Suppose f:R⊃E→R and g:R⊃E→R 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,y∈R such that |x−y|<δ
Sum f+g
Now to to prove f+g is uniformly continuous;
∙ Choose δ1>0 such that ∀ x,y∈R |x−y|<δ1 ⟹ |f(x)−f(y)|<ϵ2
∙ Choose δ2>0 such that ∀ x,y∈R |x−y|<δ2 ⟹ |g(x)−g(y)|<ε2
∙ Now take δ:=min{δ1,δ2} Then we obtain for all x,y∈R
|x−y|<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:E→R and g:E→R must be bounded , if not it doesn't hold.
∙ $\exists \ \ M>0 \ \ such \ that \ \ |f(x)|
∙ Choose δ1>0 such that ∀ x,y∈R |x−y|<δ1 ⟹ |f(x)−f(y)|<ϵ2M
∙ Choose δ2>0 such that ∀ x,y∈R |x−y|<δ2 ⟹ |g(x)−g(y)|<ϵ2M
∙ Now take δ:=min{δ1,δ2}. Then, |x−y|<δ implies for all x,y∈R, 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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