Is the following set of real-valued continuous functions on $[0,1]$ equicontinuous? $$ A = \{t\mapsto tx(t)\mid x\colon [0,1]\to\mathbb{R}\text{ and }(\forall t\in [0,1])\,\lvert x(t)\rvert \leq 1\} $$

My attempt:

Let $t_0\in [0,1],\varepsilon>0$.

Then $\lvert tx(t)-t_0x(t_0)\rvert=\lvert tx(t)-t_0x(t)+t_0x(t)-t_0x(t_0)\rvert \leq\lvert t-t_0\rvert+\lvert x(t)-x(t_0)\rvert$

But $\lvert t-t_0\rvert+\lvert x(t)-x(t_0)\rvert$ is not equicontinuous, however I can not find a counterexample of not equicontinuous functions in $A$ or another bound for $\lvert tx(t)-t_0x(t_0)\rvert$.


  • $\begingroup$ Are you assuming the functions themselves are continuous? $\endgroup$ – Almentoe Nov 12 '15 at 0:29
  • $\begingroup$ yes, they are... $\endgroup$ – José Nov 12 '15 at 0:30

I don't think they are equicontinuous. Near $1$, the influence of $t$ is negligible.

For instance, if we let $x_n(t)=t^n$, then this family will not be equicontinuous at $t=1$. Because for any $t<1$ you can get $n$ big enough so that $t^n$ is arbitrarily close to zero.

  • $\begingroup$ Thats right, I was confused for and exaple I did and since the se $A$ is actualy equicontinuous just for $t_0=0$. $\endgroup$ – José Nov 12 '15 at 0:52
  • $\begingroup$ Yes, that's the only point where they are equicontinuous. $\endgroup$ – Martin Argerami Nov 12 '15 at 1:13

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