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As shown in Strauss: Existence of solitary waves in higher dimensions, Strauss introduces the Stauss lemma. Precisely speaking, we have the following theorem:

Theorem Let $N \ge 2$, every radial function $u \in H^1(\mathbb{R}^N)$ is almost everywhere equal to a function $U(x)$, continuous for $x \not = 0$, such that \begin{equation} |x|^{\frac{N-1}{2}} U(x) \lesssim \| u \|_{H^1}, \end{equation} where the constant only depends on $N$.

In fact, we can get more precise estimate. Concretely, by the equation \begin{equation} -(r^{N-1} u^2)_r = -(N-1) r^{N-2} u^2 - 2 r^{N-1} u u_r \le -2 r^{N-1} uu_r, \end{equation} we integrate over $[r,+\infty)$ to obtain that \begin{equation} r^{N-1}u^2(r) \lesssim \int_r^\infty s^{N-1} |u(s)| |u_s(s)| ds \lesssim \| u \|_{L^2} \| \nabla u \|_{L^2}, \end{equation} thus $|x|^{\frac{N-1}{2}} U(x) \lesssim \| u \|_{L^2}^\frac{1}{2} \| \nabla u \|_{L^2}^\frac{1}{2}$.

My question is that intuitively speaking, it seems that the weight $|x|^{\frac{N-1}{2}}$ on the LHS can be controlled by the "half-gradient" on RHS. However, if the power of weight $|x|^{\alpha}$ becomes smaller, can we expect the less gradient on the right? Precisely speaking, if $\alpha <\frac{N-1}{2}$, can we have \begin{equation} |x|^{\alpha} U(x) \lesssim \| u \|_{L^2}^{1-\beta} \| \nabla u \|_{L^2}^\beta, \; |x| \ge 1 \end{equation} for some $\beta<\frac{1}{2}$?

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First, you got the scaling wrong. The correct scaling for

$$ |x|^\alpha u(x) \lesssim \|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta $$

would be $\alpha = \frac{N}{2} - \beta$ where $N$ is the spatial dimension. So for smaller $\alpha$ you need more $\beta$, not less.

For $\beta \in [\frac12, 1]$ ($\beta = 1$ only works in $N > 2$) the desired inequality can be proven using essentially the same argument as what you gave for Strauss's Lemma.

Set $$ p := \frac{\beta + \frac{N}2 - 1}{\frac{N}2 - \beta} $$ Note that when $\beta = \frac12$ you have $p = 1$. And when $\beta = 1$ you have $p = \frac{N}{N - 2}$. And within this region you have $p \geq 1$.

and run the argument using instead of $ r^{N-1} u^2$, the function $ r^{N-1} |u|^{p+1} $ instead. The same argument you gave shows that

$$ r^{N-1} |u|^{p+1} \lesssim \int r^{N-1} |u|^{p} |\partial_r u| $$

Cauchy-Schwarz the RHS you get

$$ \lesssim \| u\|_{L^{2p}(\mathbb{R}^N)}^p \|\nabla u\|_{L^2} $$

This gives

$$ r^{\frac{N-1}{p+1}} |u| \lesssim \|u\|_{L^{2p}}^{\frac{p}{p+1}} \|\nabla u\|_{L^2}^{\frac{1}{p+1}} $$

The first term has $2p \in [2,\frac{2N}{N-2}]$ so by Gagliardo-Nirenberg-Sobolev inequality, can be bounded by

$$ \|u\|_{L^{2p}} \lesssim \|u\|_{L^2}^{\theta} \|\nabla u\|_{L^2}^{1-\theta} $$ for some $\theta \in [0,1]$. If you plug in the formula for $p$ in terms of $\beta$ above, you will find, after some routine algebra, the expression listed at the beginning of this answer.


On the other hand, there cannot be any estimate with $\beta < \frac12$. This can be seen by the following counterexample.

Let $u$ be a pulse around $r = 1$, with thickness $\epsilon$ and height 1.

$$ \|u\|_{L^2} \approx \epsilon^{1/2} $$

$$ \|\nabla u\|_{L^2} \approx \epsilon^{- 1/2} $$

Thus

$$ \|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta \approx \epsilon^{\frac12 - \beta} $$

If you take $\beta < \frac12$ and $\epsilon \searrow 0$, you get a sequence of functions with $\|u\|_{L^2}^{1-\beta} \|\nabla u\|_{L^2}^\beta \searrow 0$ but unit height, contradicting any possible control of $L^\infty$.

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  • $\begingroup$ Incidentally, in your question you restricted to $|x| > 1$... in this case smaller $\alpha$ is strictly dominated by larger $\alpha$. If your goal is to get stronger decay condition at infinity, you want larger $\alpha$ and not smaller. I am pretty sure there are counterexamples for the estimate when $\beta < \frac12$; will post a second answer if I figure it out. $\endgroup$ Commented Apr 5, 2022 at 23:11
  • $\begingroup$ Thank you so much. The original question is about whether I can get less power of gradient term to control the behavior of radial function at infinity (at cost of less decay: to be honest, in my setting, I do not care much about how much decay it will give us even if $|x|^{-\varepsilon}$ is enough), and the "$\beta< \frac{1}{2}$" part in your answer seems completely solved my question. I am also looking forward to your answer about whether we can get stronger decay condition at infinity if you have figured it out. $\endgroup$
    – Tao
    Commented Apr 6, 2022 at 5:49

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