Spectral method approximation of small Laplacian deformations on Sobolev spaces

I'll give a clean, self-contained explanation showing

and point out exactly where $G_\varepsilon\to\delta$ is used.

1) Representation by the Green kernel

On a compact Riemannian manifold $S$ the solution of

is the (symmetric, smooth off the diagonal) resolvent kernel $G_\varepsilon(x,y)$, and the solution $u$ of

has the representation

So the operator $(-\varepsilon\Delta+1)^{-1}$ is the integral operator with kernel $G_\varepsilon$​.

2) $L^\infty$-estimate (where the weak convergence to $\delta$ is relevant)

Integrate the kernel equation in $x$:

The Laplacian term integrates away (or use divergence theorem), so

Moreover the maximum principle gives $G_\varepsilon\ge 0$. Thus $G_\varepsilon(\cdot,y)$ is a probability density on $S$ for each fixed $y$. From the kernel representation

so

(Here the weak convergence $G_\varepsilon\rightharpoonup\delta$ explains that the kernel concentrates as $\varepsilon\downarrow0$, and in the limit the integral operator tends pointwise to evaluation at the point — but the inequality above does not require taking limits: it follows from positivity and unit mass of $G_\varepsilon$​.)

3) $H^k$-estimate (spectral / Fourier multipliers)

Let $\{\varphi_j\}$ be an $L^2$-orthonormal basis of eigenfunctions of $-\Delta$ with eigenvalues $\lambda_j\ge0$:

$-\Delta\varphi_j=\lambda_j\varphi_j.$

Expand $f=\sum_j f_j\varphi_j$​, then

$u=(1+\varepsilon\lambda_j)^{-1}f_j\varphi_j,$

so the Fourier coefficient of $u$ on $\varphi_j$​ is $u_j=\dfrac{f_j}{1+\varepsilon\lambda_j}$. The $H^k$-norm squared is (up to equivalent normalization)

$\|w\|_{H^k}^2 \simeq \sum_j (1+\lambda_j)^k |w_j|^2.$

Hence

$\|u\|_{H^k}^2 = \sum_j \frac{(1+\lambda_j)^k}{(1+\varepsilon\lambda_j)^2}\,|f_j|^2.$

But for every $\lambda_j\ge0$ and every $\varepsilon>0$,

$\frac{(1+\lambda_j)^k}{(1+\varepsilon\lambda_j)^2}\le (1+\lambda_j)^k\cdot 1 = (1+\lambda_j)^k,$

because $(1+\varepsilon\lambda_j)^2\ge1$. Therefore

$\|u\|_{H^k}^2 \le \sum_j (1+\lambda_j)^k |f_j|^2 = \|f\|_{H^k}^2,$

so

$\|u\|_{H^k}\le \|f\|_{H^k}.$

(Equivalently, the multiplier $(1+\varepsilon\lambda)^{-1}$ has modulus $\le 1$, so $(-\varepsilon\Delta+1)^{-1}$ is a contraction on every Sobolev space $H^k$.)

4) Combine to get the $X^k$-estimate

Recall $X^k := H^k\cap L^\infty$ with norm $\|v\|_{X^k}=\|v\|_{H^k}+\|v\|_{L^\infty}$. From the two bounds above we get, with $C_k=1$,

$\|u\|_{X^k} \;=\;\|u\|_{H^k}+\|u\|_{L^\infty} \le \|f\|_{H^k}+\|f\|_{L^\infty} = \|f\|_{X^k}.$

So the desired inequality holds (in fact with constant 1) for every $\varepsilon>0$.

---

Remarks / why Ricciardi mentioned $G_\varepsilon\to\delta$

• The convergence $G_\varepsilon\rightharpoonup\delta$ explains the limiting behaviour of the solution as $\varepsilon\downarrow0$: $u=(G_\varepsilon * f)\to f$ pointwise (and in suitable norms) — consistent with the fact that $(-\varepsilon\Delta+1)^{-1}\to\mathrm{Id}$.
• But to prove the operator norm bound $\|u\|_{X^k}\le C_k\|f\|_{X^k}$​ you do not actually need the weak convergence; you only need the two facts used above:
  1. $G_\varepsilon(\cdot,y)\ge0$ and $\int_S G_\varepsilon(\cdot,y)=1$ for each $y$ (gives the $L^\infty$ bound),
  2. the spectral multiplier $(1+\varepsilon\lambda)^{-1}$ has modulus $\le1$ (gives the $H^k$ bound).

So the convergence to the delta is a helpful intuition and gives pointwise limit information, but the inequality itself follows from the kernel mass + positivity and from the spectral (Fourier multiplier) estimate.