Question on equicontinuity Am I correct in thinking that conditions 1 and 2 are equivalent to ${f_n}$ being equicontinuous on $K$?

Let $K$ be a compact set in $\mathbb{R}$.
Let $\{ f_n \}$ be a sequence that converges pointwise on $K$ to $f$
  where each $f_n$ is continuous.
Consider the following conditions:
  
  
*
  
*$f$ is continuous on $K$.
  
*For given $\epsilon>0$, there are $m$, $\delta >0$ such that if $n>m$ and $|f_k(x)-f(x)| < \delta$ then $|f_{(k+n)}(x)-f(x)| <
 \epsilon$ for all $x \in \mathbb{R}$, $k=1,2,...$.
  

 A: Assuming (1), (2), we show that $\{f_n\}$ converges uniformly to $f$ (This shows in particular that they are equicontinuous. Let $\epsilon >0$. Let $\delta$ be given as in (2). Let $x\in K$ be arbitrary. Then as $f_k \to f$ pointwisely, there is $k = k(\delta, x)$ so that 
$$ |f_k(x) - f(x) | < \delta /3.$$
As $f_k$, $f$ are both continuous, there is $\rho$ (depends on $x$, $k$) so that 
$$|f(y) - f(x)|, |f_k(y) - f_k(x)| <\delta /3$$
whenever $|y-x| <\rho$. Using triangle inequality, we have 
$$|f_k(y) - f(y)|<\delta$$
for all $y$ so that $|y-x|<\rho$. By (2), we have 
$$|f_{n+k}\ (y) - f(y)|<\epsilon$$
for all $n>m$. Now as $K$ is compact, there are $x_1, \cdots, x_l$ (which give rises to $k_1, \cdots , k_l$ and $\rho_1, \cdots, \rho_l$) so that 
$$K \subset  \bigcup_{j=1}^l (x_j - \rho_j, x_j + \rho_j).$$
Let $k = \max\{k_1, \cdots, k_l\}$. Then 
$$|f_p(y) - f(y)| <\epsilon$$
for all $y\in K$ and for all $p \ge m+k$. Thus $f_n$ converges uniformly to $f$. 
A: No for condition 1. Counterexample:
$$
f_n(x)=4\,n\max\bigl(x\,(1-n\,x),0\bigr).
$$
Each $f_n$ is continuous, and $\{f_n\}$ is uniformly bounded and converges pointwise to $f(x)\equiv0$, also continuous. However no subsequence converges uniformly, since
$$
\max_{0\le x\le1}|f_n(x)-f(x)|=1\quad\forall n\in\mathbb{N}.
$$
It follows that $\{f_n\}$ is not equicontinuous.
