So I understand the definition of uniform continuity, but wanted some suggestions to prove that a function is or isn't uniformly continuous.
I have looked ahead and have seen that if a function is continuous on a compact domain, then the function is also uniformly continuous. But for the purposes of the examples below, I cannot use compactness, or the fact that a function is uniformly continuous if it is continuous and has a bounded derivative.
Here's one example: prove $f(x)= \sqrt{x}$ is uniformly continuous on $(0, \infty)$. I have seen this question asked before on this forum, and the respondents used the traditional definition. Ie. they found a delta such that if $|x-y|<\delta$, then $|\sqrt{x}-\sqrt{y}|< \epsilon$.
I was wondering if I could try this in a slightly different way. Suppose I prove that $f$ is continuous at $x_0$, and the choice of $\delta$ doesn't depend on $x_0$.
After some scratch work, I proved that $f(x)$ is continuous at $x_0$ provided that $|x-x_0|< \min\{1,(\sqrt{x_0+1} + \sqrt{x_0})*\epsilon\}$. Since $x_0$ is always greater than $0$, we can say that $\delta= \min\{1, \epsilon\}$. Because I proved that our choice of $\delta$ doesn't depend on $x_0$, is this a valid proof that $f(x)$ is uniformly continuous?
Here's another example: I want to prove that $f(x)=x^2$ is not uniformly continuous on $(0, \infty)$. After some work, I proved that $f(x)$ is continuous at $x_0$ (in dom $f$), provided that $\delta< min \{1, \frac {\epsilon}{ (2x_0+1)}\}$. Here, since our choice of delta does depend on $x_0$, we can say the function is not uniformly continuous.
So am I approaching these problems correctly? I realize that I have not been using the definition of uniformly continuity. (that if $f$ is uniformly continuous, then if $|x-y|<\delta$, then$|f(x)-f(y)|< \epsilon$.
Also, sorry about the formatting. I realize I should learn proper formatting soon!. Also, I have read the following post regarding this topic: $\sqrt x$ is uniformly continuous
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$\begingroup$First of all, you make a mistake many beginners make: when talking about uniform continuity, it is essential to tell on what set you are studying it. For instance, $x^2$ is uniformly continuous on $[0,r]$ for every $r>0$, but is not so on $[0,\infty)$. Similarly, $\sin \frac 1 x$ is uniformly continuous on $[r,\infty)$ for every $r>0$, but is not so on $(0,\infty)$. The domain of definition is crucial.
Second, there are two main alternative techniques to plainly using the definition:
the first one is quick: if $f$ is defined on a compact set and is continuous, then it is uniformly continuous
the second one is subtler: if $f$ is derivable with $f'$ continuous and you manage to find a $M>0$ such that $\sup |f'(x)| \le M$ on your domain of definition, then $f$ will be Lipschitz and, in particular, uniformly continuous.
This would be a perfectly fine argument, if your $\delta$ were correct. You really are using the definition of uniform continuity, which is precisely that one can pick for any $\epsilon$ a $\delta$ independent of $x_0$. Unfortunately, your $\delta$ is wrong for $\sqrt x$. If $x=10^{-2},y=10^{-4},$ then certainly $|x-y|<10^{-2}$ and $|x-y|<1$, but $|\sqrt(x)-\sqrt(y)|>10^{-2}$. Your numbers are fine on $[1,\infty)$, but that's by far the easier part!
As for $x^2$, all you've done is show that there is some choice of $\delta$ which depends on $x_0$. That's a far cry from showing that every choice of $\delta$ depends on $x_0$, so you have not shown that $x^2$ fails to be uniformly continuous.
$\endgroup$ 2 $\begingroup$As you have an answer for the first case, I only give you a sketch for $x^2$.
I leave you to check: $f$ is uniformly continuous iff for all $x_n-y_n\to 0\implies f(x_n)-f(y_n)\to 0$. For the case $x^2$ is as simple as pick two sequences whose difference converges to $0$ but the difference of its images does not.
Consider $x_n=n+1/n$ and $y_n=n$. So $x_n-y_n\to 0$ but $x_n^2-y_n^2=2+1/n^2\not\to0$. Thus $x^2$ is not uniformly continuous.
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