Well-defined strip norm and the action of a common endpoint operator

The finite product $X_0{\oplus }_1{X}_1$ is Banach: a Cauchy sequence has Cauchy coordinates and coordinate limits, and the sum of their distances tends to zero. In the ambient topological vector space the map $(x_0,x_1)\mapsto x_0+x_1$ is continuous by continuity of the injections and addition. Hausdorffness makes $\{0\}$ closed, so its inverse image, the kernel of this map, is closed. The quotient theorem therefore makes the stated sum norm a genuine complete norm on the algebraic sum. The endpoint injections into the sum are contractive. Applying the same argument to $(Y_0,Y_1)$ gives its Banach sum and contractive endpoint injections.

The imported strip theorem makes $\mathcal{F}(X_0,X_1)$ Banach. Its vector strip maximum inequality, applied in the sum norm, gives $\mathrm{\parallel }f(s){\mathrm{\parallel }}_{X_0+X_1}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathcal{F}(X_0,X_1)}$ because endpoint injections into the sum are contractive. Thus evaluation has closed kernel. The quotient theorem applied to this kernel proves that its image with the infimum norm is Banach, independent of the choice of interpolating strip function, and the displayed evaluation bound makes its injection into the sum continuous and definite. This verifies the asserted definition rather than assuming its infimum is automatically a norm. The identical argument for $(Y_0,Y_1)$ establishes its interpolation space and continuous injection into its sum.

The specified common endpoint operator is bounded on the sums: for each decomposition $x=x_0+x_1$, $\mathrm{\parallel }Tx{\mathrm{\parallel }}_{Y_0+Y_1}\le M_0\mathrm{\parallel }{x}_0{\mathrm{\parallel }}_{X_0}+M_1\mathrm{\parallel }{x}_1{\mathrm{\parallel }}_{X_1}$, so infimizing gives a bound $\max (M_0,M_1)$ on the sum. It therefore maps a strip function to a bounded sum-continuous norm-holomorphic function, since bounded linear maps preserve norm limits and difference quotients. Endpoint continuity and decay are preserved by the endpoint bounds. For positive bounds $M_0,M_1$ apply this to $g(z)=(M_0\mathrm{/}{M}_1{)}^{z-s}Tf(z)$. Both boundary norms are at most $M_0^{1-s}{M}_1^s\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathcal{F}(X_0,X_1)}$, while $g(s)=Tf(s)$. Taking the infimum over $f$ proves the bound. If one bound is zero, use $M_j+\epsilon$ and let $\epsilon \downarrow 0$.

Apply step 3.1 to the maps $E:X_0+X_1\to Y_0+Y_1$ and $P:Y_0+Y_1\to X_0+X_1$ from the Statement, using their respective endpoint bounds $a_j$ and $b_j$ and swapping source and target couples for $P$. Their restrictions are bounded on the asserted interpolation spaces. For every $x\in [X_0,X_1{]}_s\subseteq X_0+X_1$, the assumed sum identity gives $P(Ex)=x$. Because these are the same maps restricted to the interpolation spaces, this is exactly $PE=I_{[X_0,X_1{]}_s}$. This proves the retraction assertion without replacing either endpoint map.