Two Definitions of Big-O notation when Doing Addition?

Question

I see the expression

$$ (f(x) + g(x))^k = O(f^k(x)) + O(g^k(x)) (x \in S) $$ for any set $S$.

in the Introduction section of Asymptotic Methods in Analysis by N.G. de Bruijn and wonder if the proof it provides is consistence to the definition it gave previously.

The definition it provided earlier would read the equality as:

There exists functions $h, l$ such that $h(x) = O(f^k(x))$ and $l(x) = O(g^k(x))$ $(x \in S)$ with $(f(x) + g(x))^k = h(x) + l(x)$.

However, the proof that it provides only shows

$|(f(x) + g(x))^k| \leq 2^k |f(x)|^k + 2^k |g(x)|^k$ for $x \in S$.

But what would be the functions $h$ and $l$ in this case following the previous definition?

Doubts follow next: When one says $f(x) = O(g(x)) + O(h(x))$ for $x \in S$, then is it always enough to simply show $|f(x)| \leq A|g(x)| + B|h(x)|$ for some constants $A, B$? Or do we have to find explicitly a decomposition of $f$ with two functions that are $O(g)$ and $O(h)$ respectively? Or are these two ways somehow the same?

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Screenshots from the book:

Previous Definition:

Proof Later on:

Answer 1

Let $f, g, h : S \to \mathbb{R}$. We show that the followings are equivalent:

1. There exist two constants $A, B \geq 0$ such that $|f| \leq A|g| + B|h|$ on $S$.
2. There exist two functions $a, b : S \to \mathbb{R}$ such that $f = a + b$, $a = \mathcal{O}(g)$, and $b = \mathcal{O}(h)$.

Proof. $(2) \implies (1)$ is trivial. So it suffices to prove $(1) \implies (2)$. To this end, assume

$$ |f(x)| \leq A|g(x)| + B|h(x)| \tag{*}$$

holds for all $x \in S$. Then we choose a constant $C$ so that $C > A$, and define $a, b : S \to \mathbb{R}$ by

$$ a(x) = \begin{cases} f(x), & |f(x)| \leq C|g(x)| \\ 0, & |f(x)| > C|g(x)| \end{cases}, \qquad b(x) = \begin{cases} 0, & |f(x)| \leq C|g(x)| \\ f(x), & |f(x)| > C|g(x)| \end{cases} $$

From the definition, it is clear that $f = a + b$ and $a = \mathcal{O}(g)$. To show $b = \mathcal{O}(h)$, it suffices to investigate the case $|f(x)| > C|g(x)|$. Assuming this and invoking the assumption $\text{(*)}$, we get $|g(x)| \leq \frac{B}{C - A}|h(x)|$. Plugging this back to $\text{(*)}$, we get

$$ |b(x)| = |f(x)| \leq \left(\tfrac{AB}{C-A} + B\right)|h(x)|. $$

This is enough to conclude $b = \mathcal{O}(h)$.

Answer 2

Here we follow closely the text given in N.G. De Bruijn's book.

We want to show \begin{align*} \color{blue}{(f(x)+g(x))^k=\mathcal{O}((f(x))^k) + ((g(x))^k)\qquad (x\in S)}\tag{1} \end{align*}

Looking at OPs screenshot and the text which follows \begin{align*} x^{-1}\mathcal{O}(1)=\mathcal{O}(1)+\mathcal{O}(x^{-2})\qquad (0<x<\infty) \end{align*} we see:

We have to show $\left(f(x)+g(x)\right)^k$ can be split into two parts $h(x)$ and $l(x)$ such that \begin{align*} h(x)&=\mathcal{O}\left((f(x))^k\right)\qquad (x\in S)\\ l(x)&=\mathcal{O}\left((g(x))^k\right)\qquad (x\in S) \end{align*} The author has already shown after (1.2.12) that \begin{align*} \color{blue}{|(f+g)^k|}&\leq \left(|f|+|g|\right)^k\leq (2\max\left(|f|,|g|\right))^k\\ &\leq 2^k\max(|f|^k,|g|^k)\leq \color{blue}{2^k(|f|^k+|g|^k)}\tag{2} \end{align*} and we are done, since we can take $\color{blue}{h(x)=(f(x))^k}$ and $\color{blue}{l(x)=(g(x))^k}$.

Conclusion: We have found in (2) a relation \begin{align*} |(f(x)+g(x))^k|\leq A|f(x)|^k+B|g(x)|^k\qquad (x\in S) \end{align*} with $A=B=2^k$ a constant not depending on $x$. This is consistent with the definition of formula (1.2.5) given by the author, namely

Definition: If $S$ is any set, and if $f$ and $\varphi$ are real or complex functions defined on $S$, the formula \begin{align*} f(x)=\mathcal{O}(\varphi(s))\qquad (s\in S)\tag{1.2.5} \end{align*} means that there is a positive number $A$, not depending on $S$, such that \begin{align*} |f(x)|\leq A|\varphi(s)|\qquad \forall s\in S\tag{1.2.6} \end{align*}

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A final note with respect to your doubt: When you have found $|f(x)| \leq A|g(x)| + B|h(x)|$ for some constants $A, B$, then you are done. This is already an explicit decomposition since $g=\mathcal{O}(g)$ and $h=\mathcal{O}(h)$ where we can take constants equal to $1$.