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Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...

Proof

$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| $$

By the triangle inequality

$$\frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| \leq \left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \tag{$\star$}$$

Note that for all $x\in \mathbb{R}$

$$\left|x\right| < 1 +x^2 \implies \left|x\right| < \left(1 +x^2\right)\left(1+y^2\right)$$

Therefore

$$\frac{\left|x\right|}{\left(1 +x^2\right)\left(1+y^2\right)} \leq 1$$

Applying this fact to $(\star)$ we see that

$$\left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \leq \left(1 + 1\right)\left|x-y\right| \leq 2\left|x - y \right|$$

Therefore $f$ is a Lipschitz function, which implies $f$ is uniformly continuous on $\mathbb{R}$.

Please comment on validity and/or readability, thank you.

Zduff
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    It looks fine to me. Perhaps the very last inequality would require a little more explantion...but nice, indeed. +1 – DonAntonio Feb 26 '18 at 21:58

3 Answers3

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Lipschitz-continuity implies uniform continuity, we cannot disagree on that.
On the other hand there is a slightly more efficient way for proving that $|f'|$ is bounded: $$ f'(x) = f(x)\cdot \frac{d}{dx}\log f(x) = \frac{1}{1+x^2}\cdot \frac{-2x}{1+x^2}$$ where $\left|\frac{1}{1+x^2}\right|\leq 1$ is trivial and $\left|\frac{2x}{1+x^2}\right|\leq 1$ just a bit less (AM-GM), so $|f'|\leq 1$.
By using the weighted AM-GM inequality we have the optimal bound $|f'|\leq\frac{3}{8}\sqrt{3}$.

Jack D'Aurizio
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  • That's pretty cool, I haven't seen that. We're supposed to be doing this without the use of derivatives but I didn't realize how simple the connection was until this. Thanks. – Zduff Feb 26 '18 at 22:15
  • To the proposer: One of the nice things about calculus is that a horde of individual Q's can be easily answered by applying one or two general results – DanielWainfleet Feb 27 '18 at 03:18
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Apply mean value theorem in an interval $[x,y], x,y\in \mathbb R$. So, $\exists \xi \in [x,y]: f(x)-f(y)=f'(\xi)(x-y)$. Also we know that $|f'(\xi)|<1$.

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Big Hint: Use this and the fact that $\vert f'(x)\vert <1$.

Vincenzo Zaccaro
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