I'm trying to figure out whether the set of functions $$\{ g_a : a \in [1, \infty)\; in\; C[-1,1], \ g_{a}(x) = \frac {1}{(x-a)^{2}+1} \}$$ is equicontinuous.
I have tried to manipulate the inequality $|g_{a}(x) - g_{a}(y)|$ and i got that $|g_{a}(x) - g_{a}(y)| \le |y-x| |2a+2|$ but this doesn't really give me a value of $\delta$ that works for any $a$. I'm stuck. I also don't really have an intuition for whether it is equicontinuous or not.. is there anything that should hint this to me?
Choose an $\epsilon > 0$ and take $\delta=\frac{\epsilon}{2}$. Let $|y-x|<\frac{\epsilon}{2}$. Then: \begin{align} |g_a(x)-g_a(y)|&=\Big|\frac{1}{(x-a)^2+1}-\frac{1}{(y-a)^2+1}\Big| \\ &=\frac{|y-x|~|y+x-2a|}{[(x-a)^2+1]~[(y-a)^2+1]} \\ & \leq \frac{|y-x|~(|y-a|+|x-a|)}{[(x-a)^2+1]~[(y-a)^2+1]} \\ &=|y-x|~\Big[\frac{|x-a|}{[(x-a)^2+1]~[(y-a)^2+1]}+\frac{|y-a|}{[(x-a)^2+1]~[(y-a)^2+1]}\Big] \end{align} Now if $|x-a|<1$ then $(x-a)^2+1-|x-a|\geq0 \Rightarrow \frac{|x-a|}{[(x-a)^2+1]} \leq 1 \Rightarrow \frac{|x-a|}{[(x-a)^2+1]~[(y-a)^2+1]} \leq 1$.
If $|x-a|\geq 1$ then $|x-a| \leq (x-a)^2 \leq (x-a)^2 +1 \Rightarrow \frac{|x-a|}{[(x-a)^2+1]} \leq 1 \Rightarrow \frac{|x-a|}{[(x-a)^2+1]~[(y-a)^2+1]} \leq 1$.
Thus in any case we have $ \frac{|x-a|}{[(x-a)^2+1]~[(y-a)^2+1]} \leq 1$. Similar for $y$ related exdpression. Now from above we get \begin{equation} |g_a(x)-g_a(y)| \leq 2~|y-x|<\epsilon \end{equation} for all $a$.