0
$\begingroup$

When $Z$ is an interpolation space between two Banach spaces $X$ and $Y$ (say real / complex method), we have a norm inequality $$ \| x \|_Z \le C \| x \|_X^\theta \| x \|_Y^{1-\theta} $$ My question is somehow for the converse : assume that $T$ is a bounded linear operator to $X$, assume $Y$ to be a subspace of $X$ (to make it easy) and we assume the above inequality holds for all $x$ in the range of $T$. Can we then conclude that the image of $T$ sits between any interpolation space with exponent $\theta\pm \varepsilon$? Or even between the real interpolation spaces of exponent $\theta$ with fine indices 1 and $\infty$ respectively ??

$\endgroup$
  • 1
    $\begingroup$ What exactly is meant by "assume $Y$ is a subspace of $X$"? Do you mean $Y$ embeds in $X$? If you use the induced norm on $Y$, then isn't $\|\cdot\|_X = \|\cdot\|_Y$? What do you mean by $\|x\|_Y$ if $x\in \mathrm{Image}(T)$? Are you assuming the image always sit inside $Y$? $\endgroup$ – Willie Wong Jul 25 '16 at 15:07
  • $\begingroup$ if $Y$ embeds in $X$ then there is no reason that $Y$ with induced norm equals $X$. For example $c_0$ is a subspace (or embeds identically) into $\ell_\infty$ without them being equal. $\endgroup$ – Eric Jul 25 '16 at 22:14
  • $\begingroup$ For the second question: Yes, I assume $Tx \in Y \subseteq X$. But due to the norm inequality, $Tx$ may always be contained in a larger (interpolation) space, between $Y$ and $X$, and that is precisely the question. $\endgroup$ – Eric Jul 25 '16 at 22:21
  • $\begingroup$ I misunderstood the first question maybe? The concrete situation I have, $Y$ is the domain of an operator as well, you may give it the graph norm, and then $Y$ embeds indeed into $X$. Is that what you meant? $\endgroup$ – Eric Jul 25 '16 at 22:28
  • $\begingroup$ So essentially $T$ plays no role in your question. You have as sets $Z\subset Y\subset X$ and norms $\|\cdot\|_Z$, $\|\cdot\|_Y$ and $\|\cdot\|_X$ making them into Banach spaces. You are asking if $\|z\|_Z \leq C\|z\|_X^\theta \|z\|_Y^{1-\theta}$, equivalently $Z$ embeds into some interpolation space of $X$ and $Y$, whether $Z$ embeds into "something else". (I am not 100% sure what you mean by the something else in your question.) $\endgroup$ – Willie Wong Jul 26 '16 at 13:13

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.