Infinite matrix which cannot be represented by bounded linear operator

Question

Let ${\cal V}$ be n Hilbert space over $\mathbb{R}$ or $\mathbb{C}$, with an orthonormal basis $(e_n)_{n=1}^{\infty}$. For every bounded linear operator $A: {\cal V} \to {\cal V}$, we can associate $A$ with an matrix $(a_{ij})_{i,j=1}^{\infty}$ where $a_{ij} = \langle Ae_j,e_i\rangle$.

Is there a matrix $(b_{ij})_{i,j=1}^{\infty}$ such that:
1. $\sum_{i=1}^{\infty}|b_{ij}|^2 < \infty$
2. $\sum_{j=1}^{\infty}|b_{ij}|^2 < \infty$
3. $\sup_{i,j \ge 1} |b_{ij}| < \infty$

and there is no bounded linear operator $B : {\cal V} \to {\cal V}$ such that $b_{ij} = \langle Be_j,e_i\rangle$

I am really struggling with this problem, I will appreciate any help!

Answer 1

Try this matrix... $$ B = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0\\ \vdots \end{bmatrix} $$ Each column has a single $1$ and all the rest zeros.
The first row has one $1$ and all the rest zeros.
The second row has two $1$s and all the rest zeros.
The third row has three $1$s and all the rest zeros.
The fourth row has four $1$s and all the rest zeros.
...
The $k$th row has $k$ $1$s and all the rest zeros.

The three conditions are satisfied.

Think of using Hilbert space $l_2$, with its canonical orthonormal basis $(e_k)$. Think of the elements as infinite column vectors.

Fix $k$. Let vector $x$ be the transpose of row $k$. Then $Bx = k e_k$. But $\|x\| = k^{1/2}$ and $\|Bx\| = k$, so for the operator norm of $B$ we conclude $$ \|B\| \ge \frac{k}{k^{1/2}} = k^{1/2}. $$ This holds for all $k$. Thus $B$ is not a bounded operator.

Answer 2

This is only a partial answer, which shows that (1)-(3) do not imply the continuity of $B$. I.e., for all $b$ satisfying (1)-(3) there is a fitting unbounded operator $B$. I could not succeed with explicitly constructing a bounded $B$ from given $b$. All attempts failed as $\sum_{i,j}|b_{ij}|^2$ may be infinite.

Let me show the existence of unbounded $B$ given $b$ satisfying (1)-(3). The idea is to construct an unbounded linear functional $f$ on $V$ such that $f(e_i)=0$ for all $i$.

For $n\in \mathbb N$ define the geometric sequences $d_n$ by $$ \langle d_n,e_i\rangle = (n+1)^{-i}. $$ Then the infinite set $M:=\{ e_i, d_n: \ i,n\in\mathbb N\}$ is linearly independent (see Vandermonde matrix).

Now define the linear map $f:span(M)\to \mathbb R$ by $$ f(e_i)= 0, \ f(d_n) = n, $$ which is possible as $M$ is a basis of $span(M)$. Then extend $f$ to $V$ by Zorn's lemma.

Define $$ Bx:=f(x)e_1. $$ It follows $Be_i=0$ and $b_{ij}=0$ for all $i,j$. But obviously, $B$ is unbounded.