Injection of the intersection of Banach spaces on the real interpolation space $(A_0,A_1)_{\theta,\infty}$.

Question

Let $A_0$, $A_1$ be two Banach spaces, both injected continuously in a Hausdorff topological vector space $\mathcal{A}$. Then we can consider the normed spaces $A_0 \cap A_1$ and $A_0 + A_1$ with the norms

$$ ||a||_{A_0 \cap A_1} = \max\{ ||a||_{A_0}, ||a||_{A_1}\}, $$ $$ ||a||_{A_0 + A_1} = \inf \{ ||a_0||_{A_0} + ||a_1||_{A_1} : a=a_0+a_1, a_j \in A_j\}, $$

and following this, define the real interpolation space $(A_0,A_1)_{\theta,\infty}$ consisting of those elements $a \in A_0 + A_1$ with finite norm:

$$ ||a||_{\theta,\infty} = \sup_{t>0} \{ t^{-\theta} K(t,a) \}, $$

where $K(t,a) = \inf\{ ||a_0||_{A_0} + t||a_1||_{A_1}: a=a_0+a_1, a_j \in A_j\}$ is Peetre K-functional.

My task now is to prove that this is an actual intermediate space, that is,

$$ A_0 \cap A_1 \hookrightarrow (A_0,A_1)_{\theta,\infty} \hookrightarrow A_0 + A_1, $$

are both continuous injections.

In my tries to prove the continuity of the first one I'm struggling quite a bit. Here is what I have so far:

Let $a \in A_0 \cap A_1 \hookrightarrow A_0$ and $s>0$, then:

$$ ||a||_{A_0 \cap A_1} = \max\{ ||a||_{A_0}, ||a||_{A_1}\} \geq ||a||_{A_0} \geq \inf \{ ||a_0||_{A_0} + s||a_1||_{A_1}: a=a_0+a_1, a_j \in A_j\} = K(s,a), $$

since $a = a + 0$ would be a feasible decomposition in $A_0 + A_1$. Given this, I have

$$ s^{-\theta} ||a||_{A_0 \cap A_1} \geq s^{-\theta} K(s,a), $$

for any $s>0$. Now, for any $\delta>0$, there exists $s_\delta>0$ so that

$$ s_{\delta}^{-\theta} ||a||_{A_0 \cap A_1} \geq s_{\delta}^{-\theta} K(s_{\delta},a) \geq \sup_{t>0} \{ t^{-\theta} K(t,a) \} - \delta = ||a||_{\theta,\infty} - \delta, $$

This is almost what I want, since

$$ ||a||_{\theta,\infty} \leq s_{\delta}^{-\theta} ||a||_{A_0 \cap A_1} + \delta. $$

However, since $s_\delta$ is not fixed and it depends on $\delta$, I can't make $\delta \rightarrow 0$ and finish the proof. Any help going forward with this would be greatly appreciated.

Answer 1

So after distancing myself from that unfruitful path, I think I finally found a much cleaner way of proving this.

Since $a \in A_0 \cap A_1$, I can use two particular decompositions of $a$:

$$ a = a + 0, \; a\in A_0, 0\in A_1 $$ $$ a = 0 + a, \; 0\in A_0, a\in A_1 $$

First, if $0<t<1$:

$$ t^{-\theta} K(t,a) \leq t^{-\theta} \|a\|_{A_0} \leq \|a\|_{A_0}. $$

On the other hand, if $t>1$:

$$ t^{-\theta} K(t,a) \leq t^{1-\theta} \|a\|_{A_1} \leq \|a\|_{A_1}. $$

So, for any $t>0$, we have that

$$ t^{-\theta} K(t,a) \leq \max\{\|a\|_{A_0}, \|a\|_{A_1}\} = \|a\|_{A_0 \cap A_1} $$

And taking the supremum over all $t>0$, it follows that

$$ \|a\|_{\theta,\infty} \leq \|a\|_{A_0 \cap A_1}. $$

Finally proving that $A_0 \cap A_1 \hookrightarrow (A_0, A_1)_{\theta,\infty}$.