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Let $f:(a,c) \to \mathbb R$. Prove that if $f$ is uniformly continuous on $(a,b)$ and $(b,c)$, then $f$ is uniformly continuous on $(a,c)$.

My attempt at a solution:

I am having doubts if it is correct, is it right there has no condition that $f$ is continuous at $b$?

Let $ϵ>0$, we know that there exist $δ_1$, $δ_2$ such that if

$$x,y∈(a,b),\; |x−y|<δ_1 \Rightarrow|f(x)−f(y)|<ϵ_2$$

$$x,y∈(b,c),\;|x−y|<δ_2\Rightarrow|f(x)−f(y)|<ϵ_2$$

Let $δ=\min(δ_1,δ_2)$, if $|x−y|=|x−b+b−y|≤|x−b|+|b−y|<δ \Rightarrow$ $|f(x)−f(y)|≤|f(x)−f(b)|+|f(b)−f(y)|≤ϵ_2+ϵ_2=ϵ$.

But is it right? I think if $f$ is not continuous at $b$, it is not true.

Kyan Cheung
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jolly
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    Two points: (1) You write "Let $\epsilon>0$" and then use $\epsilon_1$ and $\epsilon_2$. What are they? (2) You state the epsilon-delta estimates for $x,y \in (a,b)$ and $(b,c)$ but use them for $x,y=b$ as well. – Gary Jun 07 '21 at 07:09

2 Answers2

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This is indeed clearly not true if

$$\lim\limits_{x \to b^-} f(x) \neq \lim\limits_{x \to b^+} f(x)$$

as $\lim\limits_{n \to \infty} (y_n - x_n) =0$ and $\lim\limits_{n \to \infty} (f(y_n) - f(x_n)) \neq0$ where $x_n = b -1/n$, $y_n = b +1/n$ are defined for $n$ large enough. This contradict uniform continuity criterion.

However, the result holds if $f$ is supposed to be continuous at $b$.

For the proof, you can separate the cases $x \lt y \lt b$, $x \lt b \lt y$ and $b \lt x \lt y$ and use "a third $\epsilon$-$\delta$" dealing with continuity of $f$ at $b$.

mathcounterexamples.net
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See the following function

$$ \quad f(x) = \begin{cases} x & x \in (0, 5)\cup(5,\infty) \\ 0 & \text{$x=5$} \end{cases} $$

Clearly this function is uniformly continuous on $(0,5)$ and on $(5,\infty)$ but not continuous on $(0,\infty)$

Lalit Tolani
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