Compatible compact Banach triples

Let $X,B,Y$ be real Banach spaces, or complex Banach spaces over the same field, in the sense of Banach space. A compatible compact Banach triple is a pair of injective linear maps $j:X\to B$ and $k:B\to Y$ such that $j$ is compact in the sense of Compact linear operator and $k$ is bounded. We identify $X$ with $j(X)$ and $B$ with $k(B)$, using $k\circ j$ for the map from $X$ into $Y$; the notation is $X⋐B\hookrightarrow Y$. Bounded means that there are finite constants $C_j,C_k$ with $\mathrm{\parallel }jx{\mathrm{\parallel }}_B\le C_j\mathrm{\parallel }x{\mathrm{\parallel }}_X$ and $\mathrm{\parallel }kb{\mathrm{\parallel }}_Y\le C_k\mathrm{\parallel }b{\mathrm{\parallel }}_B$. No density, reflexivity or Hilbert pivot is part of this definition. Compact means that the image of every norm-bounded subset of $X$ has compact closure in $B$.

Given: The spaces and maps in the Definition.

1.1 A compact linear map is bounded by the final assertion of Compact linear operator, so $C_j$ exists. The composite is injective because both maps are injective, and $\mathrm{\parallel }kjx{\mathrm{\parallel }}_Y\le C_k{C}_j\mathrm{\parallel }x{\mathrm{\parallel }}_X$. Thus the three spaces can consistently be identified along these maps. [given, algebra]

1.2 Completeness is already required of each space by Banach space. The definition adds specified embeddings and compactness; it does not assert existence of such maps for arbitrary Banach spaces. Therefore its notation has exactly the stated hypotheses and no implicit Hilbert-space assumption. [given] ∎