Interpolation spaces

Question

Hy everybody

I am curious about the definition; A compatible couple $(X_0, X_1)$ of Banach spaces consists of two Banach spaces $X_0$ and $X_1$ that are continuously embedded in the same Hausdorff topological vector space $Z$.

Why we need that $Z$ let be a Hausdorff topological vector space?

I know that an Hausdorff space separate points but in this case, why is the weight or strong reason?

Thank you

Answer 1

At least one reason why you'd want $Z$ to be a Hausdorff space, is so that you can prove a theorem which states that $X_0 + X_1$ and $X_0 \cap X_1$ are Banach spaces for some explicitely given norms.

Specifically, in the proof of Theorem 1.3 from Chapter 3 of Interpolation of Operators by R. Sharpley, C. Bennett (1988), the fact that $X_0$ and $X_1$ are embedded in a Hausdorff space is used specifically to show completeness of $X_0 \cap X_1$ for its norm given by : $$\lVert x \rVert_{X_0 \cap X_1} = \max\lbrace \lVert x \rVert_{X_0}, \lVert x \rVert_{X_1} \rbrace$$ A brief sketch of this completeness proof is as follows :

If $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $X_0 \cap X_1$, then it is Cauchy in $X_0$ and $X_1$ by construction of the norm. As $X_0$ and $X_1$ are themselves Banach spaces, we can use completeness of the spaces to determine that there are elements $y_0 \in X_0$ and $y_1 \in X_1$ such that : $$\lVert x_n - y_0 \rVert_{X_0} \to 0, \qquad \lVert x_n - y_1 \rVert_{X_1} \to 0$$ As $X_0$ and $X_1$ are continuously embedded in $Z$, we have $x_n \to y_0$ and $x_n \to y_1$ in $Z$, and of course, as $Z$ is a Hausdorff space, we deduce from this that $y_0 = y_1$.

Hope this helps!