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Suppose $H$ is a Hilbert space with orthonormal basis $\{e_i\}_{i\in \mathbb N}$. To every operator $T$, we associate a infinite matrix $[T_{ij}]$, where $T_{ij}=\left<Te_j,e_i\right>$. We know that for any trace class operator $T$, the trace norm is $||T||_1=\operatorname{Tr}(|T|) $.

Q). Suppose $T$ is a trace class operator and $S$ is such that its matrix entries are either equal to the matrix entries of $T$ or they vanish (possibly at infinite number of points). Can I say that $||S||_1\leq ||T||_1$? If not, is there any finite upper bound to such $S$ obtained from $T$?

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    $\begingroup$ @SinaBaghal fixed. It was a mistake $\endgroup$
    – NewB
    Commented Jul 13, 2021 at 16:28
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    $\begingroup$ If I remember well there is a counterexample in the paper The main triangle projection in matrix spaces and its applications by A. Pelczynski and S. Kwapien; Studia Math. 34 (1970), 43-67. $\endgroup$
    – Dirk Werner
    Commented Jul 13, 2021 at 16:52
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    $\begingroup$ Since the trace class norm and the operator norm are in duality, wouldn't your multiplier have the same norm in either case? $\endgroup$
    – Dirk Werner
    Commented Jul 13, 2021 at 21:58
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    $\begingroup$ This is clearly so when $T$ has a finite support and therefore also when $T$ does not have finite support. (The finitely supported matrices / operators are dense in the trace class; and $S_n= Q_n T Q_n$ where $Q_n$ is the orthogonal projection onto $\{e_1,\dots,e_n\}^\perp$.) $\endgroup$
    – Dirk Werner
    Commented Jul 18, 2021 at 10:07
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    $\begingroup$ `Also trace class can be identified with projective tensor product of Hilbert space': Once you know that, it follows that $H_0\otimes H_0$ is dense, with $H_0=$ linear span of the basis vectors $e_1, e_2, \dots$. $\endgroup$
    – Dirk Werner
    Commented Jul 19, 2021 at 19:34

1 Answer 1

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Here's an algorithm for testing an ad-hoc conjecture $C$ about Hilbert space operators. :-)

  1. Set up the runtime environment correctly by loading the information "Most conjectures are false" into short term memory.

  2. Test $C$ against the zero and the identity operator.

  3. Test $C$ against finite-dimensional diagonal matrices.

  4. Test $C$ against multiplication operators on $\ell^2$ and $L^2$.

  5. Test $C$ against the following $2 \times 2$-matrices: $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. $$

  6. Test $C$ against simple modifications (appropriate to the setting of $C$) of the matrices from Step 4.

  7. Write a computer programm to test $C$ against randomly generated $2 \times 2$-matrices; make sure to restrict the matrices that your random generator creates to the set of matrices that occur in $C$.

  8. Repeat Step 6 with $3 \times 3$-matrices.

  9. If you have not found a counterexample yet, there might be a reason to believe that $C$ holds.

Of course this should not be taken completely seriously - but often it works. In my experience, for many ad-hoc conjectures the algorithm stops at Step 5 or earlier. The question from the OP adds another data point to this pattern:

The matrix $$ T = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$ has trace norm $2$, but the matrix $$ S = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} $$ has trace norm $$ \frac{1 + \sqrt{5}}{2} + \frac{\lvert 1 - \sqrt{5} \rvert}{2} = \sqrt{5} > 2. $$

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    $\begingroup$ +2 if I could, for the answer, and the humour. $\endgroup$
    – Matthew Daws
    Commented Jul 14, 2021 at 8:30
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    $\begingroup$ Also, note that taking suitably large direct sums of this counter-example shows that there is no hope of the conjecture "holding up to a constant" either (which the OP did ask about). $\endgroup$
    – Matthew Daws
    Commented Jul 14, 2021 at 8:30
  • $\begingroup$ @MatthewDaws: Thanks for your comments! Yepp, I agree. The direct sum construction is also quite a useful way to built easy counterexamples. :-) I should probably add this to the list, together with a few typical infinite-dimensional candidates for counterexamples. But at the moment I urgently need to go to bed, so I'll postpone it to a moment when I'm really awake... $\endgroup$
    – Jochen Glueck
    Commented Jul 16, 2021 at 1:34
  • $\begingroup$ @JochenGlueck Thank you! that is a very nice algorithm :) . $\endgroup$
    – NewB
    Commented Jul 16, 2021 at 12:45
  • $\begingroup$ I understand the direct sum argument. Now lets us fix a trace class operator $T$ and define $S_n$ for each $n$ entrywise as $(S_n)_{ij}=T_{ij}$ if $i,j>n$ and 0 else where. Now do you think Is it possible in general to choose $n$ large enough such that $||S_n||_1$ is as small as we wish? $\endgroup$
    – NewB
    Commented Jul 16, 2021 at 18:55

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