TITLE: is $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ compact?
QUESTION [9 upvotes]: Is $T$ a compact operator?
$T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ where $t\in[0,1]$
with supremum norm.

REPLY [3 votes]: Slightly different approach: 
Assume to the contrary that $T$ is compact. Clearly $\|T\|=1$, in fact $\|T^n\|=1,$ so spectral radius is equal to 1.
 Since the ambient space in infinite dimensional then there exists a sequence of eigenvalues that converges to zero . in particular we can find eigenvalue $\lambda$ with $|\lambda|<1$ and  $g\in C[0,1]$ such that 
$Tg=\lambda g\Rightarrow g(x^2)=\lambda g(x)\,\forall x\in[0,1]\Rightarrow $ $\|g\|=|\lambda|.\|g\|<\|g\|$ which is a contradiction.