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This is probably not a research level question but I am struggling with the geometry. My question is related to whether some monotonicity can increase the range of exponents in the Sobolev embedding. For instance on the unit ball, nonnegative radially symmetric functions which are nondecreaing in the radial direction should satisfy a Sobolev embedding for an improved range of exponents: Indeed such functions must be large on the full boundary of the ball, yet the trace theorem prevents this from happening.

So I will ask the question on a finite cone; let $ S \subset \subset S^{N-1}$ be some nice spherical cap. For explicitness let us assume that the $x_1$ axis cuts through the center of $S$ (assume e.g. that $S$ is a ball in $S^{N-1}$ centered at some $y\in S^{N-1}$ lying on the $x_1$ axis). Let $ \Omega:=\{x=r \theta: 0<r<1, \theta \in S\}$ denote the finite cone. We now look at nonnegative functions which are zero on the side of the cone and moreover nondecreasing in the radial direction, i-e $ x \cdot \nabla u(x) \ge 0$ in $\Omega$.

So my question is:

Can we expect an improved Sobolev embedding for this class of functions? (here I am using the $H^1$ norm so I am asking about possible embedding of $H^1$ into some $L^p$ space for some improved exponents $p>2^*=\frac{2N}{N-2}$)

I suspect the answer is `no' and I am attempting to disprove it considering the first strategy that comes to mind: Take $ 0\le \phi \in C_c^\infty(B_1)$ a smooth radially nonincreasing function and then translate this function so that its support is centered at $y$, and scale it in order to concentrate its support at $y$. This sequence of functions will presumably violate any alleged improved Sobolev embedding, up to proving that these functions really have the correct monotonicity. Geometrically it looks like to me that this is so, but my geometric intuition almost always fails me now. Any comments would be great.

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  • $\begingroup$ I am confused.... or maybe i don't understand your comment. On the unit ball if you just ask for radial functions then there is no improved range. $\endgroup$
    – Math604
    Aug 3, 2020 at 18:13
  • $\begingroup$ I misunderstood your question. Are you only interested in the $H^1 \to L^p$ embedding? // Also, when you say "radial increasing functions" do you mean functions that increase in the radial direction, or do you mean radially symmetric functions that also increase in the radial direction? $\endgroup$ Aug 3, 2020 at 18:19
  • $\begingroup$ ya, i was not overly precise in my question. So let $X$ denote the functions with $H^1(\Omega)$ norm; the functions are zero on side of cone and the functions are nonnegative and satisfy $u_r \ge 0$ or $ x \cdot \nabla u(x) \ge 0$. So they are radial increasing but there is no radial symmetry assumption. $\endgroup$
    – Math604
    Aug 3, 2020 at 18:25

1 Answer 1

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The answer is indeed "no." To see this, instead of considering translations of radial bump functions, one needs to use bump functions whose level sets are deformed e.g. to ellipsoids. More precisely, denote $x \in \mathbb{R}^n$ by $(x_1,\,x')$ and let $h$ be any decreasing function on $\mathbb{R}$. Then for $$H(x) := h(\Lambda^2 x_1^2 + |x'|^2)$$ we have that $x \cdot \nabla [H(x-y)] \geq 0$ in the ellipsoid $$\left\{\frac{(x_1 - 1/2)^2}{(1/2)^2} + \frac{|x'|^2}{(\Lambda/2)^2} \leq 1\right\},$$ which contains a neighborhood of $y$ in $B_1$ when $\Lambda$ is large (but not $1$). Now fix $\Lambda$ large and choose $h$ smooth with $h(s) = 1$ for $s \leq 0 $ and $h(s) = 0$ for $s \geq 1$. Then the family $$\lambda^{\frac{n-p}{p}}H(\lambda (x-y))$$ shows as $\lambda \rightarrow \infty$ that the exponent cannot be improved.

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  • $\begingroup$ @ Connor Mooney. Thank you very much for your answer. Helpful as always. $\endgroup$
    – Math604
    Aug 4, 2020 at 5:03

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