We refer to the 'reproducing convolution formula with a kernel' for an open bounded domain $U$ of $R^n$, $n \geq 2$ discussed in the paper of G. Talenti (Annali de Matematica, Dec 1976) on Best constant in Sobolev inequality. This gives $u$ in terms of its gradient $\nabla u$ for a bounded open set $U$. My question is,

For an appropriate pointwise relation from a subsequence, does the reproducing kernel formula for a bounded open set $U$ define an equivalent norm on the Sobolev space $H^1_0(U)$ ?

Note that the smooth functions with compact support are dense in $H^1_0(U)$ but not in $H^1(U)$,so the closure in regard of the test space applies on both sides of the kernel formula only for $H^1_0(U)$, while the closure applies to $L^2$ as well.

Note that on a suitable subsequence, pointwise relations hold almost everywhere; in fact, the limits can be considered in $L^p(U)$ when the support vanishes outside $U$, due to integrability of the kernel in $dim.n$ for $n \geq 2$. I also wish to add that the Atiyah-Singer index is zero for $n > 2$. Interestingly, it is awkward to show membership for both the spaces using the same $H^1$ norm,so the equivalence desired produces different norms.

Since Poincare inequality can be uniformly used for all $L^q(U)$, Hedberg estimates (Lars Inge Hedberg, Proc. AMS (1972) on convolution inequalities) appears to give the estimate for the $L^p$ norm of $\nabla u$ (not $u$), which I wish to confirm.

Note also that participation of the kernel is actually the reason for increased Lebesgue index $p*$

My research interests are applied analysis, PDE, Microlocal Analysis, Infinity Laplacian and Pseudo-differential operators.


1 Answer 1


(I want to post a potential answer that I have researched over the past months; if this is better placed as a comment, please advise)

With reference to the paper On certain convolution inequalities Lars Inge Hedberg, Proc AMS, Vol 36, No.2, Dec 1972, see the pointwise estimate (3) in terms of the Maxmimal function on pp 506 along (1) on pp 505, taking $\alpha = 1$ and replacing $f$ by $grad(f)$.

Now estimate the $p*$-norm over $\mathbb{R}^d,d \geq 2$, using the reproducing formula, noting $p* = (p*-1) + 1$, use Holder estimate for this split. Keeping aside the 'link constants' c, A, the Holder conjugate corresponding to ($p* - 1$) (for the $p*$ power) is estimated using the maximal inequality for $p > 1$ in integration over $\mathbb{R}^d$ by $|\nabla f|^{g(p)}, g(p): n - \frac{n}{p} + 1, 1 < g(p) < p < n;$ note that $p \leftrightarrow g(p)$ for $p$ in $(1,n)$ is one-one and onto, as can be seen from the graph of $g(p)$. The trajectory $g(p) = n - \frac{n}{p} + 1, p \in (1,n)$ corresponds to the 'limiting case' of the staircase constructed in Charles J Arick's paper, Proc. LMS (1978). As for the standard 'counter examples' to RK compactness at $p*$, with a choice of any $q$ on the trajectory, 'gain' is seen over and above $1 - n/p + n/p* = 0$.

Thus, equivalence of norm is seen and in this norm, identification of $H^1_0(U)$ as a Nuclear space is compact.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.