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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.

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    $\begingroup$ 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 $$\lim_{|(x,y)| \to \infty} f(x,y)=0$$ $\endgroup$ – Crostul Aug 18 '19 at 19:43
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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.

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  • $\begingroup$ I believe you are missing some squared terms in your denominator. Other than that I believe I follow your reasoning. $\endgroup$ – Walt Aug 20 '19 at 14:07
  • $\begingroup$ @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. $\endgroup$ – mathcounterexamples.net Aug 20 '19 at 18:14
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Any continuous function $f$ on $\mathbb R^2$ that satisfies $\lim_{|z|\to \infty}f(z)=0$ is uniformly continuous on $\mathbb R^2.$

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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.

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