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Prove that $g \in o(f)$ implies $f + g \in \Theta(f)$.

I'm pretty stuck on this one. I know that there is a constant $c_2$ such that $f \le c_2 (f+g)$, but I'm not sure how to prove that there exists a constant $c_1$ such that $c_1 (f+g) \le f$. If I can prove that then I can say that $f + g \in \Theta(f)$. Anyone have some suggestions?

Daniel Fischer
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Patrick Perkins
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  • This is similar to your previous question asked a few days ago. Voted to close as a duplicate. – markvs Sep 21 '20 at 03:09
  • I don't see how they are the same question, they are both proofs yes, but not the same proof. – Patrick Perkins Sep 21 '20 at 03:26
  • The new statement easily follows from the previous one and can be proved exactly the same way. Also you have not explained in the previous question and in the new one what do you mean by $O, \Theta$ or $o$. – markvs Sep 21 '20 at 03:30
  • I just wrote my variant proof for https://math.stackexchange.com/questions/3832978/prove-that-f%e2%88%88%ce%98gif-and-only-if-of-og also and seems it's little different one. – zkutch Sep 21 '20 at 03:40
  • @PatrickPerkins: Please do not vandalize your own question (or anyone else's!), especially after you have received an answer. I've rolled your question back to its previous form. – Blue Sep 21 '20 at 04:41

1 Answers1

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Let's take definition for small-o for $x \to x_0$ $$o(f(x))=\{\phi(x): \exists \delta>0, \exists \varepsilon(x), \lim\limits_{x \to x_0}\varepsilon(x)=0, \forall x \in U_\delta(x_0),\phi(x) = f(x)\cdot \varepsilon(x) \}$$ Of course this definition works, with appropriate $U_\delta$ also for $n \to \infty$.

So $g=o(f)$ gives $g=\varepsilon f$, where $\varepsilon \to 0$. so $f+g=f(1+\varepsilon)$ and $1+\varepsilon \to 1$. This last gives you possibility to obtain $ (f+g) \le Cf$, which you want.

zkutch
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  • could you explain what you mean by $\varepsilon \to 0$ ? – Patrick Perkins Sep 21 '20 at 03:39
  • @Patrick Perkins. Added to answer. Also I suggest my view to your previous question https://math.stackexchange.com/questions/3832978/prove-that-f%e2%88%88%ce%98gif-and-only-if-of-og – zkutch Sep 21 '20 at 03:49
  • All dumb down voters are welcome to express constructive critique, especially on correct answer. – zkutch Sep 24 '20 at 22:24