Equicontinuous family

Question

Is the family of functions $f_n:[0,1]\rightarrow \mathbb{R}$ defined by $f_n(x)=\sqrt{x+\frac{1}{n}}$ equicontinuous? If so how to prove it?

Note that $f'_n$ is not uniformly bounded

Answer 1

Note that for $a,b>0$ function $t \to |\sqrt{b+t} - \sqrt{a+t}|$, $(t>0)$ is decreasing (clearly $\sqrt{b+t} - \sqrt{a+t} = \frac{b-a}{\sqrt{b+t} + \sqrt{a+t}}$ ).

Then you have: $|f_n(y) - f_n(x)| = |\sqrt{y+\frac{1}{n}}-\sqrt{x+\frac{1}{n}}| \le |\sqrt{y} - \sqrt{x}|$.

So all that's left, is to show that function $f:[0,1] \to \mathbb R$, $f(x) = \sqrt{x}$ is uniformly continuous, which is due to Cantor (continouos on compact).

Answer 2

Note that $f_n$ converges uniformly to some continuous function $f$ on $[0,1]$ (which one?).

Since $[0,1]$ is compact and $f$ is continuous, then $f$ is actually uniformly continuous. Try to use these facts to prove that $f_n$ is uniformly equicontinuous (i.e., that for any $\epsilon>0$ there exist $\delta>0$ such that $|x-y|<\delta$ implies $|f_n(x)-f_n(y)|<\epsilon$ for all $n$).

Of course uniform equicontinuity implies (pointwise) equicontinuity.