function $f$ is continuous in $[0,\infty)$ and $\displaystyle\lim_{x \to \infty} f(x) = f(0)$, prove that $f$ is not injective in $[0,\infty)$

Question

function $f$ is continuous in $[0,\infty)$ and $\displaystyle\lim_{x \to \infty} f(x) = f(0)$, prove that $f$ is not injective in $[0,\infty)$.

$f$ is continuous in $[0,\infty) \implies$ $f$ is continuous at $x_0=0$ $\implies \displaystyle\lim_{x \to 0} f(x) = f(0) \implies \displaystyle\lim_{x \to 0} f(x) = \displaystyle\lim_{x \to \infty} f(x)$

How do i get to here?

$\exists a,b \ge 0$ such that $a\ne b$ for which $f(a)=f(b)$.

Can you please help?

Answer 1

Such a function must admit a global maximum on $[0,+\infty[$, say on the point $x_0\ge0$. If $x_0>0$ the conclusion follows immediately. What does happen if $x_0=0$?

Answer 2

Suppose that $f$ is not constant (if it is, then it is not injective). Then there is a $x_0$ such that $f(x_0)\neq f(0)$. But then we can assume without loss of generality that $f(x_0)>f(0)$. Since $\lim_{x\to\infty}f(x)=f(0)$, there is a $x_1>x_0$ such that $f(0)\leqslant f(x_1)\leqslant f(x_0)$. By the Intermadiate Value Theorem, there is a $x_2\in[0,x_1]$ such that $f(x_2)=f(x_1)$.

Answer 3

saying a continuous function is injective is equivalent to saying it's strictly monotonic.

so WLOG say it's strictly increasing, then for some $x, y$ such that $ 0\leq x < y ,\,$ you have $f(0)\leq f(x) < f(y)$

let $y \to \infty$ and you have $f(0) < f(0)$, contradiction.

$f$ can't be injective