Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function on all its domain. The following results hold for any $a, b \in \mathbb{R}^{+}$:
- $a=b \Leftrightarrow f(a)=f(b)$.
- $a>b \Leftrightarrow f(a)<f(b)$.
- $a<b \Leftrightarrow f(a)>f(b)$.
The three results above clearly mean $f$ is strictly decreasing. Keeping this apart, I would like to convey these 3 results and wonder whether there is a more compact way of doing this, rather than having to write the three double implications above.
One possibility would be of course showing one of the three results implies the other two. If I take (3), for example, by contraposition I can express ($\Rightarrow$) equivalently as $f(a)\leqslant f(b) \Rightarrow a\geqslant b$, and analogously I can write ($\Leftarrow$) equivalently as $a\geqslant b \Rightarrow f(a)\leqslant f(b)$. Plugging these two results leads to
\begin{equation} \label{eq1} a \geqslant b \Leftrightarrow f(a) \leqslant f(b). \quad\quad Eq.(4) \end{equation}
However, because of the "greater/smaller OR than" form above, Eq.(4) can be true if $a=b$ and $f(a)<f(b)$. This means that presenting the result (3) alone (i.e., without (1) and (2)) wouldn't be conveying the "strictness" form of (1) and (2), and I don't want this to happen.
The same would occur if I took (1). Following the same “contraposition” reasoning, I would end up having $a\neq b \Leftrightarrow f(a)\neq f(b)$, which isn’t as precise as (2) and (3).
What can I do? Is there any way of writing the results (1), (2) and (3) more compactly?
As noted in a comment, 2. and 3. are equivalent. [proof: in either 2 or 3, interchange the letters a,b and use the fact that $x<y \Leftrightarrow y>x.$]
We claim that 1. is a consequence of 2. and 3. First, the forward direction of 1. holds simply by logic, given that $a=b$ means that $a,b$ denote the same object. For the reverse direction, we assume that $f(a)=f(b)$ and apply the trichotomy law to $a$ and $b$ : If it happens that either $a>b$ or $a<b$ then since we are assuming 1. and 2. we arrive at $f(a) \neq f(b),$ a contradiction. The only possibility of the trichotomy which remins is that $a=b,$ finishing the proof of the reverse direction of 1.