# Intersection of the kernel with the interpolation space

$$\DeclareMathOperator\Ker{Ker}$$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 $$0 and $$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 $$f(X)$$.

• 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. Feb 22 at 22:07
• 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? Feb 23 at 2:04
• 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. Feb 23 at 3:22

(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.

• 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.) Feb 23 at 8:58
• Thank you @Willie Wong. Its very useful, Feb 23 at 10:32
• @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 $f:L^1\to \ell^\infty$.) Feb 23 at 14:20
• 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 $\Pi_f(f) = f$. Feb 23 at 14:30
• 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. Feb 24 at 6:53