9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Are you working with the usual supremum norm? $\endgroup$ – Thorben May 19 '14 at 22:34
  • $\begingroup$ yes,it is sup norm indeed $\endgroup$ – 104078 May 19 '14 at 22:35
  • 3
    $\begingroup$ $x(t) \mapsto x(\sqrt{t})$. $\endgroup$ – Daniel Fischer May 19 '14 at 22:42
  • 1
    $\begingroup$ As Daniel Fischer indicated, $T$ is an isomorphism, so it can hardly be compact. $\endgroup$ – Peter Franek May 19 '14 at 23:13
  • 1
    $\begingroup$ Consider $X$, $Y$ infinite-dimensional normed spaces, $B_X$ the unit ball of $X$, and let $T: X \to Y$ be an isomorphism. If $T$ was compact, then $T \, B_X \subseteq Y$ had compact closure. But this means $B_X$ has compact closure and this is absurd, from the Riesz' theorem. $\endgroup$ – Federico May 20 '14 at 14:08
25
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ But $TM$ bounded and looks like equicontinuous, that is, I couldn't prove otherwise. Can you explain how? $\endgroup$ – user108539 May 20 '14 at 9:50
  • 1
    $\begingroup$ @147263 Can you give me the proof or where I can find it? Thanks $\endgroup$ – user108539 May 20 '14 at 10:19
  • 2
    $\begingroup$ @OliverEmerson No, the image of unit ball is not equicontinuous. It contains $x(t)=t^{n}$ for every positive integer $n$. This sequence is not equicontinuous. $\endgroup$ – user147263 May 20 '14 at 20:16
7
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The arguments rests on the fact that in $C[0,1]$ each compact operator can be approximated in norm by finite ranked ones. Is it true, and how do you justify that? Anyway, we only have to show that an isometry cannot be compact in an infinite dimensional space, for example considering here the sequence $x_n(t):=t^n$ or any bounded sequence which is not convergent. $\endgroup$ – Davide Giraudo Jul 13 '14 at 12:34
  • $\begingroup$ @DavideGiraudo: I realise that there are far simpler approaches, but thought this would be entertaining (albeit my proof was initially incorrect). From what I gather, any Banach space with a Schauder basis satisfies the 'approximation property'. $\endgroup$ – copper.hat Jul 13 '14 at 17:35
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.