# Is this family of functions equicontinuous in $\mathbb{R}$?

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

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.
• @Crostul yes, you are right. I had $\lambda \in \mathbb R \cup \{\pm \infty\}$ in mind. – daw Jan 18 at 12:41
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 $$\lambda$$.
• 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! – gal16 Jan 18 at 12:48