3
$\begingroup$

Let $X,Y,Z$ be Banach spaces such that $X,Y\hookrightarrow Z$. We know that if $X\Subset Y$ (The symbol "$\hookrightarrow $" means continuous embedding), then $(Z,X)_{\theta,p}\hookrightarrow (Z,Y)_{\theta,p}$ for all $\theta\in (0,1)$ and $p\in[1,\infty]$. How about the reverse results? Let us say, if $(Z,X)_{\theta,p}\hookrightarrow (Z,Y)_{\theta,p}$ for some $\theta\in (0,1)$ and $p\in (1,\infty)$, then what can we say about the inclusion relations about $X$ and $Y$? Can one show that $X\hookrightarrow Y$. If not, is it possible to make it valid under some additional requirements? References are appriciated.

$\endgroup$
4
  • $\begingroup$ I assume you meant "intermediate," not "immediate." $\endgroup$ May 30, 2017 at 17:18
  • $\begingroup$ @MichaelRenardy Thank you. Yes, I were careless $\endgroup$
    – Ice sea
    May 30, 2017 at 17:20
  • $\begingroup$ For clarification: what do you intend with the symbol $\Subset$? For the actual question, you seem to assume a symmetric statement $(Z,X)_* = (Z,Y)_*$ and conclude an asymmetric one $X \Subset Y$. Is that intended? Finally, what if you take $X, Y, Z$ to be Lorentz spaces? Does that tell you something about what you want to know? $\endgroup$ May 30, 2017 at 21:09
  • $\begingroup$ @WillieWong I have edited my question. It should be clear now. $\endgroup$
    – Ice sea
    May 31, 2017 at 6:37

1 Answer 1

4
$\begingroup$

In general, you cannot give a definite relation between $X$ and $Y$, at least as long as you do not require a strict inclusion between your interpolation spaces $(Z,X)_{\theta,p}$ and $(Z,Y)_{\theta,p}$ (then it could work, but I'm not sure how to prove it at the moment). This is essentially because there are quite fine scales of function spaces where interpolation acts too "rough" in the following sense:

Choose the Besov spaces $Z = B^0_{p,r}$, $X = B^s_{p,q_0}$ and $Y = B^s_{p,q_1}$, all defined on $\mathbb{R}^n$, for some $s > 0$ and $p,q_0,q_1,r \in (1,\infty)$. See Triebels book "Interpolation theory, function spaces, differential operators", Ch. 2.3 for the definitions if necessary; all my references will be from there. Then you have $X,Y \hookrightarrow Z$ (continuous embedding, Ch. 2.3.2 (3)) and even $$(Z,X)_{\theta,q} = (B^0_{p,r},B^s_{p,q_0})_{\theta,q} = B^{\theta s}_{p,q} = (B^0_{p,r},B^s_{p,q_1})_{\theta,q} = (Z,Y)_{\theta,q}$$ for $\theta \in (0,1)$ and $q \in (1,\infty)$, see Ch. 2.4.1 (3); note how the interpolation parameter $q$ "overrides" the scale parameters $r,q_0,q_1$. However, again by Ch. 2.3.2 (3), $$X \hookrightarrow Y \quad \text{if}~q_0 \leq q_1$$ and $$Y \hookrightarrow X \quad \text{if}~q_1 \leq q_0.$$ There is quite an analogous phenomenon for the Triebel-Lizorkin spaces (Ch. 2.4.2 (2), and also for the Lorentz spaces (Ch. 1.18 (16)) mentioned by Willie Wong in the comments.

PS: I think the the symbol "$\Subset$" usually stands for a compact inclusion.. :)

$\endgroup$
1
  • $\begingroup$ Thx. Also, I have changed the symbol to reduce confusion. $\endgroup$
    – Ice sea
    May 31, 2017 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.