g(a)
$$|f(x)g(x) - f(x)g(a) \;+\; f(x)g(a) - f(a)g(a)| \leq$$

We use the triangle inequality
$$\leq |f(x)g(x) - f(x)g(a)| \;+\; |f(x)g(a) - f(a)g(a)| =$$

We factor out f(x) and g(a)
$$|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)|$$

From the definition pf continuity we have
$\delta_1$ such that $|f(x) - f(a)| < \frac{\epsilon}{2}$ for $|x-a|$

$\delta_2$ such that $|g(x) - g(a)| < \frac{\epsilon}{2}$ for $|x-a|$

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

From the definition, we can deduce [*] - This part I'm not really sure about.
$$|f(x) - f(a)| < \frac{\epsilon}{2} \;\Rightarrow\; |f(x)| < \frac{\epsilon}{2} + |f(a)|$$

$$|g(x) - g(a)| < \frac{\epsilon}{2} \;\Rightarrow\; |g(a)| < \frac{\epsilon}{2} + |g(x)|$$

And we can choose new delta-values:
$$\delta_3$$ such that $$|f(x) - f(a)| \;<\; \frac{\epsilon}{2(\frac{\epsilon}{2}+ |f(a)|)}$$ for $$|x-a|$$

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

In conclusion, we take $\delta_5 = \min(\delta_3, \delta_4)$
And get:
$$|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)| <$$

$$(\frac{\epsilon}{2} + |f(a)|\cdot \frac{\epsilon}{2(\frac{\epsilon}{2}\cdot |f(a)|)}) + (\frac{\epsilon}{2} + |g(x)|\cdot \frac{\epsilon}{2(\frac{\epsilon}{2}\cdot |g(x)|)}) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

And the proof is concluded. Quad Erad Demonstrandum... if it is right, of course.
Hope it wasn't too long!

 

[andytoh]

You factorized properly. Take a delta such that |f(x)-f(a)| < 1 as x -> a. Then you can get a bound on |f(x)|, say M. Then you choose another delta so that |f(x)-f(a)| < epsilon/(2|g(a)|+1). Then choose another delta such that |g(x)-g(a)| < epsilon/(2M). Then take the smaller of all three deltas, and you get your result.

 

[HallsofIvy]

You do need the fact that, since f is continuous, |f(x)| has an upper bound in order to handle the fact that |f(x)| is variable, not a constant like |g(a)|.

 

[MrGandalf]

We now choose three $\delta$'s, so
$$\delta_1$$ such that $$|f(x) - f(a)| < 1 \;\Rightarrow\; |f(x)| < M$$ (since f is continous, |f(x)| has an upper bound).

$$\delta_2$$ such that $$|g(x) - g(a)| < \frac{\epsilon}{2M}$$

Obviously, $|g(a)| < |g(a)| + 1$
$$\delta_3$$ such that $$|f(x) - f(a)| < \frac{\epsilon}{2(|g(a)| + 1)}$$

In conclusion, we take $\delta = \min(\delta_1, \delta_2, \delta_3)$
And get:
$$|f(x)||g(x) - g(a)| \;+\; |g(a)||f(x) - f(a)| <$$

$$(M\cdot \frac{\epsilon}{2M}) + (|g(a)| + 1\cdot \frac{\epsilon}{2(|g(a)| + 1)}) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

Quad Erad Demonstrandum.
Thanks for the help!

 

[andytoh]

To make the proof complete, you should explain that the +1 in epsilon/(2|g(a)|+1) is used because |g(a)| might be zero.

Also, you said "since f is continous, |f(x)| has an upper bound". Continuity alone is not enough, |f(x)| has an upper bound on what set? It certainly does not necessarily have an upper bound on R.
You can remove any doubt by getting an explicit expression for M in terms of f(a) by using the fact that |f(x)-f(a)| < 1 (which is easy). You see, |f(x)| is bounded not only because f is continuous but also because (a - delta1, a + delta1) is a subset of [a - delta1, a + delta1] and [a - delta1, a + delta1] is compact, but I don't think you should talk about this for this proof. Getting an explicit expression for M will remove any questions.