Sum of Banach spaces

Question

Let $H^2(\mathbb{R}^3)$ the usual Sobolev space and consider the following set $$X=\bigg\{u\bigg|u=\phi+\frac{Q}{|x|},\phi\in H^2,\,\, Q\in\mathbb{C}\bigg\}$$ I observe that the decomposition is unique; in fact if $$\phi_1+\frac{Q_1}{|x|}=\phi_2+\frac{Q_2}{|x|}$$ then $$(Q_1-Q_2)\frac{1}{|x|}\in H^2$$ implies $$Q_1=Q_2$$ I define in $X$ the norm $$\Vert u\Vert_X:=(\Vert\phi\Vert_{H^2}^2+|Q|^2)^{\frac{1}{2}}$$ Is X a banach space? According to me yes, using the same demonstration of the direct sum of Banach spaces.

Answer 1

As a vector space, $X$ is isomorphic to the direct product $H^2\oplus \mathbb C$. The isomorphism is given by $u\mapsto (\phi,Q)$ which is well-defined, as you just showed. The direct product of two Hilbert spaces has a canonical structure of a Hilbert space, obtained by adding two inner products: $$\langle (\phi_1,Q_1),(\phi_2,Q_2)\rangle = \langle \phi_1,\phi_2\rangle_{H^2}+\operatorname{Re}(Q_1 \overline{Q_2}) $$ Completeness of $H^2\oplus \mathbb C$ is a straightforward exercise.

The norm you put on $X$ was simply transferred from $H^2\oplus \mathbb C$; in other words, you declared the vector space isomorphism to be an isometric isomorphism. Of course, this makes $X$ a Hilbert space as well: it has exactly the same structure (vector operations, inner product) as the Hilbert space $H^2\oplus \mathbb C$. The difference between $X$ and $H^2\oplus \mathbb C$ is in the way we write down their elements: $\phi+Q/|x|$ versus $(\phi,Q)$ $-$ but this is a notational difference, not a mathematical one.