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!

  • $\begingroup$ What have you tried? $\endgroup$ – MSobak 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

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.

  • $\begingroup$ 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

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