# Show that this operator is not compact using Arzela-Ascoli

Let $$T:C[0,1]\longrightarrow C[0,1]$$ defined as $$Tx(t) =tx(t)$$. I need to prove that this operator is not compact using Arzela-Ascoli (using the Sup norm). I already prove that if X is a bounded subset of C[0,1] then AX is bounded too, so in order to T to be not compact the equicontinuity must fail. Here is my attempt: $$|Tx(u) - Tx(v) |=|ux(u) - vx(v) |=|x(u) (u-v) +v(x(u) - x(v)) |<=|x(u)||u-v|+(|x(u)|+|x(v)|)|v|<=\|x\| |u-v| + 2\|x\||v|< \text{ (since X is bounded) } M|u-v|+2M|v|$$.

So I think that the last expression that depend on $$|v|$$ is the key to find an $$\epsilon>0$$ such that no matter which $$\delta>0\;$$ I choose if $$|u-v|<\delta$$, then $$|Tx(u) - Tx(v) |>\epsilon$$.

I would appreciate if someone could help me finding such $$\epsilon$$. Thank you (sorry I don't learn Latex yet).

## 2 Answers

First of all, let's fix $$X$$ to be the closed unit ball in $$C[0,1]$$. Our task is to show that $$T(X)$$ is not equicontinuous.

Now we need to sort out our quantifiers. We must find an $$\epsilon > 0$$ such that for all $$\delta > 0$$, there exists an $$x \in X$$ and $$t_1, t_2 \in [0,1]$$ such that $$| t_1 - t_2 | < \delta$$ but $$|t_1x(t_1) - t_2x(t_2) | \geq \epsilon$$.

I claim that $$\epsilon = \frac 1 2$$ works. For a given $$\delta > 0$$, I take $$t_1 = 1 - \tfrac{\delta} 2, \ \ \ \ \ t_2 = 1,$$ $$x(t) = \begin{cases} 0 & {\rm if \ } t \leq 1 - \tfrac \delta 2 \\ 1 - \frac{2}{\delta} (1 - t) & {\rm if \ } t > 1- \frac \delta 2 \end{cases},$$

and you'll see that $$| t_1 - t_2 | = \frac \delta 2$$ but $$|t_1 x(t_1) - t_2 x(t_2) | = 1 \geq \epsilon$$.

Let $$B$$ be the unit ball of $$C[0,1]$$ endowed with the uniform norm. We have to prove that there exists a positive $$\eta$$ such that for all $$\delta$$, $$\sup_{x\in B}\sup_{\left\lvert u-v\right\rvert\lt\delta}\left\lvert Ax(u)-Ax(v)\right\rvert\gt \eta.$$ Using the beginning of the computation mentioned in the opening post, namely, $$|Ax(u) - Ax(v) |=|ux(u) - vx(v) |=|x(u) (u-v) +v(x(u) - x(v)) |\gt\left\lvert v\left(x(u)- x(v)\right)\right\rvert -\delta,$$ we get that $$\sup_{x\in B}\sup_{\left\lvert u-v\right\rvert\lt\delta}\left\lvert Ax(u)-Ax(v)\right\rvert\geqslant \frac 12 \sup_{x\in B}\sup_{\left\lvert u-v\right\rvert\lt\delta,u,v\geqslant 1/2}\left\lvert x(u)-x(v)\right\rvert -\delta.$$ It is possible to prove that $$\sup_{x\in B}\sup_{\left\lvert u-v\right\rvert\lt\delta,u,v\geqslant 1/2}\left\lvert x(u)-x(v)\right\rvert=2$$ by taking functions of the form $$x\mapsto \sin\left(2\pi x/\delta\right)$$.