How to obtain $\theta=0$?

Question

Let $\{a_\epsilon\}_{\epsilon\in(-1,1)}\subset\mathbb{R}$, $\theta\in\mathbb{R}$, suppose that: $$ a_\epsilon=a_0+\epsilon\theta+o(\epsilon),\quad \epsilon\to 0,\tag{1}$$ and: $$ \inf_{\epsilon\in(-1,1)}a_\epsilon=a_0, \tag{2}$$ i want to prove that: $\theta=0$. My idea is the following, i pass to the inf in (1) to obtain, taking into account (2): $$ \pm \theta +\inf_{\epsilon\in(-1,1)}o(\epsilon)=0,$$ but i don't know how to treat: $\inf_{\epsilon\in(-1,1)}o(\epsilon)$, for get: $\theta=0$. Any help would be appreciated!

Answer 1

When $\epsilon$ tends to $0$ (being non equal to zero), you have $$\frac{a_{\epsilon}-a_0}{\epsilon} = \theta + o(1)$$

If $\epsilon$ tends to $0^+$, then the LHS is always non-negative because $a_{\epsilon} \geq a_0$ and $\epsilon > 0$. So you deduce that $\theta \geq 0$.

If $\epsilon$ tends to $0^-$, then the RHS is always non-positive because $a_{\epsilon} \geq a_0$ and $\epsilon < 0$. So you deduce that $\theta \leq 0$.

You get finally that $\theta =0$.

Answer 2

1. We can write $f(x)=o(g(x))$ as $x\rightarrow c \iff f(x)=g(x)q(x)$, with $q$ some function defined on $(-1,1)$ such that $q(x)\rightarrow 0$ for $x\rightarrow c$ (this can be taken as a definition, if you wish)
2. Now, $a_\epsilon=a_0+\theta\epsilon +\epsilon q_\epsilon\geq a_0$, for some $\{g_\epsilon\}_{\epsilon \in (-1,1)}$, so that $\epsilon(\theta+q_\epsilon)\geq 0$ for all $\epsilon \in (-1,1)$
3. For $\epsilon <0$, divide and get $\theta+q_\epsilon\leq 0$. Pass to the limit as $\epsilon \rightarrow 0^-$ and you have $\theta \leq 0$. Proceed analogously for $\epsilon >0.$