Let < $f_n$> be a functional sequence on $A\subset \mathbb{R}$ and let $f:A\rightarrow\mathbb{R} $ then < $f_n$>is not uniform convergence iif there exists $\epsilon_0$ such that there exists the subsequnce of $<f_n>$, $<f_{n_k}>$ and a sequance $<x_k>$ for all $k\in \mathbb{N}$
$$|f_{n_k}(x_k)-f (x_k)|\geq \epsilon_0$$
I don't know how to prove this from the defintion of uniform convergence.
Actually the definition of uniform convergence is
Let $<f_n>$ be a functional sequence on $A\subset\mathbb {R}$ and $f:A\rightarrow \mathbb {R} $ then for all $\epsilon>0$ there exist $K (\epsilon)\in \mathbb {N}$ such that $n \geq K (\epsilon ), x\in A$ imply
$$|f_n(x)-f (x)|<\epsilon$$
Then we say < $f_n$> uniform convergent to $ f$ on $A$
Best Answer
It is just the negation of the definition
Uniform Convergence $$f_n \rightarrow f, \text{ uniformly if } \forall \epsilon>0, \exists N, \forall n > N, \forall x\in A, \ |f_n(x) - f(x)| <\epsilon$$ Not Uniform Convergence $$f_n \not\rightarrow f, \text{ uniformely if } \exists \epsilon >0, \forall N>0, \exists n>N, \exists x\in A, |f_n(x) - f(x)| \geq\epsilon $$
In words: Suppose $f_n\not\rightarrow f$ uniformly, then there exists $\epsilon_0 >0$ such that for each $N\in \mathbb{N}$ there exists some $n_N>N$ and $x_N\in A$ such that $$|f_{n_N}(x_N) - f(x_N)| \geq\epsilon_0 $$
This is exactly the statement.