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$f(x) = \frac{x}{1 +x^2}$ on $\mathbb{R}$ to comment about its uniform continuity, I tried to figure out the definition " A function $f(x)$ is said to be uniformly continuous on a set $S$, if for given $\epsilon > 0$, there exists $\delta > 0$ such that $x$, $y \in S$, $|x − y| < \delta ⇒ |f(x) − f(y)| < \epsilon$" but couldn't get where to start.

then i thought about breaking down the interval into some closed and open intervals and as $f(x)$ is continuous in closed bounded interval so it would be uniformly continuous as well. also i observed that limit $x$ tending to infinity and minus infinity exist and is equal to zero but don't know how to use it.

any hint would be highly appreciated

Gary
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Svidi Runs
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    Any continuous function on $\mathbb R$ which has finite limits at $\pm \infty$ is uniformly continuous. – Kavi Rama Murthy Feb 13 '24 at 05:06
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    Not as neat as geetha's solution, but you could also look to show that the derivative of $f$ is bounded. You can then use the mean value theorem to show that $\frac{|f(x) - f(y)|}{|x - y|}$ is bounded by the same constant, and uniform continuity follows readily from there. – Theo Bendit Feb 13 '24 at 05:23
  • See : https://math.stackexchange.com/questions/75491/how-does-the-existence-of-a-limit-imply-that-a-function-is-uniformly-continuous – Kavi Rama Murthy Feb 13 '24 at 06:26
  • @geetha290krm so similarly sinx/x can be solved as its limit at ±∞ exist and function is continuous on R – Svidi Runs Feb 13 '24 at 13:43

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It is enough to prove that $f$ is Lipschitz continuous. By simple calculation, $$ \left| {f(x) - f(y)} \right| = \frac{{\left| {xy - 1} \right|}}{{(x^2 + 1)(y^2 + 1)}}\left| {x - y} \right|. $$ Observe that $$ \left| {xy - 1} \right| \le |xy|+1=\sqrt {x^2 y^2 } + 1 \le \frac{{x^2 + y^2 }}{2} + 1 = \frac{{(x^2 + 1) + (y^2 + 1)}}{2}. $$ Whence $$ \frac{{\left| {xy - 1} \right|}}{{(x^2 + 1)(y^2 + 1)}} \le \frac{{(x^2 + 1) + (y^2 + 1)}}{{2(x^2 + 1)(y^2 + 1)}} = \frac{1}{{2(y^2 + 1)}} + \frac{1}{{2(x^2 + 1)}} \le \frac{1}{2} + \frac{1}{2} = 1. $$ Therefore, $|f(x)-f(y)|\le |x-y|$ for all $x,y\in \mathbb R$.

Gary
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  • thanks, i got it now. although i am new to the topic, i find difficulty in having intuitive idea about any function's uniform continuity. – Svidi Runs Feb 13 '24 at 10:08
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Hint

By MVT,

$f(x)-f(y)=f^\prime(c)(x-y)$

with $|f^\prime(c)|=\bigg{|}\dfrac{1-c^2}{(1+c^2)^2} \bigg{|}\leq \ \dfrac{1+c^2}{(1+c^2)^2}\ \leq \dfrac{1}{(1+c^2)} \leq 1$

follows $|f(x)-f(y)|\leq |x-y|$ for all $x,y \in \mathbb{R}$.

Akalanka
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