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I posted this on Math StackExchange, but I figured it couldn't hurt to ask here as well.

I'm trying to decipher a particular claim in a paper I'm reading, but I just can't seem to figure it out.

The M. Riesz Interpolation Theorem says:

Let $T:L^{p_0}\cap L^{p_1} \to L^{q_0} \cap L^{q_1}$ be a linear map satisfying: $$||Tf||_{q_j}\leq M_{j}||f||_{p_j}, \quad j=0,1$$ with $1\leq p_j,q_j\leq\infty$. Then, if for $0< t <1$, we define $$\frac{1}{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1}, \quad \frac{1}{q_t}=\frac{1-t}{q_0}+\frac{t}{q_1},$$ we have $$||Tf||_{q_t}\leq (M_0)^{1-t}(M_1)^{t}||f||_{p_t}, \quad f\in L^{p_0}\cap L^{p_1}.$$

Note that $p_0 \leq p_t \leq p_1$ and $q_0 \leq q_t \leq q_1$.

The paper I'm reading gives the following:

$$||e_\lambda||_p \leq C \lambda^{\delta(p)}||e_\lambda||_{2}, \quad 2\leq p \leq 6$$

$$\delta(p) = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{p} \right)$$

Here $e_\lambda$ is an eigenfunction (with eigenvalue $\lambda$) of $\sqrt{-\Delta}$ on a fixed, compact Riemannian manifold $M$, with no boundary, and $C$ is a fixed constant depending only on the manifold. Note in particular that $\delta(2)=0$ and $\delta(6)=\tfrac{1}{6}$.

It then proceeds to claim that using the above result for $p=2$ and $p=6$, along with the interpolation theorem, we have the following:

$$\limsup_{k\to\infty} \lambda_{k}^{-\delta(p)}||e_{\lambda_k}||_{L^{p}(M)} = 0$$ is true for $2<p<6$ if and only if it is true for $p=4$, where the eigenfunctions are all taken to have unit $L^{2}(M)$-norms, and $\lambda_1 \leq \lambda_2 \leq \cdots.$

If we let $T=E_{k}$ be the projection onto the $k^{\text{th}}$ eigenspace, that is, $$E_{k}f = \langle f,e_{\lambda_k}\rangle e_{\lambda_k}(x), \quad \langle f,e_{\lambda_k}\rangle = \int_{M}f(x)\overline{e_{\lambda_k}(x)}dx, \quad \forall f\in L^{2}(M)$$ then $E_{k}:L^{2} \to L^{q_0}\cap L^{q_1}$, for any $2\leq q_0,q_1\leq 6$, and it seems like we should be able to use the interpolation theorem in some manner.

I'm lost though when it comes to showing the above claim is true for $2<p<6$, assuming it is true for $p=4$. Moreover, I feel like I've seen this kind of trick a lot, where we only have to prove a statement for some particular value of $p$ in order for it to be true for some larger range of values. What am I missing?

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My answer is for $p \in (2,4)$, the other case should follow similar.

Let $t \in (0,1)$ be given, such that $$\frac1p = \frac{1-t}{2}+\frac{t}{4}.$$ I will go to use the following consequence of Hölder (this is similar to the Riesz Interpolation Theorem, but somewhat simpler to use in your case): $$\lVert f \rVert_p \le \lVert f\rVert_2^{1-t} \lVert f\rVert_4^t$$ Using $\delta(p) = (1-t)\;\delta(2) + t \; \delta(4)$ you find $$\lambda_k^{-\delta(p)} \lVert e_k \rVert_p \le \Big(\lambda_k^{-\delta(2)}\lVert e_k\rVert_2\Big)^{1-t}\Big(\lambda_k^{-\delta(4)}\lVert e_k\rVert_4\Big)^t = \Big(\lambda_k^{-\delta(4)}\lVert e_k\rVert_4\Big)^t.$$ This shows, that your lim-sup holds for all $p \in (2,4)$ if it holds for $p = 4$.

Edit: A intuition for the statement is the following: You have $$\lambda_k^{-\delta(p)} \lVert e_k \rVert_p$$ is bounded for $p \in [2,6]$ as $k \to \infty$. If for some $p \in (2,6)$, the $\limsup$ is zero, than you can interpolate and obtain a zero $\limsup$ for all $p \in (2,6)$. Note there is nothing special about $p = 4$.

A similar argument is: if a sequence of functions $f_k$ is bounded in $L^p$ and converges towards zero in $L^q$, $q < p$, then it converges to $0$ in $L^r$ for all $r \in [q,p)$.

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  • $\begingroup$ Thanks. That really helps. This is the [Hölder interpolation][1] theorem you're referencing, correct? In the case when $p_t\in(4,6)$ though, we end up with $$\lambda_k^{-\delta(p_t)}||e_{\lambda_k}||_{p_t} \leq \left( \lambda_k^{-\delta(4)}||e_{\lambda_k}||_4 \right)^{1-t}\left( \lambda_k^{-\delta(6)}||e_{\lambda_k}||_6 \right)^{t}$$ and we can no longer rely on the unit-norm in $L^{2}(M)$ nor the fact that $\delta(2)=0$. How do we proceed from here? [1]: en.wikipedia.org/wiki/… $\endgroup$
    – Patch
    Mar 28, 2013 at 10:19
  • $\begingroup$ You can use $\lVert e_k \rVert_6 \le C \, \lambda_k^{\delta(6)} \, \lVert e_k \rVert_2$. $\endgroup$
    – gerw
    Mar 28, 2013 at 10:41
  • $\begingroup$ Oh, of course! Sorry. Thanks again. $\endgroup$
    – Patch
    Mar 28, 2013 at 10:44
  • $\begingroup$ The value $p=4$ might not be special, but it is the only instance in which we are able to prove the $\limsup$ convergence explicitly. $\endgroup$
    – Patch
    Mar 28, 2013 at 10:49

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