Do bounded derivatives imply equi-continuity of function sequence?

Let $f_n: [0, 1] \rightarrow \mathbb{R}$ $\forall n \in \mathbb{N}$

If $f_n \in C^1([0, 1])$ and $\vert\vert f_n'\vert\vert_\infty \le 3 \Longrightarrow \{f_n\}$ are equicontinuous

I know that bounded derivative for $f \in C^1([a, b])$ implies $f$ is Lipschitz, which implies uniform continuity. I'm sure that it isn't true for the $\{fn\}$ sequence, because the previous one is an answer between other three, and it can't be the correct one. But I can't find a counter-example.

If $\vert\vert f_n'\vert\vert_\infty \le 3$, $\{fn\}$ should be equi-Lipschitz (Lipschitz $\forall$ $n$), because $\exists$ $L > 0: \vert\vert f_n(x_1) - f_n(x_2) \vert \vert \le L \vert \vert x_1 - x_2 \vert \vert$ because of Lagrange theorem (we can use the "biggest" $L$ that is good $\forall$ $n$).

But in that case, $\{f_n\}$ are equicontinuous, so I can't understand where I am wrong. Maybe I can't take the "biggest" $L$ because I could have infinite $L$s ($n \in \mathbb{N})$?

Any help is appreciated, thanks!

EDIT: The answer I checked as correct is: $f_n \in C^0([0, 1])$ $\Longrightarrow$ they are equibounded because of Weierstrass theorem. Indeed, $\exists \max f_n, \exists \min f_n \Longrightarrow f_n$ are bounded $\forall$ $n$

EDIT: So, Weierstrass lost, and bounded derivatives imply equicontinuity in my case!

• If the derivatives are uniformly bounded, then the family is equicontinuous. Otherwise choose for example $x \mapsto \sin (nx)$. – Marko Karbevski Jul 8 '17 at 10:50
• I'll add the answer I considered correct. I could have gone wrong there. – moonknight Jul 8 '17 at 10:59

yes your answer is not the correct one. If $f_n$ is continuous, there exists $\max_{x\in [0,1]}|f_n(x)|=m_n$ but $m_n$ could go to $\infty$. Equi-bounded means that $\max_{x\in [0,1]}|f_n(x)|\le M$ for every $n$ and for some constant $M$ independent of $n$. Take $f_n(x)=n$ for every $x\in [0,1$, then each $f_n$ is continuous but the sequence is not equi-bounded.