1. ## Uniform Continuity Proof

Let f, g be uniformly continuous functions on (a,b). Show that the function f+g is uniformly continuous on (a,b) by using the epsilon - delta definition.

I'm pretty sure it will involve something along the lines of |x-y| => |f(x)+g(x)-f(y)-g(y)|< epsilon and then some triangle inequality of something.

Thanks

2. Originally Posted by tbyou87
Let f, g be uniformly continuous functions on (a,b). Show that the function f+g is uniformly continuous on (a,b) by using the epsilon - delta definition.

I'm pretty sure it will involve something along the lines of |x-y| => |f(x)+g(x)-f(y)-g(y)|< epsilon and then some triangle inequality of something.

Thanks
$\displaystyle |f(x)+g(x) - f(y) - g(y)| \leq |f(x) - f(y)| + |g(x) - g(y)|$.
And each one can be made arbitrary small.

3. Complete Proof:
Since f is uniformly continuous on (a,b),
for every eps1>0, there exists delt1>0 s.t. x,y ele (a,b) and |x-y|< delt1 => |f(x)-f(y)|< eps1
Since g is uniformly continuous on (a,b),
for every eps2>0, there exists delt2>0 s.t. x,y ele (a,b) and |x-y|< delt2 => |g(x)-f(y)|< eps2
Let eps = eps1 + eps2 >0 and let delt = delt1 + delt2.
Then |x-y|<delt1+delt2<delt => |f(x)-f(y)+g(x)-g(y)|<= |f(x)-f(y)|+|g(x)-g(y)|< eps1+eps2=eps
Thus f+g is uniformly continuous.

Is this ok?

4. Originally Posted by tbyou87
Complete Proof:
Since f is uniformly continuous on (a,b),
for every eps1>0, there exists delt1>0 s.t. x,y ele (a,b) and |x-y|< delt1 => |f(x)-f(y)|< eps1
Since g is uniformly continuous on (a,b),
for every eps2>0, there exists delt2>0 s.t. x,y ele (a,b) and |x-y|< delt2 => |g(x)-f(y)|< eps2
Let eps = eps1 + eps2 >0 and let delt = delt1 + delt2.
Then |x-y|<delt1+delt2<delt => |f(x)-f(y)+g(x)-g(y)|<= |f(x)-f(y)|+|g(x)-g(y)|< eps1+eps2=eps
Thus f+g is uniformly continuous.

Is this ok?
Sorry for being a little late.

No let $\displaystyle \delta = \min \{ \delta_1 , \delta_2 \}$.