Suppose that $f: \mathbb{R} \to \mathbb{R}$ such that it is continuous everywhere.
Define $f_o(x) := \frac{f(x) - f(-x)}{2}$.
Fact:
Every function can be written in this form, and is the odd part of $f$.
I am trying to find a function such that $|f_o(x)|>0$ for all $x>0$, but f is not injective (or vice versa).
My proof is as follows:
Proof:
If $f_o$ is injective, then $|f_0(x)|>0$ for all $x>0$:
Suppose $f_o$ is injective.
Claim: Let $x_0 \in \mathbb{R}$. $f_o(x)=0$ iff $x=0$.
Proof: $f_o$ is continuous everywhere as well since it is a linear combination of two everywhere-functions: $f(x)$ and $f(-x)$. Now,
$$\begin{align*} f_o(x_0) = 0 &\iff f_o(x) = \frac{f(0) - f(0)}{2} = \frac{f(0) - f(-0)}{2}\\ &\iff f_o(x_0) = f_o(0) \\ &\iff x_0 = 0 \end{align*}$$
Therefore, if $x>0$, then $f_0(x) \ne 0$. Hence, $|f_o(x)| > 0$.
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If $|f_o(x)|>0$ for all $x>0$, then $f_o$ is injective:
Proof by contraposition:
Suppose that $|f_0(x)| > 0$ for all $x>0$. First, let's proof that $|f_o(x)|$ is strictly increasing, for all $x>0$.
Claim: $|f_o(x)|$ is strictly increasing. Proof:
Notice that either $\lim_{x \to \infty} |f_o(x)| = \infty$ or $\lim_{x \to \infty} |f_o(x)| = M > 0$. Which means that $|f_o(x)|$ must go upward. Hence, $|f_0(x)|$ has to be at least monotonically increasing, e.g., non-decreasing. So, if $\exists x_o \in \mathbb{R}^+$ such that $f_o(x) = f_o(x_0)$, then $0 = |f_o(x) - f_o(x_0)| \ge \bigg | |f_o(x)| - |f_o(x_0)| \bigg | > 0$, an impossibility. Hence, $|f_o(x)|$ is strictly increasing. Q.E.D.
Suppose that $x \ne y$ and both $x,y>0$. We either have: $0<x<y$ or $0<y<x$. In either case, $|f_o(x)|$ is strictly increasing. So, $0 < \bigg | |f_o(x)| - |f_o(y)| \bigg | \le |f_o(x) - f_o(y)|$. This implies that $f_o(x) - f_o(y) \ne 0$. Hence, $f_o(x) \ne f_o(y)$.
By contraposition, this implies that if $f_o(x) = f_o(y)$, then $x=y$.
Q.E.D.
Comments
Is the proof correct?
