Functional Analysis – Proof of Non-Compactness of Linear Operator

Question

Let we have the linear operator in $l_2$: $Ax = (x_1, (x_1 +x_2)/2, (x_1 +x_2 + x_3)/3, … )$
how to proof the non-compactness of this operator?$$$$
My idea is to find the inverse operator, thus the non-compactness will be obvious. But the problem is that I cannot show the surjectivity of this operator, which is why I cannot apply Banach’s theorem on the inverse operator.

Answer 1

We will apply the theorem: if $P_n$ is a sequence of bounded operators such that $P_nx\to Px$ for every $x\in \ell^2$ and $A$ is a compact operator then $\|P_nA-PA\|\to 0.$ Let $$P_n(x)=\sum_{k=1}^nx_ke_k$$ where $\{e_k\}_{k=1}^\infty$ denotes the standard basis in $\ell^2.$ Then $P_nx\to x$ for every $x\in \ell^2.$ We will show that $\|(I-P_n)A\|$ does not tend to $0,$ hence $A$ cannot be compact. To this end let $$x^{(N)}=\sum_{k=1}^Ne_k$$ Then $\|x^{(N)}\|_2^2 = N.$ On the other hand $$[(I-P_n)Ax^{(N)}]_k =\begin{cases}0, & 1\le k\le n\\ 1, & n+1\le k\le N \\ Nk^{-1} &k>N \end{cases}$$ Hence $$\|(I-P_n)Ax^{(N)}\|^2\ge N-n$$ Thus $${\|(I-P_n)Ax^{(N)}\|^2\over \|x^{(N)}\|_2^2 }\ge 1-N^{-1}n$$ which implies $\|(I-P_n)A\|\ge 1-N^{-1}n.$ As $N$ is arbitrary we get $\|(I-P_n)A\|\ge 1.$

Remark 1 By considering $$x^{(N)}=\sum_{k=1}^Nk^{-1/2}e_k$$ we can get a better estimate $\|(I-P_n)A\|\ge 2,$ which is based on $$n^{-1}\sum_{k=1}^nk^{-1/2}\approx 2n^{-1/2}$$ Remark 2 The same proof can be applied to the generalized operator of the form $$(Ax)_n=\sum_{k=1}^n a_{nk}x_k,\ \sum_{k=1}^na_{nk}=1$$ In OP case we the case $a_{nk}=n^{-1}$ is considered.