Let $X$ and $Y$ be sub-spaces of a large vector space, and both formed Banach spaces with associated norm.

If $X\cap Y$ with norm $\|u\|_{X\cap Y}=\|u\|_X+\|u\|_Y$, is dense in $X$ and $Y$ , it's easy to see $(X\cap Y)^*\ge X^*+Y^*$ by natural embedding.

But can we have $(X\cap Y)^*\le X^*+Y^*$? That means every bounded linear functional $X\cap Y\to\mathbb{R}$ can be written as a form of $f+g,f\in X^*,g\in Y^*$

The original problem come from a book of PDE, saying that "The space $H^{-1}(\Omega)+L^{p'}(\Omega)$ is given a strong topology, and is the dual space of $H_0^{1}(\Omega)\cap L^{p}(\Omega)$", $\Omega\subset\mathbb{R}^n$ is bounded open set with $C^{\infty}$ boundary. It is just a footnote without further statement. So I want to proof this statement with the word of abstract norm space. If the title question is wrong, can we correct it?

  • $\begingroup$ To clarify the situation: We do not assume any relation between the norms $\lVert \,\cdot\,\rVert_X$ and $\lVert\,\cdot\,\rVert_Y$? $\endgroup$ – Daniel Fischer Mar 1 '16 at 14:46
  • $\begingroup$ @DanielFischer $X\cap Y$ is dense in $X$ and $Y$, is that work? $\endgroup$ – yaoliding Mar 1 '16 at 14:48
  • $\begingroup$ We can always come into that situation by replacing $X$ with $\operatorname{cl}_X(X\cap Y)$ and ditto for $Y$ if need be. So we may assume it from the start. $\endgroup$ – Daniel Fischer Mar 1 '16 at 14:53
  • $\begingroup$ I don't think your question can be answered in the affirmative, but I don't have a counterexample. :o( $\endgroup$ – Friedrich Philipp Mar 1 '16 at 14:56
  • $\begingroup$ It is certainly possible to have that situation, even excluding the trivial cases when e.g. $\lVert u\rVert_Y \leqslant C\cdot \lVert u\rVert_X$ for some constant $C$. Is your question whether that equality always holds, or what the conditions are under which it holds, or just whether there are situations in which it holds? $\endgroup$ – Daniel Fischer Mar 1 '16 at 15:05

We always have $(X \cap Y)^{\ast} = X^{\ast} + Y^{\ast}$, viewing $X^{\ast}$ and $Y^{\ast}$ as subspaces of $(X\cap Y)^{\ast}$.

Consider the space $Z = X \times Y$, endowed with the norm $\lVert (x,y)\rVert_Z = \lVert x\rVert_X + \lVert y\rVert_Y$. This is a Banach space with dual $Z^{\ast} = X^{\ast} \oplus Y^{\ast}$, where $\lVert (\lambda,\mu)\rVert_{Z^{\ast}} = \max \{ \lVert \lambda\rVert_{X^{\ast}}, \lVert\mu\rVert_{Y^{\ast}}\}$.

Now $\Delta \colon X\cap Y \hookrightarrow Z$ given by $\Delta(u) = (u,u)$ is an isometric embedding, let $F := \operatorname{im}\Delta$. By the Hahn-Banach theorem, the restriction

$$\rho \colon Z^{\ast} \to F^{\ast} \cong (X\cap Y)^{\ast}$$

is surjective, and $\rho(f,g) \colon \Delta(u) \mapsto f(u) + g(u)$.

  • $\begingroup$ Thanks, I need to think for a while. Why Hahn-Banach is needed? $\endgroup$ – yaoliding Mar 1 '16 at 16:25
  • $\begingroup$ Hahn-Banach ensures that every continuous linear form on $F$ has a representation $f+g$ with $f\in X^{\ast},\; g \in Y^{\ast}$. If we had a continuous linear form on $F$ without a continuous extension to $X\times Y$, it would not be of the desired form. $\endgroup$ – Daniel Fischer Mar 1 '16 at 16:33

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