# Counterexample of: "$|f_o(x)|>0$, for all $x>0$ iff $f_o(x)$ is injective" and validity of proof.

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: $$0 or $$0. 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.

• "Which means that $|f_o(x)|$ must go upward. Hence, $|f_o(x)|$ has to be at least monotonically increasing," -- The first statement is too vague to interpret, but the second statement is definitely false. The theorem that says "if $f$ is monotonic, then $\lim_{x\to\infty} f(x)$ exists as an extended real number" doesn't work in reverse, which is what you seem to be doing. Commented Sep 13, 2022 at 22:43
No, your claim is wrong. Take e.g. $$f(x)=xe^{-x^2}$$. Then, $$f_0(x)=f(x)>0$$ forall $$x>0$$ but $$f_0$$ is not injective.
There are a whole bunch of continuous functions $$f$$ such that $$f(x)$$ is nowhere equal to $$f(-x)$$ except at $$x=0$$. There is no reason to expect that they are all injective. For instance, any non-injective function that is positive for $$x>0$$ and negative for $$x<0$$ will serve as a counterexample.
• "bounded" is useless, and to "serve as a counterexample", your $f$ has to be chosen non-injective. Commented Sep 13, 2022 at 23:00