# $(X,Y)$ interpolation couple, then $(X^\prime,Y^\prime)$ interpolation couple

Let $(X,Y)$ be an interpolation couple, i.e. there exists a Hausdorff topological vector space $U$ s.t. both $X$ and $Y$ embed continuously into $U$. We define $X\cap Y$ and $X+Y$ in $U$ and turn them into Banach spaces using the maximum of the respective norms and a norm defined in terms of the infimum over decompositions. (for further information, take any book on interpolation theory)

My question is: Under the hypothesis that $X\cap Y$ is dense in both $X$ and $Y$, why is $(X^\prime, Y^\prime)$ again an interpolation couple.

Background of this question is that there is a duality theorem for complex interpolation which states $$[X^\prime,Y^\prime]_\theta=[X,Y]_\theta^\prime$$ in the above situation where additionally one of the spaces needs to be reflexive. I have looked in several sources, e.g. the original article by Calderon and the book by Bergh/Löfström, but all of them don't address this issue.

There is an article by Watbled (http://www.ams.org/mathscinet-getitem?mr=1795744) where the question is at least raised, but necessary identifications for sum and intersection spaces are only claimed.

• As $V=X\cap Y$ is dense in $X$ and $Y$ the restriction mappings $X'\to V'$, $\varphi\mapsto \varphi|_V$ and $Y'\to V'$ are injective and continuous. – Jochen Aug 28 '17 at 11:25
• $X\cup Y$ dense in $X$ actually means density in the embedding of $X$. The identification with this in $X$ doesn't need to be continuous. – Sebastian Bechtel Aug 28 '17 at 12:22
• I don't understand. $V=X\cap Y$ is endowed with the norm $\|v\|_V=\max\{\|v\|_X,\|v\|_Y\}$ so that the inclusion $V\to X$ is clearly continuous. The restriction is the transposed operator which is thus also continuous. The locally convex space $\tilde U$ for $(X',Y')$ can be taken as $V'$ endowed with the dual norm. – Jochen Aug 28 '17 at 12:28
• First of all: the $\cup$ in my comment should be a $\cap$... You are right, I didn't took the concrete choice of norm on the intersection space into account! Let $i$ be the inclusion of $X$ into $U$ and $W=i^{-1}(V)$. Then the restriction is actually $\varphi \mapsto (\varphi\circ i^{-1})|_V$ which is continuous by your previous argument. We don't know whether $W$ is dense in $X$ (correct?), but since the extension of $\varphi\circ i^{-1}$ is unique from $V$ to $i(X)$, we derive unique extensions of $\varphi$ from $W$ to $X$? – Sebastian Bechtel Aug 28 '17 at 13:44
• Seems to be okay. – Jochen Aug 28 '17 at 13:49