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If $f,g$ are uniformly continuous, then is $\alpha f+\beta g$ uniformly continuous?
So far, I looked at here If $f,g$ are uniformly continuous prove $f+g$ is uniformly continuous, but I didn't understand something.
I know that from what I've been told,:
1. $\forall\epsilon >0$ $\exists\delta_1 >0$, if $|x-y|<\delta_1$ then $|f(x)-f(y)|<\epsilon$
2. $\forall\epsilon >0$ $\exists\delta_2 >0$, if $|x-y|<\delta_2$ then $|g(x)-g(y)|<\epsilon$

I need to show that for every $\epsilon$ there exists some $\delta$ such that if $|x-y|<\delta$ then $|(\alpha f+ \beta g)(x) - (\alpha f+ \beta g)(y)|<\epsilon$. So I know the method, I need to show that from $|(\alpha f+ \beta g)(x) - (\alpha f+ \beta g)(y)|<\epsilon$ - I need to do some manipulations and get $|x-y|<\delta$ (I need to find that $\delta$). So, why in the answers there it just shows that $|(\alpha f+ \beta g)(x) - (\alpha f+ \beta g)(y)|<\epsilon$?

Thanks!

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You need to point out $\delta$ before proving $|\alpha f(x)+\beta g(x)-\alpha f(y)+\beta g(y)|<\varepsilon$.

Indeed, suppose $\alpha\beta\neq 0$ (the cases $\alpha=0$ or $\beta=0$ are trivial). There exists $\delta_1, \delta_2$ such that $$\tag{1}|f(x)-f(y)|<\frac{\varepsilon}{2\alpha}, \forall x,y: |x-y|<\delta_1$$ $$\tag{2}|g(x)-g(y)|<\frac{\varepsilon}{2\beta}, \forall x,y: |x-y|<\delta_2$$

Choose $\delta=\min(\delta_1,\delta_2)$, then for all $x,y$ such that $|x-y|<\delta$: $$|\alpha f(x)+\beta g(x)-\alpha f(y)+\beta g(y)|\le \alpha|f(x)-f(y)|+\beta|g(x)-g(y)|<\varepsilon$$

Thus $\alpha f(x)+\beta g(x)$ is uniformly continuous.

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  • $\begingroup$ this is exactly what i talked about mate. i don't understand why in this proof we need to show that |(αf+βg)(x)−(αf+βg)(y)|<ϵ, but for example here f(x)=1/x in $[a,\infty)$ for $a>0$ you doing :$|f(x)−f(y)|=∣y−x/xy∣≤|y−x|/a2 $ and then you "choose" $\delta = a^2\epsilon$ and you finish. $\endgroup$ Jan 24, 2015 at 16:27
  • $\begingroup$ Yes, that's the way to find $\delta$! But when you write the solution, you have to choose $\delta$ before proving $|f(x)-f(y)|<\varepsilon$, based on the definition. So sometimes $\delta$ seems very unnatural. $\endgroup$ Jan 24, 2015 at 16:46
  • $\begingroup$ how can you find δ in your proof so? tnx a lot for answering by the way..! $\endgroup$ Jan 24, 2015 at 16:48
  • $\begingroup$ In your example, it's easy to choose $\delta$ since $f(x)-f(y)$ has the factor $x-y$. But in general, we just need to choose $\delta$ which satisfies both $(1)$ and $(2)$ (as I've just edited), so $\delta=\min(\delta_1,\delta_2)$ is the best choice. $\endgroup$ Jan 24, 2015 at 17:04

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