This website should help.
go by the definition. there may be a shorter way, but i can't remember now.
Defintion: Let be a real valued function defined on a set . Then is uniformly continuous on if
for each there exists such that for all and we have .
I will start you off in the proof.
Proof: Let be uniformly continuous on some interval .
Then we have that for each there exists such that for all and we have .
A similar thing holds true for if we choose some .
Now let , fix such an and choose .
Then, for all and , we have:
finish it off
that's what i originally thought of using, but i saw that the first theorem (if applicable) would get the job done a lot faster, with a lot less work.
for the benefit of the class, here's the new theorem we are proposing:
Theorem: Let be a continuous function on an interval ( may be bounded or unbounded, hehe, something my last theorem was missing). Let be the interval obtained by removing from any endpoints that may have. If is differentiable on and if is bounded on , then is uniformly continuous on .
I'm still very stuck on #1. I hate proving unif. cont.
I attempted #2, and I'd appreciate if you could look at it for me:
Let be given. Then, , thus, since is unif. convergence to on . Similarly for g, . Let M = max{R,P}. Now, for , we have:
Thus, is unif. cont. on A.
Grr, I mis-read it. I proved that they are uniformly CONVERGENT, not uniformly CONTINUOUS!!
Well, I know how to prove f is unif. continuous but not f + g...
So proving f is unif. continuous...:
Let (f_n) be the sequence of functions where are unif. cont. on A and that [tex](f_n) converges unif. to f on A. Let be given. Then, a natural number by unif. conv. of to f on A. Particularly, .
As is unif. cont on A, whenever satisfy . Then we have for this , suppose satisfy and consider:
\begin{eqnarray}
|f(x) - f(y)| &=& |f(x) - [f_{N}(x) - f_{N}(y)] - [f_{N}(y) - f_{N}(y)] - f(y)|\\
&=& |[f(x) - f_{N}(x)] + [f_{N}(x) - f_{N}(y)] + [f_{N}(y) - f(y)]|\\
&\leq& |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(y)| + |f_{N}(y) - f(y)|\\
&<& \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{eqnarray}
Thus, f is unif. cont. on A.
(NOTE: I get LaTeX syntax error on the last part (for the eqnarray stuff, not sure why) so I left it without [tex] so you can see the general jist. I think it's easy to read just in latex format.
But, I have no idea how to show f + g will be unif cont...
firstly, you do not have to prove (nor do you have enough information to do so) that f is uniformly continuous, that was given. secondly, we are not dealing with a sequence of functions here. were that the case, we would be concerned with uniform convergence, not uniform continuity.
your proof should start EXACTLY as mine did (you can neglect to define the new function as h(x) if you wish), all you have to do is complete my proof