Showing if a family of functions is equicontinuous, equibounded and/or equi-Lipschitz

Question

I don't understand how to prove this kind of problem:

Let $u_n:[0,1]\to \Bbb{R}$, $\qquad$ $u_n(x):=n+{\sqrt{\frac x n}}$ $\qquad \forall n \in \Bbb{N}$

Is $u_n$ equicontinuous, equibounded and/or equi-Lipschitz in $[0,1]$?

I know each definition but I don't understand how to apply them. Please help me, I really don't know where to start.

Answer 1

Let $v_n=u_n-n=\sqrt {\frac x n}$. Tne $v_n \to 0$ uniformly. This implies that $(v_n)$ is equi-continuous. if you write down the definition if equi-continuity of $u_n=v_n+n$ you will see immediately that $(u_n)$ is also equi-continuous.

Of course $u_n(x) \geq n$ so even point-wise boundedness is not true.

Suppose $(u_n)$ is uniformly Lipschitz. Then there exists $C$ such that $|u_n(x)-u_n(y)| \leq C|x-y|$ for all $n$ for all $x,y$. Put $y=0$ to get $\sqrt {\frac x n} \leq C x$. I will let you find an $x$ for which this fails. [Take $x$ close to $0$].

Answer 2

Here is an intuitive account of how to think about these properties. Not really an answer, more of a long comment. Just my two cents, hope it helps.

Any "equi"-property is a guarantee that says something about what happens with $u_n$ as $n\to \infty$, and that these functions, in some sense, never stray too far away from one another in terms of the relevant property.

If $u_n$ is equibounded, then each $u_n$ is bounded, sure. But not only that, we also require that the same bound applies to all the $u_n$.

If $u_n$ is equicontinuous, then each $u_n$ is continuous, sure. But not only that, we also require that given any fixed $x\in[0,1]$ and $\varepsilon>0$, then the same $\delta>0$ works to prove continuity at $x$ for all of the $u_n$.

If $u_n$ is equi-Lipschitz, then each of the $u_n$ is Lipschitz, sure. But not only that, we also require that the same Lipschitz constant $k$ works for all the $u_n$.