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Let $X,Y$ be Banach spaces, $1\leq p\leq \infty$ and $\theta\in (0,1)$. Then we want to define the real interpolation space $(X_0,X_1)_{\theta,p}$. This is possible via the trace method: Define the space $V$ of $X+Y$-valued regular distributions $u$ on $(0,\infty)$ such that $t^\theta u(t) \in \mathrm{L}^p_*(0,\infty; X_0)$ and $t^\theta u'(t) \in \mathrm{L}^{p'}_*(0,\infty; X_1)$ where $\mathrm{L}^q_*$ denotes the Lebesgue space with weight $1/t$ and $p'$ is the conjugate exponent to $p$. Then one can show that point evaluation in $0$ is meaningful and consider $V/\{u: u(0)=0\}$. This is a Banach space and via evaluation in $0$ we can define the interpolation space $(X_0,X_1)_{\theta,p}$.

One then needs to show that this construction actually is an interpolation method. Most books do this by showing equivalence to some other real interpolation method, e.g. the K-method.

I want to show this directly. It seems reasonable to try to show that $Tu\in V$ for $T: X_0+X_1\to Y_0+Y_1$. I have difficulties to show first that $Tu$ is a regular distribution and would like to have $(Tu)'=T(u')$ which would show the claim.

Does anybody have ideas on how to resolve those two problems or a reference where this calculation is done?

The only source I have found that directly shows the interpolation property is by Lions and Peetre (http://www.ams.org/mathscinet-getitem?mr=165343, in frensh), however their definitions are different from mine and I was not able to translate them to my setting.

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You should use the fact that $u:(0,\infty)\mapsto X_0+X_1$ satisfies the fundamental theorem of calculus, that is, $u(t)=u(t_0)+\int_{t_0}^tu'(r)\,dr$. Then since $T:X_0+X_1\mapsto Y_0+Y_1$ is linear and continuous you get \begin{align}T(u(t))&=T\left(u(t_0)+\int_{t_0}^tu'(r)\,dr\right)\\&=T(u(t_0))+T\left(\int_{t_0}^tu'(r)\,dr\right)=T(u(t_0))+\int_{t_0}^t T(u'(r))\,dr,\end{align} where in the last equality you are using the fact that the Bochner integral commutes with linear continuous functions $T$. Since this is true for every $t$ and $t_0$ you get that the function $T(u(t))$ is differentiable for a.e $t$ and $(T(u(t)))'=T(u'(t))$. To prove this you have to use Lebesgue points. The lemma that you want is that if $f:I\mapsto Z$ is Bochner integrable, then for a.e $t$,$$\frac{1}{h} \int_{t}^{t+h}f(r)\,dr\to f(t)$$ as $h\to 0$. This follows by applying the result for real valued functions to $t\mapsto\Vert f(t)-z_n\Vert$ where $\{z_n\}$ is dense in $f(I)$. I am skipping a lot of details but I hope this helps.

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  • $\begingroup$ Thanks for your answer! The point that I missed is, that actually $u$ and $T$ are continuous into $X+Y$... I think we can then replace your argument by an easier one using difference quotients, then we don't have to deal with properties of the Bochner integral. $\endgroup$ – Sebastian Bechtel Jun 7 '17 at 6:37
  • $\begingroup$ I am not sure. Differentiability is not enough. You need the fundamental theorem of calculus for $T\circ u$ to conclude that $T\circ u$ is absolutely continuous. $\endgroup$ – Gio67 Jun 7 '17 at 17:34
  • $\begingroup$ That $T\circ u$ is absolutely continuous is a byproduct in the proof that the trace method actually makes sense. That $t^\theta (T\circ u)'(t)$ belongs to $\mathrm{L}^p_*(0,\infty;X_1)$ should conclude the proof, I would say? $\endgroup$ – Sebastian Bechtel Jun 7 '17 at 19:26
  • $\begingroup$ Nevertheless I will reward the bounty to you. I missed to realize the continuity into $X_0+X_1$ and your answer pointed me to this issue. $\endgroup$ – Sebastian Bechtel Jun 7 '17 at 19:28
  • $\begingroup$ In the trace method that I know the function $u$ is in the $L^p$ space you wrote and it has a weak or distributional derivative in $L^p$ space above. But this is just a definition. So if you are trying to prove that $T\circ u$ is admissible for the trace method, you need to show that it has a weak derivative. It is not enough that it is differentiable a.e. The strong derivative is not the weak one in general (think about the Cantor function). So I am confused on how else you are proving that $T\circ u$ has a weak derivative. The difference quotient will give you strong but not weak diff. $\endgroup$ – Gio67 Jun 7 '17 at 20:17

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