$f: (a, b] \to \mathbb{R}$ continuous. Prove: $f$ continuously resumable in $a \iff f$ is uniformly continuous

Question

Let $f: (a, b] \to \mathbb{R}$. Prove that: $f$ is continuously resumable in $a \iff f$ is uniformly continuous.

My approach:

"$\implies$:"

Since $a$ is continuously resumable, then $a$ has to be an accumulation point and $\lim_{x \to a} f(x)$ exists. I define another function $g$:

$$g: [a, b] \to \mathbb{R}, \left\{\begin{array}{lr} x \to f(x), & \text{for } x \in (a, b]\\ \lim_{z \to x} f(z), & \text{for } x = a\\ \end{array}\right\}$$

$g$ is continuous on $[a, b]$ so $g$ is also uniformly continuous on $[a, b]$ and therefore also uniformly continuous on $(a, b]$. This implies $f$ is also uniformly continuous on $(a, b]$.

"$\impliedby$:"

I think this direction is quite trivial to prove. Edit: This direction is not so trivial, as Danny Pak-Keung Chan's proof shows.

My question is: Is my proof correct? Did I made a too big jump from "$g$ is uniformly continuous in $[a, b]"$ to "$g$ is uniformly continuous in $(a, b]$"?

Answer 1

$\Leftarrow:$ Suppose that $f$ is uniformly continuous on $(a,b]$. We go to prove that we can define $f(a)$ such that $f:[a,b]\rightarrow\mathbb{R}$ is continuous. Let $(x_{n})$ be an arbitrary sequence in $(a,b]$ such that $x_{n}\rightarrow a$. We go to show that $(f(x_{n}))_{n}$ is a Cauchy sequence. Let $\varepsilon>0$ be given. Choose $\delta>0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $x,y\in(a,b]$ with $|x-y|<\delta$. Choose $N\in\mathbb{N}$ such that $|x_{n}-a|<\delta/2$ whenever $n\geq N$. For any $m,n\geq N$, we have that $|x_{m}-x_{n}|\leq|x_{m}-a|+|a-x_{n}|<\delta$, so $|f(x_{m})-f(x_{n})|<\varepsilon$. Therefore $(f(x_n))_n$ is a Cauchy sequence. By completeness of $\mathbb{R}$, $\lim_{n}f(x_{n})$ exists.

We go to show that $\lim_{n}f(x_{n})$ is independent of the choice of $(x_{n})$. For, let $(x_{n}')$ be another sequence in $(a,b]$ such that $x_{n}'\rightarrow a$. Choose $N_{1}$ such that $|x_{n}-a|<\delta/2$ and $|x_{n}'-a|<\delta/2$ whenever $n\geq N_{1}$. Note that for $n\geq N_{1}$, we have $|x_{n}-x_{n}'|\leq|x_{n}-a|+|a-x_{n}'|<\delta$. Therefore $|f(x_{n})-f(x_{n}')|<\varepsilon$. Letting $n\rightarrow\infty$ yields $|\lim_{n}f(x_{n})-\lim_{n}f(x_{n}')|\leq\varepsilon$. Hence, $\lim_{n}f(x_{n})=\lim_{n}f(x_{n}')$ because $\varepsilon>0$ is arbitrary.

Define $f(a)=\lim_{n}f(x_{n})$, where $(x_{n})$ is any sequence in $(a,b]$ such that $x_{n}\rightarrow a$. From the above discussion, $f(a)$ is well-defined. For the function $f:[a,b]\rightarrow\mathbb{R}$, we already know that it is continuous on $(a,b]$, so it remains to show that it is continuous at $a$. By Heine Theorem, it suffices that for any sequence $(x_{n})$ in $(a,b]$ such that $x_n\rightarrow a$, $f(x_{n})\rightarrow f(a)$, but this follows from the very definition of $f(a)$. Therefore, $f$ is continuous at $a$.

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$\Rightarrow:$ This direction is trivial. By assumption, we can define $f(a)$ such that $f:[a,b]\rightarrow\mathbb{R}$ is continuous. Since $[a,b]$ is compact, $f$ is uniformly continuous on $[a,b]$ (a result due to Cantor), and hence trivially uniformly continuous on $(a,b]$.

Answer 2

Suppose $f$ extends to a continuous function $g$ on $[a, b]$. Since $[a, b]$ is compact, the function $g$ is uniformly continuous. Therefore, so is its restriction $$ f = g|_{(a, b]}. $$

If $f$ is uniformly continuous on $(a, b]$, suppose $\{x_{n}\}_{n}$ is an arbitrary sequence in $(a, b]$ that converges to $a$. The uniform continuity of $f$ implies that the sequence $\{f(x_{n})\}_n$ is Cauchy, hence has a limit $y$, and also that this limit is the same for all sequences $\{x'_{n}\}_{n}$ in $(a, b]$ that converge to $a$. Therefore, letting $g$ agree with $f$ on $(a, b]$ and putting $g(a) = y$ gives us a continuous extension $g$ of $f$ to $[a, b]$.

Answer 3

For another approach to the problem of proving that if $f : (a, b] \to \mathbb{R}$ is uniformly continuous, then $\lim_{x \to a^+}$ exists, we can adapt a common proof that any Cauchy sequence in $\mathbb{R}$ has a limit.

First, observe that the uniform continuity implies that $f$ is "Cauchy as $x \to a^+$"; in other words, for every $\epsilon > 0$, there exists $\delta > 0$ such that whenever $x, y \in (a, a + \delta) \cap (a, b]$, then $|f(x) - f(y)| < \epsilon$. To see this, apply the uniform continuity to get $\delta > 0$ such that whenever $|x - y| < \delta$, we have $|f(x) - f(y)| < \epsilon$; and then note that whenever $x, y \in (a, a + \delta) \cap (a, b]$ we certainly have $|x - y| < \delta$.

Now, the fact that $f$ is Cauchy as $x \to a^+$ implies that $$\limsup_{x\to a^+} f(x) \overset{def}{=} \inf_{\delta > 0} \sup \{ f(x) | x \in (a, a + \delta) \cap (a, b] \}$$ and $$\liminf_{x\to a^+} f(x) \overset{def}{=} \sup_{\delta > 0} \inf \{ f(x) | x \in (a, a + \delta) \cap (a, b] \}$$ are both finite, and that the two numbers are equal. Furthermore, that implies that $\lim_{x\to a^+} f(x)$ exists and is equal to the common value of $\limsup_{x\to a^+} f(x) = \liminf_{x\to a^+} f(x)$. Here, the proofs of these facts will be analogous to the proofs mentioned previously for sequences.