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Consider the $L^p$ estimate of the Laplacian on a compact boundaryless Riemannian manifold, suppose that $-\Delta_ge_{\lambda}=\lambda^2e_\lambda(x), x\in M$. C.D. Sogge proved that we have the following $$ \|e_{\lambda}\|_{L^p(M)}\leq C\lambda^{(\frac{1}{2}-\frac{1}{p})}\|e_{\lambda}\|_{L^2(M)}, \quad 2\leq p\leq 6,\\ \|e_{\lambda}\|_{L^p(M)}\leq C\lambda^{2(\frac{1}{2}-\frac{1}{p})-\frac{1}{2}}\|e_{\lambda}\|_{L^2(M)}, \quad 6\leq p\leq \infty. $$ Such estimates are sharp because for the round sphere $S^2$, the first one is sharp because of the highest weight spherical harmonics and the second one is sharp due to the zonal functions on $S^2$. A natural question is if we can get improved estimates under some additional assumption, indeed, if the manifold is everywhere nonpositive, then again Sogge can show that, when $2<p< 6$, the above bound can improve to be $o(\lambda^{(\frac{1}{2}-\frac{1}{p})})$, but the endpoint case $p=6$ is not known, however, it's valid for the standard torus $\mathbb T^2$ due to a result by Zygmund who showed that one has $\frac{\|e_{\lambda}\|_{L^4(\mathbb T^2)}}{\|e_{\lambda}\|_{L^2(\mathbb T^2)}}=o(1)$, and interpolation with $p=\infty$.

My question is why is it hard to prove the improved bound for $p=6$ in general (with nonpositive curvature)? The improved bound is of interested since one of its application can lead to the better lower bounds of the measure of the nodal sets, which is still open for the $C^\infty$ case.

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    $\begingroup$ This is very interesting: it is an appearence of $L^6$ in nature! $\endgroup$ Oct 28, 2013 at 5:58
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    $\begingroup$ I have a doubt about the exponents. Because we may always interpolate by using Hölder, the exponents should match at $p=6$, while yours are $\frac13$ (for $p=6-0$) and $\frac16$ (for $p=6+0$). $\endgroup$ Dec 8, 2020 at 10:42
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    $\begingroup$ I have a doubt about Zygmund’s result. The paper matwbn.icm.edu.pl/ksiazki/sm/sm50/sm50112.pdf proves L4-L2 bounds with universal constant (instead of little o of 1) which is sharp for monomials. Also can you provide the reference of Sogge’s bounds? $\endgroup$ Dec 8, 2020 at 13:04
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    $\begingroup$ Apparently the exponent in the first case should be $\frac{1}{2}(\frac{1}{2}-\frac{1}{p})$. With this correction they agree for p=6. $\endgroup$ Dec 8, 2020 at 16:12

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