In the linear sigma model, the Lagrangian is given by
L=12∑Ni=1(∂μϕi)(∂μϕi)+12μ2∑Ni=1(ϕi)2−λ4(∑Ni=1(ϕi)2)2
(for example see Peskin & Schroeder page 349).
When perturbatively computing the effective action for this Lagrangian the derivative δ2Lδϕk(x)δϕl(x) needs to be computed. (for instance, Eq. (11.67) in P&S):
δ2Lδϕk(x)δϕl(x) = −∂2δkl+μ2δkl−λ[ϕiϕiδkl+2ϕkϕl].
My question: where are two delta functions?
If you don't understand why there need them, I write full calculation:
δ2L[ϕ]δϕa(x)δϕb(y)=δ2δϕa(x)δϕb(y){12N∑i=1(∂μzϕi(z))(∂μzϕi(z))+...}=δδϕa(x){N∑i=1((∂μzϕi(z))(∂μzδδϕb(y)ϕi(z)))+...}=δδϕa(x)z{N∑i=1((∂μzϕi(z))(∂μzδibδ(z−y)))+...}=δδϕa(x){(∂μzϕb(z))(∂μzδ(z−y))+...}=(∂μzδabδ(x−z))(∂μzδ(z−y))+...=−δab(∂μz∂μzδ(x−z))δ(z−y)+...
You may see two delta functions there.