# Construction of real interpolation Banach space

Let $$(A,\| \cdot \|_A)$$, $$(B,\| \cdot \|_B)$$ and $$(E,\| \cdot \|_E)$$ real Banach spaces such that $$A\subseteq E$$ and $$B\subseteq E$$ with continuous injections. Let $$0 < \theta < 1$$. For $$x\in E$$ and $$C\geq 0$$ consider the following property on the pair $$(x,C)$$: $$(P) \hspace{10mm} \forall \epsilon > 0, \,\exists (a,b)\in A\times B \text{ such that } x = a + b, \, \|a\|_A\leq C\epsilon^{-\theta} \text{ and } \|b\|_B \leq C\epsilon^{1-\theta}.$$ Let $$[A,B]_{\theta,\infty}$$ the set of all $$x\in E$$ such that there exists $$C\geq 0$$ such that $$(x,C)$$ verifies the property (P). It is easy to prove that if $$(x_1,C_1)$$ and $$(x_2,C_2)$$ verifies (P), then $$(x_1 + x_2,C_1+C_2)$$ verifies (P) and that for every real $$\alpha$$, $$(\alpha x_1, |\alpha|C_1)$$ verifies (P), thus $$[A,B]_{\theta,\infty}$$ is a vector subspace of $$E$$.

For every $$x\in [A,B]_{\theta,\infty}$$ define $$\|x\|_{[A,B]_{\theta,\infty}} = \inf\{C\geq 0 : (x,C) \text{ verifies (P)}\}.$$ Then $$\| \cdot\|_{[A,B]_{\theta,\infty}}$$ is a norm on $$[A,B]_{\theta,\infty}$$.

My question is: Is this construction equivalent to the construction of the interpolation space $$[A,B]_{\theta,\infty,K}$$ obtained with the $$K$$-method?

I can prove the inequality $$\|x\|_{[A,B]_{\theta,\infty,K}}\leq 2\|x\|_{[A,B]_{\theta,\infty}},$$ where the norm in the left is the one obtained with the $$K$$-method.

How can I prove that the norms $$\| \cdot\|_{[A,B]_{\theta,\infty,K}}$$ and $$\| \cdot\|_{[A,B]_{\theta,\infty}}$$ are equivalent without requiring the open mapping theorem?

Thank you in advance!

• I am not sure I understand your question. If in (P) you take $\epsilon=\frac1t$ you get (P)' for $t$ and vice-versa, so the two properties seem to be the same. It does not say in property (P) that $\epsilon$ is small, but it is for every $\epsilon>0$. – Gio67 Nov 4 '18 at 18:35
• You are right. I did't realize that! Thank you! But the question actually is if the construction presented here is equivalent to the construction of the interpolation space using the $K$-method. If it is the case, how to prove that the norms are aquivalent without using the open mapping theorem. – Albert Nov 4 '18 at 21:10

\begin{align*} \Vert x\Vert_{\lbrack A,B]_{\theta,\infty,K}} & =\sup_{t>0}t^{-\theta }K(x,t)\\ K(x,t) & =\inf\{\Vert a\Vert_{A}+t\Vert b\Vert_{B}:\,a+b=x\}. \end{align*} By the definition of infimum, given $$t>0$$ and $$\delta>0$$ there exist $$a\in A$$, $$b\in B$$ such that $$a+b=x$$ and $$\Vert a\Vert_{A}+t\Vert b\Vert_{B}\leq K(x,t)+\delta t^{\theta}%$$ and so$$t^{-\theta}\Vert a\Vert_{A}+t^{1-\theta}\Vert b\Vert _{B}\leq t^{-\theta}(K(x,t)+\delta t^{\theta})\leq\Vert x\Vert_{\lbrack A,B]_{\theta,\infty,K}}+\delta,$$ which shows that \begin{align*} t^{-\theta}\Vert a\Vert_{A} & \leq\Vert x\Vert_{\lbrack A,B]_{\theta ,\infty,K}}+\delta=:C_{0}\\ t^{1-\theta}\Vert b\Vert_{B} & \leq\Vert x\Vert_{\lbrack A,B]_{\theta ,\infty,K}}+\delta=:C_{0}. \end{align*} By replacing $$t$$ with $$1/\epsilon$$ we have that property P holds and $$C_{0}$$ is an admissible constant $$C$$ in the definition of $$\Vert x\Vert_{\lbrack A,B]_{\theta,\infty}}$$. Thus,$$\Vert x\Vert_{\lbrack A,B]_{\theta,\infty}}\leq C_0=\Vert x\Vert_{\lbrack A,B]_{\theta,\infty,K}}+\delta$$ for every $$\delta$$. Letting $$\delta\rightarrow0^{+}$$ gives the inequality.