Show that if the restriction f : [0, ∞) → R of f to the interval [0, ∞) is strictly decreasing, then f : R → R is strictly decreasing

Question

A function $f: \Bbb R\to \Bbb R$ is odd if $f(−x) = −f(x)$ for all $x\in\Bbb R$. Let $f:\Bbb R\to\Bbb R$ be an odd function. Show that if the restriction $f\vert_{[0,\infty]}: [0, \infty)\to\Bbb R$ of $f$ to the interval $[0, \infty)$ is strictly decreasing, then $f:\Bbb R\to\Bbb R $ is strictly decreasing.

My thought process is as follows:

Proof. Suppose there exist real numbers u and v such that v>u. We must consider three cases:

$\quad (\textrm{i})$ Suppose $v>u \geq 0$ then $f(v)<f(u)$, as $f\vert_{[0,\infty]}:[0, \infty)→ \mathbb{R}$ is strictly decreasing.

$\quad (\textrm{ii})$ If $u<v<0$, then $-u>-v>0$, and therefore $f(-u)>f(-v)$. Since $f$ is odd, $\quad f(v)=f(u)$.

$\quad (\textrm{iii})$ Suppose $v \geq 0$ and $u<0$. Since $f$ is odd, $f(0)=f(-0)$, and as $f:[0, \infty)→ \mathbb{R}$ is strictly $\quad$ decreasing, $f(v) \geq 0$ and $f(-u)>0$. Thus, because $f(-u)>0$, we have $-f(u)>0$, i.e. $\quad f(u)<0$. Therefore, $f(v) \geq 0 >f(u)$.

In all three cases, $f(v) \geq f(u)$, and so we see that $f$ is strictly decreasing. QED

I was hoping for proof verification, specifically when it comes to all these inequalities, which I've had a hard time wrapping my head around. I did this work last night, but I am not confident in it. Today it seems to me that it is incorrect.

Answer 1

That's it.

You have a typo at case ii: we don't have $f(v) =f(u) $ rather $$f(v) =-f(-v)\ <\ -f(-u) =f(u) $$

Answer 2

The proof is substantially good, with at least a typo.

Let me try to rephrase it.

Suppose $u,v\in\mathbb{R}$, with $u<v$. We have to show that $f(u)>f(v)$.

Case 1: $0\le u<v$. This follows directly from the assumption.

Case 2: $u<v\le0$. Then $0<-v<-u$, so by assumption $f(-v)>f(-u)$. As the function is odd, $f(-x)=-f(x)$; thus $-f(v)>-f(u)$ that implies $f(u)>f(v)$.

Case 3: $u<0<v$. For an odd function, $f(0)=f(-0)$, so $f(0)=0$. Since $f$ is decreasing over $(-\infty,0)$ by case 2, we have $f(u)>0$; since $f$ is decreasing over $(0,\infty)$, $f(v)<0$. Hence $f(u)>f(v)$.