Let Ψ ε ( − Δ ) = ( 1 + ε ( − Δ ) ) − 1 \Psi_\varepsilon(-\Delta)=(1+\varepsilon(-\Delta))^{-1} Ψ ε ( − Δ ) = ( 1 + ε ( − Δ ) ) − 1 and let { φ j } j ≥ 0 \{\varphi_j\}_{j\ge0} { φ j } j ≥ 0 be an orthonormal L 2 L^2 L 2 -basis of eigenfunctions of − Δ -\Delta − Δ with eigenvalues λ j ≥ 0 \lambda_j\ge0 λ j ≥ 0 :
− Δ φ j = λ j φ j . -\Delta\varphi_j=\lambda_j\varphi_j. − Δ φ j = λ j φ j .
The spectral functional calculus gives
Ψ ε ( − Δ ) f = ∑ j ≥ 0 Ψ ε ( λ j ) f j φ j , f j = ⟨ f , φ j ⟩ L 2 , \Psi_\varepsilon(-\Delta)f=\sum_{j\ge0}\Psi_\varepsilon(\lambda_j)\,f_j\,\varphi_j, \qquad f_j=\langle f,\varphi_j\rangle_{L^2}, Ψ ε ( − Δ ) f = ∑ j ≥ 0 Ψ ε ( λ j ) f j φ j , f j = ⟨ f , φ j ⟩ L 2 ,
with multiplier Ψ ε ( λ ) = 1 1 + ε λ \Psi_\varepsilon(\lambda)=\dfrac{1}{1+\varepsilon\lambda} Ψ ε ( λ ) = 1 + ε λ 1 . Hence the integral kernel of Ψ ε ( − Δ ) \Psi_\varepsilon(-\Delta) Ψ ε ( − Δ ) is the series
G ε ( x , y ) = ∑ j ≥ 0 1 1 + ε λ j φ j ( x ) φ j ( y ) . \boxed{\,G_\varepsilon(x,y)\;=\;\sum_{j\ge0}\frac{1}{1+\varepsilon\lambda_j}\,\varphi_j(x)\varphi_j(y)\,.} G ε ( x , y ) = j ≥ 0 ∑ 1 + ε λ j 1 φ j ( x ) φ j ( y ) .
This series converges in the sense of distributions and in C ∞ C^\infty C ∞ away from the diagonal x = y x=y x = y . Applying the operator to f f f gives the usual kernel formula
( Ψ ε ( − Δ ) f ) ( y ) = ∫ S G ε ( x , y ) f ( x ) d V x . (\Psi_\varepsilon(-\Delta)f)(y)=\int_S G_\varepsilon(x,y)\,f(x)\,dV_x. ( Ψ ε ( − Δ ) f ) ( y ) = ∫ S G ε ( x , y ) f ( x ) d V x .
One checks immediately that
( − ε Δ x + 1 ) G ε ( x , y ) = ∑ j 1 + ε λ j 1 + ε λ j φ j ( x ) φ j ( y ) = ∑ j φ j ( x ) φ j ( y ) = δ y ( x ) (-\varepsilon\Delta_x+1)G_\varepsilon(x,y)=\sum_j \frac{1+\varepsilon\lambda_j}{1+\varepsilon\lambda_j}\varphi_j(x)\varphi_j(y)=\sum_j\varphi_j(x)\varphi_j(y)=\delta_y(x) ( − ε Δ x + 1 ) G ε ( x , y ) = ∑ j 1 + ε λ j 1 + ε λ j φ j ( x ) φ j ( y ) = ∑ j φ j ( x ) φ j ( y ) = δ y ( x )
(in the distributional sense), so indeed G ε G_\varepsilon G ε is the Green kernel for − ε Δ + 1 -\varepsilon\Delta+1 − ε Δ + 1 .
2) Convergence as ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0 : distributional and strong operator limits
From the spectral formula,
G ε ( x , y ) = ∑ j 1 1 + ε λ j φ j ( x ) φ j ( y ) . G_\varepsilon(x,y)=\sum_j \frac{1}{1+\varepsilon\lambda_j}\varphi_j(x)\varphi_j(y). G ε ( x , y ) = ∑ j 1 + ε λ j 1 φ j ( x ) φ j ( y ) .
For each fixed j j j we have 1 1 + ε λ j → 1 \dfrac{1}{1+\varepsilon\lambda_j}\to 1 1 + ε λ j 1 → 1 as ε → 0 \varepsilon\to0 ε → 0 . Thus the kernel coefficients tend to those of the formal series ∑ j φ j ( x ) φ j ( y ) \sum_j \varphi_j(x)\varphi_j(y) ∑ j φ j ( x ) φ j ( y ) , which equals δ y ( x ) \delta_y(x) δ y ( x ) in the sense of distributions. Concretely, for any test functions ϕ , ψ ∈ C ∞ ( S ) \phi,\psi\in C^\infty(S) ϕ , ψ ∈ C ∞ ( S ) ,
∬ G ε ( x , y ) ϕ ( x ) ψ ( y ) d x d y = ⟨ Ψ ε ( − Δ ) ϕ , ψ ⟩ L 2 ⟶ ⟨ ϕ , ψ ⟩ L 2 = ∬ δ ( x − y ) ϕ ( x ) ψ ( y ) d x d y , \iint G_\varepsilon(x,y)\,\phi(x)\,\psi(y)\,dx\,dy = \langle \Psi_\varepsilon(-\Delta)\phi,\psi\rangle_{L^2} \longrightarrow \langle\phi,\psi\rangle_{L^2} = \iint \delta(x-y)\phi(x)\psi(y)\,dx\,dy, ∬ G ε ( x , y ) ϕ ( x ) ψ ( y ) d x d y = ⟨ Ψ ε ( − Δ ) ϕ , ψ ⟩ L 2 ⟶ ⟨ ϕ , ψ ⟩ L 2 = ∬ δ ( x − y ) ϕ ( x ) ψ ( y ) d x d y ,
G ε ⇀ δ G_\varepsilon\rightharpoonup\delta G ε ⇀ δ as measures/distributions.
You can also state convergence in function spaces:
Strong L 2 L^2 L 2 convergence on functions. For any fixed f ∈ L 2 ( S ) f\in L^2(S) f ∈ L 2 ( S ) ,
∥ Ψ ε ( − Δ ) f − f ∥ L 2 2 = ∑ j ( 1 − 1 1 + ε λ j ) 2 ∣ f j ∣ 2 → ε → 0 0 \|\Psi_\varepsilon(-\Delta)f - f\|_{L^2}^2 = \sum_j\Big(1-\frac{1}{1+\varepsilon\lambda_j}\Big)^2 |f_j|^2 \xrightarrow{\varepsilon\to0}0 ∥ Ψ ε ( − Δ ) f − f ∥ L 2 2 = ∑ j ( 1 − 1 + ε λ j 1 ) 2 ∣ f j ∣ 2 ε → 0 0
by dominated convergence (the pointwise factors are ≤ 1 \le1 ≤ 1 and tend to 0 for each fixed j j j ). So Ψ ε ( − Δ ) → I \Psi_\varepsilon(-\Delta)\to I Ψ ε ( − Δ ) → I strongly on L 2 L^2 L 2 (and similarly on H k H^k H k if f ∈ H k f\in H^k f ∈ H k , because the multipliers are bounded by 1 uniformly).