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I want to show $f$ is uniformly continuous on $(a,b)$ if and only if it is continuous and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist. I saw the if statement on Show f is uniformly continuous on $(a,b)$ if it is continuous and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist but I want to show if and only if. Here is what I've done:

Assuming $f$ is continuous on $(a,b)$ and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist, we want to show that $f$ is uniformly continuous on $(a,b)$.

Proof from the right

Let $\{x_n\}$ be an arbitrary sequence in $(a,b)$ and $x_0\in(a,b)$ be arbitrary s.t. $\{x_n\}\to x_0$ as $n \to \infty$.

$\lim\limits_{x\to a^+}f(x):=l\;\;\text{for some}\;\; l\in\Bbb{R}$ exists $\implies$ If $\{a_n\}$ is an arbitrary sequence in $(a,a+\delta)$ s.t. $a_n\to a$ as $n \to \infty$, then $f(a_n)\to l$ as $n \to \infty$.

$\lim\limits_{x\to b^-}f(x):=k\;\;\text{for some}\;\; k\in\Bbb{R}$ exists $\implies$ If $\{b_n\}$ is an arbitrary sequence in $(b-\delta,b)$ s.t. $b_n\to b$ as $n \to \infty$, then $f(b_n)\to k$ as $n \to \infty$.

By continuity of $f$, we have that $x_n\to x_0$ as $n \to \infty\implies f(x_n)\to f(x_0)$ as $n \to \infty$. Hence, $f$ is uniformly continuous on $(a,b)$

My question is: I'm I right? Different kinds of sequential proofs, even $\epsilon-\delta$ proofs and criticisms are welcome!

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Let $f:(a,b)\to\mathbb R$ be a function such that $\lim_{x\to a+}f(x)$ does not exist.

It is our aim to prove that $f$ is not uniformly continuous on $(a,b)$.

The absence of the limit implies the existence of a strict monotonically decreasing sequence $(x_k)_k$ in $(a,b)$ with $\lim_{k\to\infty} x_k=a$ such that $\lim_{k\to\infty} f(x_k)$ does not exist.

Then $(f(x_k))_k$ is not a Cauchy-sequence, so some $\epsilon>0$ exists such that for every $n$ we can find integers $m,k$ with $m,k>n$ and $|f(x_k)-f(x_m)|>\epsilon$.

This tells us that $f$ is not uniformly continuous on $(a,x_n]$ hence is not uniformly continuous on $(a,b)$.

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I would prove that statement as follows: define$$\begin{array}{rccc}F\colon&[a,b]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\lim_{x\to a^+}f(x)&\text{ if }x=a\\f(x)&\text{ if }x\in(a,b)\\\lim_{x\to b^-}f(x)&\text{ if }x=b.\end{cases}\end{array}$$Then $F$ is continuous and therefore uniformly continuous. Since $f=F|_{(a,b)}$, $f$ is uniformly continuous too.

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  • $\begingroup$ It is unclear to me what the OP wants. This part was proved allready in an answer to the question in the link provided by the OP. So you would expect that he/she wants a proof for the other side. $\endgroup$ – drhab May 10 '18 at 8:41
  • $\begingroup$ @drhab Let us wait and see what the OP has to say about this. I will delet my answer if I misinterpreted his or her intentions. $\endgroup$ – José Carlos Santos May 10 '18 at 8:45
  • $\begingroup$ Okay. I provided an answer for the other side. $\endgroup$ – drhab May 10 '18 at 9:04
  • $\begingroup$ Yes, you both are rightly answered my question but I wanted to see if I could prove it in my own way. Sir José Carlos Santos used unique extension while Sir drhab used contradiction. I like both approach but is it impossible that I move forward from where I stopped? $\endgroup$ – Omojola Micheal May 10 '18 at 9:14
  • $\begingroup$ @Mike I don't see how. Actually, I don't know what you are trying to do. Besides, there's a problem with approach: you mention $f(a)$ and $f(b)$, but $f$ is not defined there. $\endgroup$ – José Carlos Santos May 10 '18 at 9:20
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If you need a more direct proof, you can use the following argument.

Let $(x_n)$ be a sequence converging to $a$. We claim that $(f(x_n))$ is a Cauchy sequence. Indeed, take $\varepsilon > 0$. By uniform continuity, there is $\delta > 0$ such that for any $x, \tilde{x} \in (a, b)$, if $\lvert x - \tilde{x} \rvert < \delta$ then $\lvert f(x) - f(\tilde{x}) \rvert < \varepsilon$. The sequence $(x_n)$ is a Cauchy sequence: consequently there exists $N$ such that $\lvert x_n - x_m \lvert < \delta$ for all $m, n \ge N$, which gives that $\lvert f(x_n) - f(x_m) \lvert < \varepsilon$ for all $m, n \ge N$. Consequently, from the claim it follows that the sequence $(f(x_n))$ converges to some real number.

Now, for any two sequences, $(x_n)$, $(\tilde{x}_n)$, converging to $a$, the sequence $$ \bar{x}_n = \begin{cases} x_k & \text{ if } n = 2k - 1 \\ \tilde{x}_k & \text{ if } n = 2k \end{cases} $$ converges to $a$, and $(x_n)$, $(\tilde{x}_n)$ are its subsequences. Consequently, both $(f(x_n))$ and $(f(\tilde{x}_n))$ are subsequences of $(f(\bar{x}_n))$, so all three of them must have the same limit. This completes the proof of the fact that $\lim\limits_{x \to a^+}f(x)$ exists (as a real number).

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