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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!

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    $\begingroup$ 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 $\epsilon>0$. $\endgroup$ – Gio67 Nov 4 '18 at 18:35
  • $\begingroup$ 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. $\endgroup$ – Albert Nov 4 '18 at 21:10
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\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.

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