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)| &lt; 1 \;\Rightarrow\; |f(x)| &lt; M (since f is continous, |f(x)| has an upper bound).

\delta_2 such that |g(x) - g(a)| &lt; \frac{\epsilon}{2M}

Obviously, |g(a)| &lt; |g(a)| + 1
\delta_3 such that |f(x) - f(a)| &lt; \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)| &lt;

(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.