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Consider $f_n=\sin(nx),\,x\in[0,1]$. In order to show that this is not an equicontinuous family, take $x=0,y=\frac{1}{N}$ where $N \in \mathbb{N}$ can be arbitrarily large, so $\forall \, \delta>0 \rightarrow y<\delta$.

Now, if we consider $|f_N(x)-f_N(y)|=|\sin(N\frac{1}{N})|=|\sin(1)|$, hence there exists $\varepsilon_0 > 0$ such that $\forall \, \delta>0,\, \exists\, x=0,\, y=\frac{1}{N},\, N \in \mathbb{N} \, \text{ such that } |x-y|=y<\delta, \, \text{ but } |f_N(x)-f_N(y)| \geq \varepsilon_0$

Is there anything not right with that logic?

Ilia
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    I think it is perfect. Only for more clarity, I'd specify that $\epsilon_0=\sin 1$. You can also take $y=\pi/(2N)$ (assuming $N\ge 2$) if you prefer a number more... "normal". – ajotatxe Dec 03 '16 at 19:55
  • I think this is problematic because it needs to be for all $x,y \in [0,1]$ and here you have fixed x. – AndyDufresne Feb 09 '19 at 09:48
  • @AndyDufresne This solution is correct. For every $\delta$ one must find $x,y\in [0,1]$ and $N\in\mathbb N$. – AlvinL Apr 25 '22 at 10:16

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