This question has been posted on Math Stack exchange for a while and received no response. So I decide to move it here to get more attention.
Let Ω⊂RN be open, bounded and with smooth boundary. Then we could prove that for any u∈BV(Ω) and ω∈kerE, where E denotes the distributional symmetric derivative Eω=12(∇ω+∇ωT) from RN to RN, there exists a constant C>0 independent of u and ω such that
‖Du‖M(Ω)≤C(‖Du−ω‖M(Ω)+‖u‖L1)
The proof is not long and can be found here, Theorem 3.3, equation (4).It can also be shown that ω∈kerE iff ω=Ax+b where A=−AT, A∈RN×N and b∈RN
However, the proof is done by using contradiction, which is short and simple but can not give any information of this constant C.
I am interested in finding the best constant C>0. I am sure this constant should only depend on Ω but I really want to know how it depends on Ω. Can we make the best constant larger or smaller by changing Ω? Also, if Ω:=[0,1]×[0,1] in R2, can we explicitly compute this constant?
Thank you!
Answer
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Source : Link , Question Author : JumpJump , Answer Author : Community