Characterization of the dual of intersection of Banach spaces

Question

Let $U,V$ Banach spaces and define $X = U\cap Y$ endowed with the norm $\|u\|_X = \|u\|_U + \|u\|_V$. If we take $\varphi \in U'$ and $\psi \in V'$, I can prove that $\varphi|_X + \psi|_X \in X'$, because for $u \in X$, $$ |(\varphi|_X + \psi|_X)(u)| \leq |\varphi(u)| + |\psi(u)| \leq C_1 \|u\|_U + C_2 \|u\|_V \leq (C_1 + C_2) \|u\|_X. $$ Now take $f \in X'$. Is it true that $f = \varphi|_U + \psi_V$ for some $\varphi \in U'$ and $\psi \in V'$? I'm not sure if this makes sense. What should be the dual of $X$?

My problem: Consider $X = H^1(\mathbb{R}^N) \cap D^{1,q}(\mathbb{R}^N)$ endowed with the norm $\|u\|_X = \|u\|_{H^1(\mathbb{R}^N)} + \|u\|_{D^{1,q}(\mathbb{R}^N)}$. Now take $u_n, u \in X$ such that $u_n \rightharpoonup u$ in $H^1(\mathbb{R}^N)$ and $u_n \rightharpoonup u$ in $D^{1,q}(\mathbb{R}^N)$. I trying to prove that $u_n \rightharpoonup u$ in $X$. If what I asked is true, I would know how to prove it.

This question arose when I tried to prove that $X = H^1(\mathbb{R}^N) \cap D^{1,q}(\mathbb{R}^N)$ is reflexive. If anyone knows another approach to prove it I would be very grateful.

In the following I write a idea I had, which I don't know whether is correct:\

My attempt: Let $Y = C^\infty_0(\mathbb{R}^N)$ and $\phi \in X'$. Notice that $\phi|_Y$ is continuous from $Y$ to $\mathbb{R}$ with the usual norms from $H^1(\mathbb{R}^N)$ and $D^{1,q}(\mathbb{R}^N)$, respectively(I think I know how to prove it by showing the norms are equivalent just in $Y$). Denote by $f$ and $g$ the respective continuous extensions of $\phi$ to $H^1(\mathbb{R}^N)$ and $D^{1,q}(\mathbb{R}^N)$(using the Hahn Banach theorem). Observe that $(f|_X + g|_X)/2 \in X'$ and $(f|X + g|X)/2 = \phi$, because $(f|X + g|X)/2(u) = \phi(u)$, for all $u \in Y$ and there is just one continuous extension of $Y$ to $X$, because $\overline{Y}^{\|\cdot\|_X} = X$(is this true?).

Answer 1

In general, in order that the intersection be defined, we should assume that $(U,\|\cdot\|_U)$ and $(V,\|\cdot\|_V)$ are continuously embedded into an ambient Hausdorff TVS $E$. Then, $(U\cap V,\|\cdot\|_U+\|\cdot\|_V)$ is indeed a Banach space isometric to the subspace $\Delta:=\{(x,x): x\in U\cap V\}$ of $U\times V$ (endowed with the sum norm). The latter is reflexive, and a closed subspace of a reflexive space is reflexive.

rmk: Note that in general we can isometrically represent the dual of $(U\cap V,\|\cdot\|_U+\|\cdot\|_V)$ as $ (U^*\times V^*)/ \Delta^\perp$ (you may check the book of Rudin for the details on the dual of subspaces and quotients).

Answer 2

This answers the question in the title.

The dual of $X$ is indeed $U' + V'$. Consider the product space $W = U\times V$, endowed with the norm $\|(u,v)\|_{W} = \|u\|_{U} + \|v\|_{V}$. You can easily check that this is a Banach space and that the (diagonal) embedding $\iota:X\to W$, $\iota(x) = (x,x)$ is isometric. So, $X\cong X_W:=\operatorname{im}\iota$ and by Hahn-Banach every continuous linear functional $f\in X_W'$ can be extended to an element of $W' = U'\oplus V'$, giving a representation of $f$ as $\varphi + \psi$ for some $\varphi\in U'$ and $\psi\in V'$ (considering $U'$ and $V'$ as subspaces of $X'$).