# interpolation inequalities and embeddings

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 ??

• 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 $Y$? – Willie Wong Jul 25 '16 at 15:07
• 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. – Eric Jul 25 '16 at 22:14
• 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. – Eric Jul 25 '16 at 22:21
• 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? – Eric Jul 25 '16 at 22:28
• 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.) – Willie Wong Jul 26 '16 at 13:13