The best constant of Sobolev inequality on a bounded interval

Abstract

The best constants C m,j of Sobolev embedding of H m (0, a) into C j [0, a] (0 j m − 1) are obtained. Especially, when a = ∞, these constants can be represented in a closed form.

A Note on the Volume Growth Criterion for Stochastic Completeness of Weighted Graphs

“…We need the following Sobolev embedding W 1,2 ((0, l)) ⊂ C([0, l]) with optimal constant ([45], [49]), sup t∈[0,l] |u(t)| 2 ≤ coth(l) l 0 u 2 (t) + (u ′ (t)) 2 dt.…”

Validated bounds on embedding constants for Sobolev space Banach algebras

“…and together with |k| ≥ 2M and | | ≤ M < 2M ≤ |k|, this implies both 1 − | |∕|k| > 0 as well as the inequality 1 − | |∕|k| ≥ 1 − | |∕(2M) > 0 for all | | ≤ M. Using these estimates, the first sum in (30) can be bounded as…”

“…While Sobolev space embeddings are of general importance in the study of partial differential equations, their application in the context of computer-assisted proofs for nonlinear problems relies crucially on the explicit knowledge of the associated embedding constants. Even though there are results for general Sobolev spaces on one-dimensional domains, [28][29][30] as well as for Sobolev spaces on higher-dimensional domains subject to homogeneous Dirichlet boundary conditions, 11 in the context of problems subject to homogeneous Neumann boundary conditions, these constants cannot easily be found in the literature. This is true even for the important case of higher-dimensional rectangular domains.…”

“…The two sums on the right-hand side will now be estimated separately. We begin by considering the first sum in (30). This sum basically approximates the sum of one of the two humps in the function values of g that were mentioned earlier in this section.…”

“…and together with the factor 2 in (30), this gives precisely the first term in the right-hand side of (29), after dividing the numerator and denominator of the fraction by 4 . We now turn our attention to the second sum in (30). The summation multi-index in this sum satisfies | | ≤ |k − |, and we therefore have |k| ≤ |k − | + | | ≤ 2|k − |.…”