I am working through a text on Numerics for SPDEs and there the concept an interpolation (Hilbert-)space associated to an operator is used. To be specific:

Definition. Let $H$ be an $\mathbb{R}$-Hilbert space and $A: D(A) \rightarrow H$ a diagonal linear operator on $H$ with point spectrum $\sigma_P(A) \subseteq (0,\infty)$ and $\inf \sigma_P(A)>0$. Then the family $(H_r)_{r\in \mathbb{R}}$ of $\mathbb{R}$-Hilbert spaces with the properties

  1. $\forall$ $r\geq s$ : $H_r \subset H_s = \overline{H_r}^{H_s}$

  2. $\forall$ $r\in [0,\infty)$ : $(H_r, \langle \cdot,\cdot\rangle_{H_r}) = (D(A^r), \langle A^r(\cdot), A^r(\cdot)\rangle_H)$

  3. $\forall$ $r\in (-\infty,0]$,$v\in H$ : $\|v\|_{H_r} = \|A^r v\|_{H}$

is called a family of interpolation spaces associated to $A$.

Now, I am searching for a concrete explanation of how this concept is useful in the analysis of SPDEs. By looking through the literature I found some hints, but no concise answer:

  • It seems the interpolation spaces play a role in dealing with the nonlinearity F of the SPDE $dX_t = [AX_t+F(X_t)] dt+ B(X_t)dW_t$.
  • They are somewhat of an analogue to Sobolev spaces and reveil a certain kind of regularity.

I would be very greatful if somebody has a good explanation/motivation of this idea.

Thanks in advance!


1 Answer 1


In the definition, "diagonal" does not make much sense, you probably mean "self-adjoint", although this can be relaxed to "m-sectorial" (powers of $A$ still make sense then). Also, it is irrelevant for the definition whether or not $A$ has some essential spectrum.

Anyway, the main reason why these interpolation spaces are useful for the analysis of SPDEs (and deterministic PDEs for that matter) is that they give us good short-time control over the behaviour of the semigroup $S(t) = e^{At}$ for small times. More precisely, one has $$ \|S(t)u\|_\alpha \le C t^{-(\alpha-\beta)}\|u\|_\beta\;, $$ for any $u \in H_\beta$ and any $\alpha \ge \beta$, uniformly over $t \in (0,1]$, say. When you want to show that a given equation admits local solutions, this allows you to play a game whereby you rewrite the equation in its mild formulation and then trade some of the loss of regularity by the nonlinearity $F$ (and / or $B$) against some negative power of $t$ via the action of the semigroup. You can find more details in my lecture notes on SPDEs available here.

  • $\begingroup$ Thank you for your answer! I do, however, not understand why "diagonal" makes no sense? Having a diagonal linear operator given as $Ah=\sum_{b\in B} \lambda_b\langle h,b \rangle b$ for an orthonormal basis $B$ of $H$ we defined the powers as $A^r h=\sum_{\mu \in \sigma_P(A)} \mu^r P_{Kern(\mu-A)}h$, where $P$ is projection. Then $A:D(A)\subseteq H\rightarrow H$ should be symmetric, and so self-adjoint. Furthermore, A is bounded if and only if the function $\lambda$ is bounded, hence the essential spectrum. Am I missing something, or is the definition only less general than in your notes? $\endgroup$
    – alexlo
    Commented Jan 15, 2015 at 15:28
  • 2
    $\begingroup$ "Diagonal" only makes sense if you are already given a distinguished orthonormal basis a priori. $\endgroup$ Commented Jan 15, 2015 at 16:54

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