Uniform convergence of a sequence of operators We are given a sequence of operators $A_nx(t)=x\left(t^{1+\frac{1}{n}}\right):C[0,1]\to C[0,1]$. It is easy to show that it converges pointwise to the unit operator by the Banach-Steinhaus theorem. Judging by the answer, there is no uniform convergence. But I can't show it. It is necessary to find an example of such a sequence $x_n(t)$, for which $\frac{||A_nx_n-Ix_n||}{||x_n||}$ does not tend to zero. I just can't find such a sequence. This task stuck in my head, there is either something very elementary, or vice versa.
I've tried like this:
Let $0<a<b<1$. Consider a continuous function equal to $0$ on $[0,a]$, linear on $[a,b]$, and equal to $1$ on $[b,1]$. Its norm is $1$. For a fixed $n$, put $a=(1/2)^{1+1/n}$ and $b=1/2$. We will take this as $x_n(t)$.
 A: Let $I$ be the identity mapping on $C[0,1]$. Then
for any $x\in C[0,1]$, note that  $x$ is uniformly continuous on $[0,1]$, and
$0\leq t-t^{1+1/n}\leq \left(\frac{n}{n+1}\right)^n- \left(\frac{n}{n+1}\right)^{n+1} , \ \forall t\in [0,1]$,
we obtain
\begin{equation}\|A_n x -I  x\|  =\max_{t\in [0,1]} \left| x(t)-x\left( t^{1+1/n}\right)\right| \to 0,\text{as $n\to \infty$.} 
\end{equation}
But $\|A_n-I \|\nrightarrow 0$. Indeed, for each $n=1,2,...$, consider the function $x_n \in C[0,1]$ with $\|x_n\|=1$ defined by
\begin{equation}
x_n(t)= 
\begin{cases}
    1-2^n t, & 0\leq t \leq 2^{-n},\\
0 ,& 2^{-n} <t \leq 1.
\end{cases}
\end{equation}
Then
\begin{equation}
(I   x_n -A_nx_n) (t) =x_n(t)- x_n\left(t^{1+1/n}\right)
=\begin{cases}
    2^n t( t^{1/n}-1), & 0\leq t\leq 2^{-n},\\
    2^n t^{1+1/n} -1 , & 2^{-n} <t\leq 2^{-\frac{n^2}{n+1}},\\
    0, & 2^{-\frac{n^2}{n+1}} <t \leq 1.
\end{cases}
\end{equation}
So, $\|A_n-I   \| \geq \|(A_n -I  )(x_n)\| =\frac12$. Hence, $A_n\nrightarrow I $
in $\mathcal{L}(C[0,1] )$.
