# Given an averaging operator $A: l_p \rightarrow l_p$, why $A$ is not compact

Let $$A$$ be an operator $$A: l_p \rightarrow l_p , 1 $$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!

• What have you tried? – MSobak Dec 9 '18 at 17:11
• 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 $A$ – Anton Zagrivin Dec 9 '18 at 17:30
• I though I could have $lim sup...\ne 0$(from my oppost criterion), but I couldn’t find this limit – Anton Zagrivin Dec 9 '18 at 17:32
• Maybe my entire idea is wrong and it will not work here – Anton Zagrivin Dec 9 '18 at 17:32
• A useful keyword for request is "Cesaro" or "Cesaro mean". Related : math.stackexchange.com/q/1313738 – 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.