# Is $f(x,y)=\frac{1}{x^2+y^2+1}$ uniformly continuous?

Is \begin{align*} f(x,y)=\frac{1}{x^2+y^2+1} \end{align*} uniformly continuous?

I was able to show that $$f$$ has a global maximum at $$f(0,0)=1$$, but I can't seem to work out a proper estimate for uniform continuity. Any insights would be greatly appreciated.

• Yes. This is because its gradient is bounded, so the function is Lipschitz continuous. Another way to prove it is using the fact that it is continuous and $$\lim_{|(x,y)| \to \infty} f(x,y)=0$$ – Crostul Aug 18 '19 at 19:43

Name $$X=(x,y)$$. Then

$$f(x,y)=f(X)= \frac{1}{\Vert X\Vert^2 +1}.$$ Where $$\Vert \cdot \Vert$$ is the Euclidean norm.

We have

\begin{aligned}\vert f(X_1)-f(X_2) \vert &= \left\vert \Vert X_1\Vert - \Vert X_2 \Vert\right\vert \left\vert \frac{\Vert X_1\Vert + \Vert X_2\Vert}{(\Vert X_1\Vert +1)(\Vert X_2\Vert +1)}\right\vert\\ &\le 2\left\vert \Vert X_1\Vert - \Vert X_2 \Vert\right\vert\\ &\le 2\Vert X_1 -X_2 \Vert\end{aligned}

as $$\frac{\Vert X_i\Vert}{(\Vert X_1\Vert +1)(\Vert X_2\Vert +1)} \le 1$$ for $$i=1,2$$ and using reverse triangle inequality.

Which implies uniform continuity.

Note: we get as a bonus that proof above is valid for any normed space.

• I believe you are missing some squared terms in your denominator. Other than that I believe I follow your reasoning. – Walt Aug 20 '19 at 14:07
• @Walt You're right! Fortunately the end result still remains as $\frac{\Vert X_i\Vert}{(\Vert X_1\Vert^2 +1)(\Vert X_2\Vert^2 +1)} \le 1$ is also true. – mathcounterexamples.net Aug 20 '19 at 18:14

Any continuous function $$f$$ on $$\mathbb R^2$$ that satisfies $$\lim_{|z|\to \infty}f(z)=0$$ is uniformly continuous on $$\mathbb R^2.$$

• – irchans Aug 18 '19 at 22:36

Note that $$\displaystyle \|\nabla f(x,y)\| = 2\sqrt{\frac{x^2+y^2}{(x^2+y^2+1)^4}}$$ and the function $$z\mapsto \frac{z}{(z+1)^4}$$ is bounded for non-negative $$z$$, so $$\nabla f$$ is bounded and $$f$$ is Lipschitz, hence uniformly continuous.