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Let $\{ \varphi_{\lambda}(x,y): \lambda \in \mathbb{R} \}$ be the family of functions defined by $$\varphi_{\lambda}(x,y):= \frac{1}{1 + x^4 + \frac{1}{2}\arctan(\lambda)\sin(y^6)}$$ for all $\lambda \in \mathbb{R}$. Is this family of functions equicontinuous in $\mathbb{R}$?


It is easy to see that this family of functions is uniformly bounded by $1$, but I don't know how to see if is equicontinuous. I have thought about using subsequences and maybe Arzela-Ascoli theorem, but I do not get the point... For example, it is not equi-Lipschitz as its partial derivatives are not bounded in a convex open set, so this doesn't prove nothing neither...

Any hint will be appreciated. Thanks in advance

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Take $\lambda,\mu$. Since $\frac1a-\frac1b = \frac{b-a}{ab}$ we have $$ |\phi_\lambda(x,y)-\phi_\mu(x,y)|\le \frac1{(1-\frac\pi4)^2} \frac12 |\arctan(\lambda)-\arctan(\mu)|. $$ This implies $\phi_\mu\to \phi_\lambda$ uniformly for $\arctan(\mu) \to \arctan(\lambda)$.

Since the set $\{\arctan \lambda: \ \lambda\in \mathbb R\}$ is bounded, we can find for each sequence $(\lambda_n)$ a subsequence such that $(\arctan(\lambda_{n_k}))$ converges, and so $(\phi_{\lambda_{n_k}})$ converges uniformly. Then by Arzela-Ascoli, the family is equicontinuous.

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  • $\begingroup$ @Crostul yes, you are right. I had $\lambda \in \mathbb R \cup \{\pm \infty\}$ in mind. $\endgroup$ – daw Jan 18 at 12:41
  • $\begingroup$ Useful. Thanks!! $\endgroup$ – gal16 Jan 18 at 12:45
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A direct calculation/estimate works: The denominator is bounded below by $1 - \pi/4 > 0$, so that $$ | \varphi_{\lambda}(x_1,y_1) - \varphi_{\lambda}(x_2,y_2)| \le \frac{|x_2^4 + \frac{1}{2}\arctan(\lambda)\sin(y_2^6) - x_1^4 - \frac{1}{2}\arctan(\lambda)\sin(y_1^6)|}{(1-\pi/4)^2} \\ \le \frac{|x_2^4 -x_1^4|+ \frac{\pi}{4}|\sin(y_2^6)-\sin(y_1^6)|}{(1-\pi/4)^2} = C |f(x_2)-f(x_1)| + D|g(y_2)-g(y_1)| $$ with some constants $C, D$ and the continuous functions $f(x) = x^4$ and $g(y) = \sin(y^6)$. For given $(x_1, y_1) \in \Bbb R^2$ this becomes arbitrary small if $(x_2, y_2)$ is sufficiently close to $(x_1, y_1)$, independently of $\lambda$.

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  • $\begingroup$ Understood it. That can also be justified using the mean value theorem at the function $g$ and taking the limit when $(x_{1}, y_{1}) \to (x_{2}, y_{2})$. Thanks! $\endgroup$ – gal16 Jan 18 at 12:48

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