The discrete $J$ method is, given Banach spaces $A_0$ and $A_1$:

The interpolationn space $[A_0, A_1]_\theta$ is defined by: $a \in [A_0, A_1]_\theta$ if and only if $a$ can be written as $a=\sum_{k \geq 1}a_k$ where the series converges in $A_0 + A_1$, where each $a_k$ belongs to $A_0 \cap A_1$ and where $(2^{-j\theta} J(2^j, a_j)) \in \ell^2$.

Here $J(t,\cdot)$ is a norm and irrelevant to my question. (Eg. see page 16 of this)

Suppose I have an element $u \in [A_0, A_1]_\theta$ and I know that it can be written as $$u=\sum_j {u_j},$$ where $u_j$ is known.

I want to apply the theorem. Since $u \in [A_0, A_1]_\theta$, the theorem states it can be written as $u=\sum_{k \geq 1}a_k$ for some $a_k$. Can I pick $a_k = u_k$? I want to do this so that if so, I know the sequence $(2^{-j\theta} J(2^j, u_j))$ is in $\ell^2,$ which is the result that I want.

Or is this $a_k$ sequence not known?

Originally posted https://math.stackexchange.com/questions/1109371/discrete-j-method-of-interpolation-about-understanding-theorem-statement.


closed as unclear what you're asking by Bill Johnson, Stefan Kohl, Ryan Budney, Denis-Charles Cisinski, Peter Crooks Feb 3 '15 at 22:22

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    $\begingroup$ What do you know about the functions $u_j$ and in which sense does the series $\sum_ju_j$ converge? $\endgroup$ – Joonas Ilmavirta Jan 22 '15 at 19:56
  • $\begingroup$ @JoonasIlmavirta we have $u=\sum_j {u_j} = \sum_j (u,w_j)w_j$ where $w_j$ is an o.n basis of $A_0$ and an orthogonal basis of $A_1$. The sum converges in $A_0$. Suppose that $A_1 \subset A_0$. $\endgroup$ – GuestTwo Jan 22 '15 at 21:30