6
$\begingroup$

Let $T$ be a linear operator acting on a finite-dimensional real or complex vector space. As a direct consequence (or rather a particular case) of the Riesz-Thorin theorem, we have $$ \|T\|_2 \le \sqrt{\|T\|_1\|T\|_\infty}\,, $$ where $\|T\|_p$ are the induced operator norms of $T$.

  • Does this inequality have a name?
  • What does the ratio $\sqrt{\|T\|_1\|T\|_\infty}\,\big/\,\|T\|_2$ tell us about $T$? Has it ever been studied? Is there any standard notation or name for it?
Improve this question
$\endgroup$
3
  • $\begingroup$ I'd guess it doesn't tell us anything, because we can bound that independent of $T$ because of the equivalence of the $1$-,$2$- and $\infty$-norms on finite dimensional spaces. $\endgroup$
    – Johannes Hahn
    Commented May 24, 2014 at 14:20
  • $\begingroup$ @Johannes: the norms are equivalent, but the equivalence constants are not absolute: they depend on the dimension. And so, knowing that the ratio is small definitely provides some non-trivial information about $T$. $\endgroup$
    – Seva
    Commented May 24, 2014 at 18:53
  • $\begingroup$ This is a special case of the Schur test. en.wikipedia.org/wiki/Schur_test $\endgroup$
    – Narutaka OZAWA
    Commented May 25, 2014 at 23:26

2 Answers 2

Reset to default
4
$\begingroup$

Sorry, my answer below is only partial, but I thought that it may still be somewhat interesting.

As far as I know, this inequality does not have a distinguished name. It is ultimately a consequence of the duality between $\|\cdot\|_\infty$ and $\|\cdot\|_1$, and the submultiplicativity of $\|\cdot\|_1$.

This is certainly known to you, but I wanted to remark that $$\|T\|_2^2 = \|T^*T\|_2 = \rho(T^*T) \le \|T^*T\|_1 \le \|T^*\|_1 \|T\|_1 = \|T\|_\infty \|T \|_1.$$ Thus, this inequality is nothing but an expression of:

  1. The C*-identity of the spectral norm
  2. Dominance of the spectral radius $\rho(\cdot)$ by the $\|\cdot\|_1$-norm.
  3. Submultiplicativity of the operator $\|\cdot\|_1$-norm.

As to the second inequality in the OP, we might need some further restrictions to make it more interesting. For instance, if $T$ is symmetric, then it reduces to $\|T\|_2 \le \|T\|_1$, whereby the ratio becomes $\|T\|_1/\|T\|_2$. This ratio has been studied for vectors (diagonal $T$) in the world of "sparse coding / compressed sensing, etc." --- see e.g., the paper here, and references therein.

Improve this answer
$\endgroup$
2
$\begingroup$

In this paper, Ron DeVore refers to this inequality as simply "interpolation of operators." He uses it to prove that a particular matrix $\Phi$ satisfies the restricted isometry property, i.e., that every sufficiently small principal submatrix $A$ of $\Phi^*\Phi$ has all of its eigenvalues between $1/2$ and $3/2$, or equivalently, $\|A-I\|_2\leq1/2$. Of course, DeVore appeals to interpolation of operators since $\|A-I\|_1$ and $\|A-I\|_\infty$ are easier to access.

I don't know if the ratio you mentioned has been studied before, but it certainly measures how tight the bound is. It's worth mentioning that in the case where the diagonal of $T$ is all zeros, interpolation of operators follows from the Gershgorin circle theorem, and these are actually equivalent if $T$ is also self-adjoint. In this case, this work might be related to your ratio.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .