$\DeclareMathOperator\Ker{Ker}$Given two Banach spaces $X$ and $Y$ with a continuous inclusion $X\subset Y$, and another couple $X’ \subset Y’$ with the same properties. Take $f : Y \longrightarrow Y’$ linear continuous, such that $f_{\mid X}$ induces a linear continuous map from $X$ to $X'$. My question is for $0<s<1$ and $p>1$, is the following true:

$$(X\cap \Ker(f),\Ker(f))_{s,p}=\Ker(f) \cap (X,Y)_{s,p}\;\;?$$

Here I'm considering the K-method for the interpolation.

The inclusion $(X\cap \Ker(f),\Ker(f))_{s,p}\subset \Ker(f) \cap (X,Y)_{s,p}$ follows directly from the definition. My problem is the other inclusion.

It's clear that if we have $Z\subset Y$ then in general the following is not true:

$$(X\cap Z,Z)_{s,p}=Z \cap (X,Y)_{s,p},$$

one can take $X=H^{2}(U)$, $Z=H^{1}(U)$, $Y=L^{2}(U)$, $s=\frac{1}{2}$, and $p=2$. But this does not contradict our case (because $Z$ here is not even closed in $Y$).

If what I'm asking is not true in general, is it true under the following assumptions:

  • $f(X)$ is closed in $X'$, and $f(Y)$ is closed in $Y'$, and

  • $f$ is open onto $f(Y)$, and $f_{\mid X}$ is open onto $f(X)$.

  • $\begingroup$ Is your counterexample really a counterexample? What norms are you putting on $X\cap Z$ and on $Z$? In some sense the "correct" norm should be the $H^2$ norm on the former (consider $X\cap Z$ as a subspace of $X$) and the $L^2$ norm on the latter (considering $Z$ as a subspace of $Y$). By thinking of $Z = H^1(U)$ with its own norm, you are introduce additional data that is not available in the $\ker$ case. $\endgroup$ Feb 22 at 22:07
  • $\begingroup$ Of course just saying $Z \subset Y$ , isn't enough, we assume everything fits the interpolation functor. And here we mean that $Z$ is a Banach space with contnuous inclusion in $Y$, and $X \cap Z $ is given the max norm so that we have a continuous inclusion of $X \cap Z$ in $Z$. In the exemple, this coïncide with the standard sobolev norms. Is it clear now? $\endgroup$
    – M.Oud
    Feb 23 at 2:04
  • $\begingroup$ You miss my point entirely, so let me ask a different way: in your analogy, what is the norm on $\ker f$? My guess is it is the induced norm as a closed subspace of $Y$. But then your "counterexample" is different in that your $Z$ has a finer topology. $\endgroup$ Feb 23 at 3:22

1 Answer 1


(Making CW as this is an extended comment.)

Let's generalize slightly1: let $X\hookrightarrow Y$ be an embedding of Banach spaces, and let $S\subset Y$ be a closed subspace with the induced norm.

Your question essentially boils down to comparing the $K$ functionals $K$ and $K^{(S)}$, where $$ K(t,z) = \inf_{z = x + y} \|x\|_X + \|y\|_Y $$ and $K^{(S)}$ is analogously defined, except that instead of allowing arbitrary decompositions $z = x+y$ with $x\in X$ and $y\in Y$, you are looking at decompositions with $x \in X\cap S$ and $y\in S$. (Obviously here $z\in S$ as well.)

Clearly $K(t,z) \leq K^{(S)}(t,z)$ for $z\in S$.

Your question will have a positive answer if $K(t,z) \gtrsim K^{(S)}(t,z)$ for some implicit constant independent of $t$ and $z$.

(This is why I objected to your "$H^1$ counterexample; the $K$-functional for the $(X\cap Z,Z)$ interpolation has almost nothing to do with the functional for the $(X,Y)$ interpolation in that case. Compared to this case where it is essentially a geometry question about how $S$ is situated in $X$ and in $Y$.)

A sufficient condition is therefore that there exists a projection $\Pi:Y\to S$ with the property that $\Pi(X) \subseteq X\cap S$ and the operator norms $\|\Pi\|_{Y\to S}$ and $\|\Pi\|_{X\to X\cap S}$ are both bounded.

(The existence of such projection operators is in general a non-trivial assumption, as for Banach spaces there are generally closed subspaces which are not images of continuous projections, and for Hilbert spaces non-orthogonal projections may fail to be bounded.2)

This would be the case if, for example, $X, Y$ were Hilbert, and $S$ is such that whenever $x\in X$ is orthogonal to $S$ w.r.t. the $X$ inner product, it is also orthogonal w.r.t. the $Y$ inner product.

As Hannes noted in a comment below: the result alluded to above is given as Theorem 1 in Section 1.17.1 of Interpolation Theory, Function Spaces, Differential Operators (ed. Hans Triebel, North-Holland (1978)).

1: When $Y$ is separable, every closed subspace is a kernel, so in this case it is not more general. https://arxiv.org/abs/1811.02399

2: I don't see how to relate the existence of these projections to the additional conditions you are willing to assume, as stated in the question. But chances are this is because I don't know very much about complemented subspaces.

  • 1
    $\begingroup$ Chapter 1.17 in the Interpolation book of Triebel supports your complemented-subspace/projection suggestion. (And possibly worth a read for OP in any case.) $\endgroup$
    – Hannes
    Feb 23 at 8:58
  • $\begingroup$ Thank you @Willie Wong. Its very useful, $\endgroup$
    – M.Oud
    Feb 23 at 10:32
  • $\begingroup$ @OUDRANE a quick note, however: the complemented subspace argument is sufficient, but not necessary. An example: the Hardy spaces $H^p(\mathbb{T})$ can be described as those $L^p$ functions (necessarily $\subset L^1$) whose Fourier transforms have no negative modes. $H^1$ is not complemented in $L^1$, but for $p\in (1,\infty)$ we do have $H^p$ is complemented in $L^p$. However the sort of interpolation result you are asking for is true for Hardy spaces. (In this case you can write $H^p$ as the kernel of some $f:L^1\to \ell^\infty$.) $\endgroup$ Feb 23 at 14:20
  • $\begingroup$ The proof of that last statement as given in numdam.org/article/AIF_1992__42_4_875_0.pdf is based on relaxing the conditions on $\Pi$: instead of looking for a single projection operator, the author constructs a family of uniformly bounded operators (not necessarily projections!) $\Pi_f: L^1 \to H^1$, indexed by $f\in H^1$, such that $\Pi_f(f) = f$. $\endgroup$ Feb 23 at 14:30
  • $\begingroup$ Thank you a lot @Willie Wong , I think just the condition of the existence of projections is enough for me ( for my original probleme ). But this is very interesting!. And also I found here a paper devoted to this probleme researchgate.net/publication/… . They try to go more general on the conditions that allow the interpolation functor to be compatible with the Kernel. $\endgroup$
    – M.Oud
    Feb 24 at 6:53

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