2
$\begingroup$

Let $A$ be an operator $A: l_p \rightarrow l_p , 1 <p<\infty$ $$A(x_1, ..., x_n, ...)=\left(x_1, \frac{x_1+x_2}{2}, ..., \frac{x_1+...+x_n}{n}, ...\right)$$ I want to show that operator $A$ is not compact. I want to prove it using one of the equivalent definitions of operator compactness: I want to show that the image of unit ball $A(B_1)=\left\{ \left(x_1, \frac{x_1+x_2}{2}, ..., \frac{x_1+...+x_n}{n}\right) : ||x_n||_{l_p}\le 1 \right\} $ is not relatively compact.

I have a criterion for that: I know that the subset $K \subset l_p $ is relatively compact iff $K$ is bounded and $\lim_{N\rightarrow\infty}\sup_ {x \in K}\sum_{n=N}^{\infty}|x_n|^p=0$ so I want to show that my $A(B_1)$ doesn't satisfy this criterion.

I would be grateful for any help!

$\endgroup$
  • $\begingroup$ What have you tried? $\endgroup$ – MisterRiemann Dec 9 '18 at 17:11
  • $\begingroup$ I tried to take such elements from unit ball $(1, 1, ..., 1, 0, ..., 0, ...)$-only finite number of nonzero coordinates and to look at the image of such elements after action of operator $A$ $\endgroup$ – Anton Zagrivin Dec 9 '18 at 17:30
  • $\begingroup$ I though I could have $lim sup...\ne 0$(from my oppost criterion), but I couldn’t find this limit $\endgroup$ – Anton Zagrivin Dec 9 '18 at 17:32
  • $\begingroup$ Maybe my entire idea is wrong and it will not work here $\endgroup$ – Anton Zagrivin Dec 9 '18 at 17:32
  • $\begingroup$ A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738 $\endgroup$ – Jean Marie Dec 9 '18 at 17:56
1
$\begingroup$

Let $N$ be a fixed integer and let $v=v^{(N)}$ be the vector defined by $v_i=2^{-(N+1)/p}$ for $1\leqslant i\leqslant 2^{N+1}$ and $0$ otherwise. Then $v$ belongs to the unit ball. Moreover, for $2^{N}+1\leqslant n\leqslant 2^{N+1}$, the $n$-th coordinate of $Av$, denoted $(Av)(n)$, satisfies $$ (Av)(n)=2^{-(N+1)/p}\frac 1n\cdot n=2^{-(N+1)/p} $$ hence $$ \sum_{n=2^N+1}^{2^{N+1}}\left\lvert (Av)(n)\right\rvert^p=2^N\left(2^{-(N+1)/p}\right)^p=2^{-1}. $$ This proves, by the mentioned compactness criterion, that the set $\left\{Av^{(N)},N\geqslant 1\right\}$ is not relatively compact in $\ell^p$ hence that $A$ is not a compact operator.

$\endgroup$
  • $\begingroup$ Oh, that's what I was trying to do but couldn't achieve $\endgroup$ – Anton Zagrivin Dec 11 '18 at 7:16
  • $\begingroup$ Thank you a lot $\endgroup$ – Anton Zagrivin Dec 11 '18 at 7:16
  • $\begingroup$ You are welcome $\endgroup$ – Davide Giraudo Dec 11 '18 at 9:22

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.