Simple proof of uniform continuity

[mrchris]

If the function f:D→ℝ is uniformly continuous and a is any number, show that the function a*f:D→ℝ also is uniformly continuous.

Ok, so I am just learning my proofs so be patient with me, I'm very new at it.

take a>0, ε>0 and x,y in D. We know |x-y|<δ whenever |f(x)-f(y)|<ε.
If we take a*f:D→ℝ, we have |a*f(x)-a*f(y)|<ε → |a*[f(x)-f(y)]|<ε→
a*|f(x)-f(y)|<ε→ |f(x)-f(y)|<ε/a. Therefore, if we use ε/a, the result is proven.

This just seems a little too easy to me, plus I've only done a few of these on my own. any suggestions/advice are greatly appreciated. Also, do I need to do this separately for a<0?

 

[Office_Shredder]

|x-y|<δ whenever |f(x)-f(y)|<ε.

You have this written backwards. As written, you're saying that if you pick values of x and y such that |f(x)-f(y)| is small, then |x-y| must be small as well. But this isn't true -for example if f is a constant function - and isn't what you want with uniform continuity. The actual statement is |f(x)-f(y)|<ε whenever |x-y|<δ (note that you need to state at some point that given epsilon, you can find delta such that this is true)Your next string of logic looks good until this point:

Therefore, if we use ε/a, the result is proven.

Use ε/a for what? To prove uniform continuity, you need: for any ε that I give you, you must produce δ such that if |x-y|<δ, then |a*f(x)-a*f(y)|<ε. Unless you're suggesting δ = ε/a you need to expand upon this to have a complete proof with all the details

Also, do I need to do this separately for a<0?

The a<0 case is handled by factoring the a out of the absolute value signs and getting an |a|, not just an a. Also the case a=0 isn't covered by the proof (but that's a pretty easy case)

 

[HallsofIvy]

Assuming that by a you mean a number and a*f is just multiplication of the function f by a, yes, it really is that easy!