Prove or Disprove -

a) If $f:[0,1) \rightarrow \mathbb{R}$ is uniformly continuous in it's domain, $f$ is bounded.

b) If $f:[0,1] \rightarrow \mathbb{R}$ is uniformly continuous in it's domain, $f$ is bounded.

Attempt -

It feels like $b$ might be true and $a$ mustn't. Maybe some manipulation of the definition of uniformly continuity. I do want to say that because $f$ is bounded in both cases because we are talking about a bounded domain. But how should I proceed ? Thank you!

  • 2
    $\begingroup$ Every continuos function on a compact intervall is bounded by it's maximum/minimum. $\endgroup$ – Jannik Pitt Feb 15 '18 at 15:35

Any function which is uniformly continuous in an open bounded interval $(a,b)$ can be extended to a uniformly continuous function in $[a,b]$ (see for example Show for $f:A \to Y$ uniformly continuous exists a unique extension to $\overline{A}$, which is uniformly continuous). Hence since any continuous function on a compact set is bounded, it follows that both (a) and (b) are true.


A more general result that might help. I think this was an exercise in baby Rudin.

If $f: A \subset \mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous on $A$, and $A$ is bounded, then $f(A)$ is bounded.

One way to prove this is as follows: Given $\epsilon =1$, there exists some $\delta >0$ such that if $x, y \in A$, $$|x-y| < \delta \quad \Rightarrow \quad |f(x)-f(y)|<1$$ If $A$ is a bounded subset of $\mathbb{R}$, then there exists finitely many points $\{a_1,\dots,a_n\}$ in $A$ such that for every $x \in A$, there exists some $k =1,\dots,n$ such that $|x - a_k| < \delta$.

Let $M = \max \{|f(a_1)|,\dots,|f(a_n)|\}$. Then given any $x \in A$, $$|f(x)| \leq |f(a_k)| + 1 \leq M + 1$$


(b) Is because uniformly continuous implies continuous and this implies bounded if the domain is compact.

(a) Assume that it is unbounded. Take a sequence $x_n$ such that $f(x_n)\to\infty$. Without loss of generality $\to+\infty$. We must have $x_n\to 1$ because on compact subintervals of $[0,1)$ the function is bounded according to (b).

Therefore, $x_n$ gives a counterexample to uniform continuity. In fact, fixing say $\epsilon=1$ we can pick $n, M$ such that $|x_n-x_m|<\delta$ for all $m>M$ and then pick $m$ such that $f(x_m)>f(x_n)+2$. Therefore $|f(x_n)-f(x_m)|>\epsilon$. Since we are assuming that it is uniformly continuous, it follows that it can't be unbounded.

  • $\begingroup$ @orlp (a) is true. A counterexample doesn't exist. That function is not uniformly continuous. $\endgroup$ – user530891 Feb 15 '18 at 15:39
  • $\begingroup$ I read over the uniformly part of uniformly continuous, my bad :) $\endgroup$ – orlp Feb 15 '18 at 15:42
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    $\begingroup$ @ParamanandSingh No, if the $f(x_n)\to\infty$ the whole sequence $x_n$ must tend to $1$. Assume there is some subsequence $x_{n_m}\to a\neq 1$. Then $f(x_{n_m})$ is bounded. That contradicts that $f(x_n)\to\infty$. $\endgroup$ – user530891 Feb 15 '18 at 16:04
  • $\begingroup$ Ok got it. I will delete my previous comment then because it is no longer applicable. $\endgroup$ – Paramanand Singh Feb 15 '18 at 16:09

Both are true; the second one is trivially true since $[0,1]$ is compact and the continuous image of any compact set is still compact (in this case, since $[0,1]$ is closed and bounded, the image under $f$ will be closed and bounded).

The first one is true since $f$ can only vary by a bounded amount over a given $\Delta x$. Since $[0,1)$ is bounded, $f$ will have an upper bound.


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