证明满足极限相等的连续实值函数既非单射也非满射

Great question! Let's prove this conjecture rigorously using fundamental properties of continuous functions on $\mathbb{R}$.

Proving $f$ is not injective (1-to-1) #

We split the proof into two cases based on the value of the shared limit:

Case 1: $\lim_{x\to-\infty}f(x) = \lim_{x\to\infty}f(x) = L$ (where $L$ is a finite real number) #

By the Extreme Value Theorem, the continuous function $f$ attains a maximum $M_f$ and minimum $m_f$ on the closed interval $[-M, M]$ for some sufficiently large $M > 0$. For this $M$, we know $|f(x) - L| < 1$ whenever $|x| > M$ (by the definition of limits at infinity).

• If $M_f > L$: Since $f(x) \to L$ as $x \to \pm\infty$, there exist $x_1 < -M$ and $x_2 > M$ such that $f(x_1) = f(x_2) = \frac{M_f + L}{2}$ (by the Intermediate Value Theorem). Since $x_1 \neq x_2$, $f$ is not injective.
• If $m_f < L$: Similarly, there exist $x_1 < -M$ and $x_2 > M$ such that $f(x_1) = f(x_2) = \frac{m_f + L}{2}$, so $f$ is not injective.
• If $M_f = m_f = L$: $f$ is a constant function, which is clearly not injective (every input maps to the same output).

Case 2: $\lim_{x\to-\infty}f(x) = \lim_{x\to\infty}f(x) = +\infty$ (or $-\infty$) #

Let's use $+\infty$ first (the $-\infty$ case is symmetric):
Since $f$ is continuous and tends to $+\infty$ at both ends, it attains a global minimum $m_f$ on $\mathbb{R}$ (by the Extreme Value Theorem applied to a sufficiently large closed interval). For any value $y = m_f + 1$, there exist $x_1 < -K$ and $x_2 > K$ (for some large $K$) such that $f(x_1) = f(x_2) = y$ (again, by the Intermediate Value Theorem). Since $x_1 \neq x_2$, $f$ is not injective.

Proving $f$ is not surjective (onto) #

Again, we split into cases:

Case 1: $\lim_{x\to-\infty}f(x) = \lim_{x\to\infty}f(x) = L$ (finite) #

Take the same large $M$ as before, where $|f(x) - L| < 1$ for $|x| > M$. Let $M_f$ be the maximum of $f$ on $[-M, M]$, and $m_f$ be the minimum.

• If $M_f > L$: Choose a value $Y > \max(M_f, L+1)$. For $|x| > M$, $f(x) < L+1 < Y$; on $[-M, M]$, $f(x) \leq M_f < Y$. So $Y$ is not in the range of $f$, meaning $f$ is not surjective.
• If $m_f < L$: Choose $Y < \min(m_f, L-1)$. Similar logic shows $Y$ is never achieved by $f$.
• If $f$ is constant: Its range is just ${L}$, which clearly doesn't cover all real numbers.

Case 2: $\lim_{x\to-\infty}f(x) = \lim_{x\to\infty}f(x) = +\infty$ #

Since $f$ attains a global minimum $m_f$, any value $Y < m_f$ cannot be in the range of $f$ (because $f(x) \geq m_f$ for all $x \in \mathbb{R}$). Thus $f$ is not surjective.

For the $-\infty$ limit case: $f$ attains a global maximum $M_f$, so any $Y > M_f$ is outside the range of $f$, proving non-surjectivity.

Conclusion #

In all scenarios, a continuous function $f:\mathbb{R}\to\mathbb{R}$ with equal limits as $x\to\pm\infty$ is neither injective nor surjective.

内容的提问来源于stack exchange，提问作者Benji Altman