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 $x_n(t)$.
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 $\mathcal{L}(C[0,1] )$.