Friday, 24 February 2017

real analysis - If f,g are uniformly continuous prove f+g is uniformly continuous but fg and dfracfg are not


Suppose f:RER and g:RER are uniformly continuous. Show that f+g is uniformly continuous. What about fg and fg?




My Attempt



Firstly let's state the definition; a function is uniformly continuous if
ε>0    δ>0  such that  |f(x)f(y)|<ε    x,yR  such that  |xy|<δ




Sum f+g



Now to to prove f+g is uniformly continuous;
Choose δ1>0 such that x,yR |xy|<δ1 |f(x)f(y)|<ϵ2



Choose δ2>0 such that x,yR |xy|<δ2 |g(x)g(y)|<ε2



Now take δ:=min{δ1,δ2} Then we obtain for all x,yR
|xy|<δ|f(x)+g(x)f(y)+g(y)|<|f(x)f(y)|+|g(x)g(y)|<ε2+ε2=ε






Product fg



Now for fg for this to hold both f:ER and g:ER must be bounded , if not it doesn't hold.

$\exists \ \ M>0 \ \ such \ that \ \ |f(x)|



Choose δ1>0 such that x,yR |xy|<δ1 |f(x)f(y)|<ϵ2M



Choose δ2>0 such that x,yR |xy|<δ2 |g(x)g(y)|<ϵ2M



Now take δ:=min{δ1,δ2}. Then, |xy|<δ implies for all x,yR, that
|f(x)g(x)f(y)g(y)||g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)|

M|f(x)+f(y)|+M|g(x)+g(y)|<Mϵ2M+Mϵ2M=ϵ



Are these proofs correct?
I am not sure how to approach the fg case.

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