About the additive property of little-o(h)

Question

So I encountered this definition in Salas Hille Etgen's One and Several Variables Calculus:

(This definition is for single variable case)

Definition: Let $g:\Bbb R \to \Bbb R$ be a function defined at least in some neighbourhood of $0$. We say that $g(h)$ is $o(h)$ and write $g(h)=o(h)$ to indicate that $\lim_{h\to 0}\frac{g(h)}{h}=0$.

So, after this definition, our class soon introduced a theorem:

Theorem: The following are equivalent:

$1. f(x+h)-f(x)-f'(x)h=o(h)$

$2. f(x+h)-f(x)=f'(x)h+o(h)$

And my professor gave us such proof:

$f(x+h)-f(x)=[f(x+h)-f(x)-f'(x)h]+f'(x)h=o(h)+f'(x)h$

So I began to doubt: What makes it legal to operate $o(h)$ as a normal function? Although it seems so natural, we still added something wasn't in the original definition. It is not an equation, it is a sentence, isn't it? Since the original definition write $g(h)=o(h)$ for merely implying $\lim_{h\to 0}\frac{g(h)}{h}=0$, is it okay to treat $o(h)$ like another number or function(e.g. replace $g(h)$ with $o(h)$ in the proof)?

Just like the definition of the notation of limit, it doesn't imply that it is a number. It is a sentence(description) replaced by abbreviations through the definition.

Hence, my question are summed up in 2:

1. Is it legal to operate $o(h)$ in such manner?
2. If not, is there another way to go through this proof?

Thanks in advance.

Answer 1

I know that writing $g(h)=\mathrm o(h)$ is a thing, but I think it is one of the most misleading notations ever $-$ when used without further explanation. So I will try to give the explanations: define $\mathrm o(h)$ as a set of functions:

$$\mathrm o(h):=\left\{f:\Bbb R\to\Bbb R \mid \lim_{h\to 0} \frac{f(h)}h = 0\right\},$$

and then $g\in \mathrm o(h)$ is what formerly was denoted by $g(h)=\mathrm o(h)$. Now we add a notational convention: any expression that contains $\mathrm o(h)$ as if it where a function is to be interpreted as "there is a function $g\in\mathrm o(h)$ replacing this occurence of $\mathrm o(h)$". So e.g. $f(x+h)-f(x)-f'(x) = \mathrm o(h)$ means

$$\forall x:\exists g_x\in\mathrm o(h):f(x+h)-f(x)-f'(x)=g_x(h).$$

Note that this might be a different $g_x$ for every $x$ (the derivative is defined point-wise). Now it might be clear why we can handle this "function" $\mathrm o(h)$ as if it where a real function. The right side above is obviously equivalent to

$$\forall x:\exists g_x\in\mathrm o(h): f(x+h)-f(x)=f'(x)+g_x(h).$$

And by our notational convention, we can write this as $f(x+h)-f(x)=f'(x)+\mathrm o(h)$.

Answer 2

I believe this is your definition (assuming, some definition of $o(\cdot)$) $$ a = b + o(c) \text{ iff there exists } d \text{ such that } d=o(c), \text{ and } a-b=d$$

So the proof- let $x$ be fixed. Then note that we can always define $R(h)$ to force the following two equalities at the same time, $$f(x+h) - f(x) - f'(x) h = R(h) \iff f(x+h) - f(x) = f'(x) h+ R(h)$$ Then statement 1. makes the claim that in the LHS equation, $R(h) = o(h)$. The statement 2. (assuming you meant f'(x)h and not f'(x)) asserts again that $R(h) = o(h)$ but starting from the RHS. That these are equivalent is the above $\iff$ arrow.