is $T:C[0,1]\rightarrow C[0,1]$: $x(t)\mapsto x(t^2)$ compact?

Question

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.

Answer 1

First, a general remark. A common way to prove that some operator $T$ is not compact is to exhibit an infinite-dimensional subspace $M$ on which $T$ has a lower bound: that is, there exists $c>0$ such that $$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$ If (1) holds, then the image of unit ball under $T$ contains a ball of radius $c$ in the infinite-dimensional subspace $TM$, and therefore is not compact.

In view of the above, you should be asking yourself: for which functions $x$ can I prove an inequality of the form $\|Tx\|\ge c\|x\|$? Looking at what $T$ does, and recalling the definition of the norm, you will realize that $\|Tx\| = \|x\|$ holds for all $x\in C[0,1]$. Therefore, $T$ is not compact.

Answer 2

Here is a much less intuitive answer.

Note: Unfortunately this was not as simple as I had originally thought (thanks to Davide for catching my oversight). The result depends on the fact that $C[0,1]$ has the 'approximation property', that is, any compact operator is the limit (in the operator norm) of a sequence of finite rank operators. (See Remark 1.1.15 in "An Introduction to Nonlinear Analysis: Applications", Vol. 2, Z. Denkowski, S. Migórski, N. S. Papageorgiou.)

Note that $\|Tx\|= \|x\|$ for all $x$.

Suppose $A$ is a finite rank operator, then $\ker A $ is non-trivial (consider the effect of $A$ on $t \mapsto t^n$, for example).

Choose $x \in \ker A$ of unit norm, then $\|(T-A)x\| = \|Tx\| = \|x\| = 1$, and so $\|T-A\| \ge 1$, hence $T$ cannot be approximated by finite rank operators. It follows that $T$ is not compact.

Answer 3

Here's Daniel Fischer's comment in more explicit terms: any compact operator on an infinite dimensional Banach space (e.g. $C([0,1])$) cannot be invertible. However the map $x(t)\mapsto x(\sqrt{t})$ is the inverse map for $T$ so $T$ cannot be compact.

Answer 4

Same approach as previous answers, using limited operators:

If $T\colon X \to X$ is a compact operator and $S \colon X \to X$ is a bounded operator, then both $TS$ and $ST$ are compact.

Let $S: C^1[0,1] \to C^1[0,1]$ such that $u(t) \mapsto u(\sqrt{t})$. Readily we see that $\|S\| = 1$, so $S$ is bounded. But $TS = Id$. If $T$ is compact so is $Id$, but as $C^1[0,1]$ is infinite dimensional, $Id$ cannot be compact and therefore $T$ is not compact.

Answer 5

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.