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Problem:
Let $g : \mathbb{R} \to \mathbb{R}$ be continuous, and let \begin{align} g(x) \to a \quad &\text{for} \quad x \to -\infty, \\ g(x) \to b \quad &\text{for} \quad x \to \infty. \end{align} Prove that $g$ is uniformly continuous.

My attempt at a solution:

The first limit says that $$ \forall x \in \mathbb{R} \ \forall\varepsilon > 0 \ \exists A < 0 : x < A \ \Rightarrow \ |g(x) - a| < \varepsilon, $$ and the second limits says that $$ \forall x \in \mathbb{R} \ \forall \varepsilon > 0 \ \exists B > 0 : x > B \ \Rightarrow \ |g(x) - b| < \varepsilon. $$ The way I have attempted to solve this is by looking at the interval $[A, B]$ and diving the problem into cases. If I can prove that $g$ is uniformly continuous for wherever $x$ and $y$ are in $\mathbb{R}$, then I will be done.

  1. $g$ is uniformly continuous in $x, y \in [A, B]$ by the Heine-Cantor theorem.

  2. For $x, y < A$ or $x, y < B$ (outside of $[A, B]$), that $g$ is uniformly continuous follows from using the limits and then the triangle inequality:

\begin{align} x, y &< A \ \Rightarrow \ |g(y) - g(x)| \leq |g(y) - a| + |a - g(x)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \\ x, y &> B \ \Rightarrow \ |g(y) - g(x)| \leq |g(y) - b| + |b - g(x)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align}

  1. For $x \in [A, B]$ and $y > B$ (or $y < A$) where one is inside the interval and the other is outside, I don't know what to do.

I have tried using the triangle inequality twice:

\begin{align} |g(y) - g(x)| = |g(y) - g(B) + g(B) - g(x)| &\leq |g(y) - g(B)| + |g(B) - g(x)| \end{align} Since $x, B \in |A, B|$, $|g(B) - g(x)| \leq \frac{\varepsilon}{3}$. Using the triangle inequality on the first term, we get $$ |g(y) - g(B)| = |g(y) - b + b - g(B)| \leq |g(y) - b| + |b - g(B)|... $$ I am stuck here and have no idea if this is what I am supposed to do.

Edward Chen
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  • Remember you can control your $\delta$, so instead of taking the interval $[A,B]$, take $[A-1,B+1]$, and take $\delta>0$ small enough so that if $x$ is outside of $[A,B]$, it is still in $[A-1,B+1]$. – GBA Feb 28 '25 at 18:41
  • Would this work for arbitrary length of intervals? Say, I could choose any set $[A', B']$ where $A' \leq A$ and $B' \geq B$? And then just repeat the first part with the Heine-Cantor theorem? – Edward Chen Feb 28 '25 at 18:45

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