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!

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.