{"record_type":"pith_number_record","schema_url":"https://pith.science/schemas/pith-number/v1.json","pith_number":"pith:2017:B4DGX4BMEN7TAXPUCCOIIT5DP4","short_pith_number":"pith:B4DGX4BM","schema_version":"1.0","canonical_sha256":"0f066bf02c237f305df4109c844fa37f25bf1a0ee4b96ca81da3472d33e7a7d3","source":{"kind":"arxiv","id":"1704.02599","version":1},"attestation_state":"computed","paper":{"title":"Traces for fractional Sobolev spaces with variable exponents","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.AP","authors_text":"Julio D. Rossi, Leandro M. Del Pezzo","submitted_at":"2017-04-09T13:19:38Z","abstract_excerpt":"In this note we prove a trace theorem in fractional spaces with variable exponents. To be more precise, we show that if $p\\colon\\overline{\\Omega}\\times \\overline{\\Omega}\\to (1,\\infty)$ and $q:\\partial \\Omega \\rightarrow (1,\\infty)$ are continuous functions such that \\[\n  \\frac{(n-1)p(x,x)}{n-sp(x,x)}>q(x)\n  \\qquad \\mbox{in} \\partial \\Omega \\cap\n  \\{x\\in\\overline{\\Omega}\\colon n-sp(x,x) >0\\}, \\] then the inequality $$\n  \\Vert f\\Vert _{\\scriptstyle L^{q(\\cdot)}(\\partial \\Omega)}\n  \\leq C \\left\\{\\Vert f\\Vert _{\\scriptstyle L^{\\bar{p}(\\cdot)}(\\Omega)}+\n  [f]_{s,p(\\cdot,\\cdot)} \\right\\} $$ holds. 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Rossi, Leandro M. Del Pezzo","submitted_at":"2017-04-09T13:19:38Z","abstract_excerpt":"In this note we prove a trace theorem in fractional spaces with variable exponents. To be more precise, we show that if $p\\colon\\overline{\\Omega}\\times \\overline{\\Omega}\\to (1,\\infty)$ and $q:\\partial \\Omega \\rightarrow (1,\\infty)$ are continuous functions such that \\[\n  \\frac{(n-1)p(x,x)}{n-sp(x,x)}>q(x)\n  \\qquad \\mbox{in} \\partial \\Omega \\cap\n  \\{x\\in\\overline{\\Omega}\\colon n-sp(x,x) >0\\}, \\] then the inequality $$\n  \\Vert f\\Vert _{\\scriptstyle L^{q(\\cdot)}(\\partial \\Omega)}\n  \\leq C \\left\\{\\Vert f\\Vert _{\\scriptstyle L^{\\bar{p}(\\cdot)}(\\Omega)}+\n  [f]_{s,p(\\cdot,\\cdot)} \\right\\} $$ holds. 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