Proof 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.

Answer 2

The given linear operator is known as Cesàro operator. Aside from $\,\ell^2(\mathbb N)\,$ it can also be defined on $\,\ell^\infty$ and on many other sequence spaces.

If $A$ were compact, then $$AA^* \ =\ \begin{pmatrix} 1& \frac12& \frac13\\ \frac12& \frac12& \frac13& \dots\\ \frac13& \frac13& \frac13\\ & \vdots& & \ddots \end{pmatrix}$$ had to be compact too.
We show that it is not: Let $$\big[AA^*\big]_n \ =\ \begin{pmatrix} 1& \frac12& \dots& \frac1n&0&\\ \frac12& \frac12& \dots& \frac1n& 0&\\ \vdots& \vdots& \ddots& \vdots &\vdots\\ \frac1n& \frac1n& \dots& \frac1n &0 &\dots\\ 0& 0& \dots& 0& 0\\ & & & \vdots& & \ddots \end{pmatrix}$$ be the truncation given by the left upper $n\times n$ block of $AA^*$. It equals $\,P_nAA^*P_n\,$ where $P_n$ is the orthogonal projector onto the subspace $\langle e_1,\dots,e_n\rangle$. The sequence of finite-rank operators $\big[AA^*\big]_n$ strongly converges to $AA^*$, that is "pointwise convergence" for each point $x\in\ell^2$.
But the sequence is not norm-convergent:
Let $Q$ be the $n\times n$ matrix where every entry is $1/n$. Then $Q$ is an orthogonal projector (of rank one) and $\|Q\|=1$. On the diagonal of $AA^* - \big[AA^*\big]_n\,$, starting at ($n+1,n+1$), we have the $n\times n$ block $$B \ =\ \begin{pmatrix} \frac1{n+1}& \frac1{n+2}& \dots& \frac1{2n}\\ \frac1{n+2}& \frac1{n+2}& \dots& \frac1{2n}\\ \vdots& \vdots& \ddots& \vdots \\ \frac1{2n}& \frac1{2n}& \dots& \frac1{2n} \\ \end{pmatrix}$$ which is entrywise larger than $Q/2$. Thus, $\frac12 =\|Q/2\|\leqslant\|B\| \leqslant\left\|AA^* - \big[AA^*\big]_n\right\|\,$ for all $n$.
Hence $AA^*$ is not compact.

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Remark regarding your approach to find the inverse operator: It is known that the spectrum of $A$ is $\sigma(A)= \big\{\lambda\in\mathbb C\text{ with }|\lambda -1|\leqslant 1\big\}$, in particular $0\in\sigma(A)$ and $A$ is not invertible.
AFAIK, $A$ is injective, is not surjective, however has a dense image.