I have been wondering whether this function

$$f(x) = \sqrt{1-x^2} + \sqrt{x^2 -1} + 3$$

is uniformly continuous or not at its domain ($D = \{-1;1\}$). It's clearly a continuous function at $1$ and $-1$ since it satisfies the $\epsilon-\delta$ definition.

"$f$ is continuous at $c$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x − c| < \delta$ and $x \in A$ implies that $|f(x) − f(c)| < \epsilon$."

If $c$ is an isolated point of the domain $D$, then the continuity condition holds automatically since, for sufficiently small $\delta > 0$, the only point $x \in D$ with $|x − c| < \delta$ is $x = c$, and then $0 = |f(x) − f(c)| < \epsilon$. Thus, a function is continuous at every isolated point of its domain.

My question is whether this function is also uniformly continuous at $D$ since you could choose $\delta>2$ and it would satisfy the definition of continuity for $1$ and $-1$. Actually, the election of $\delta$ does not depend on $\epsilon$ since $|f(x) − f(c)|$ will always be $0$, as $f(1)-f(-1)=3-3=0$; $f(-1)-f(1)=3-3=0$; $f(1)-f(1)=3-3=0$; $f(-1)-f(-1)=3-3=0$.

Is it correct what I'm saying? Is $f$ uniformly continuous at its domain?

  • 1
    $\begingroup$ Yes, your reasoning is correct. It would be appropriate for you to move it to be an answer of your question and accept it. $\endgroup$ – Ceph May 5 '17 at 19:29
  • $\begingroup$ @Ceph I'm new to this site and don't know how to that. Thank you very much for your quick answer. $\endgroup$ – Facundo May 5 '17 at 19:34
  • $\begingroup$ In my old analysis texts they only considered points that are limit points of the domain, both for limits and continuity. It will depend on your textbook. $\endgroup$ – Henno Brandsma May 5 '17 at 22:18

The reasoning was correct and the function is actually uniformly continuous at its domain.


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