Extension by zero of functions of H^1

Let ω\omega be an open subset of an other bounded open set Ω\Omega of Rn,n∈N.R^n,\; n\in N. For an element ff of H1(ω)H^1(\omega) we consider the function defined by F(x)=f(x),x∈ωF(x)=f(x),\; x\in \omega and F(x)=0F(x)=0 else. Are there any conditions on ff such that F∈H1(Ω)?F\in H^1(\Omega)?

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You may want to assume some regularity at least of the boundary of ω\omega. In full generality, this is probably a very hard question (but I stand to be proven wrong).
– MaoWao
2 days ago

  

 

math.stackexchange.com/questions/1387983/…
– daw
2 days ago

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1 Answer
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Here, the following holds: if f∈H10(ω)f\in H^1_0(\omega) then F∈H1(Ω)F\in H^1(\Omega).
If the extension-by-zero FF is in H1(Ω)H^1(\Omega) and ω\omega has C0C^0-boundary (ω\omega is locally on one side of the boundary), then f∈H10(ω)f\in H^1_0(\omega) necessarily.

If ff belongs to H10(ω)H^1_0(\omega), then the extension by zero belongs to H10(Ω)H^1_0(\Omega). This is intuitively true: extending a function with zero boundary values by zero will not introduce new singularities. Since C∞cC_c^\infty-functions are dense in H10(ω)H^1_0(\omega), we can find a sequence fkf_k of such functions approximating ff. Extending fkf_k by zero yields fk∈C∞c(Ω)f_k\in C_c^\infty(\Omega), fk→Ff_k\to F in H1(Ω)H^1(\Omega) and F∈H1(Ω)F\in H^1(\Omega).

The reverse statement is much more difficult to prove. Since the boundary is C0C^0, we have a suitable covering U0…UmU_0\dots U_m of ω\omega, where
UiU_i, i=1…mi=1\dots m, contain parts of the boundary. Then we can take a parition of unity ψi\psi_i subordinate to this covering. The final step is to translate ψiF\psi_i F inside ω\omega and mollify there. This yields an approximation of FF by C∞c(ω)C_c^\infty(\omega) functions.

A proof of the latter statement for C1C^1-boundary can be found in Evans [Section 5.5, Thm 2]. There, it is proven that if f∈H1(Ω)f\in H^1(\Omega) and trace of ff is zero, then f∈H10(Ω)f\in H^1_0(\Omega).

I could not found a reference for the sketched proof above.