Theorem 12.4.19 | The Eigenspace

Theorem 12.4.19.label The distribution associated to the $L^{1}_{loc}$ function

\begin{align*}N(x)\define \frac{\abs{x}^{2-n}}{(2-n)\om_{n}}\end{align*}

is a fundamental solution for $\la$ in $\R^{n}$ for $n\geq 3$.

Proof. It is clear that $N(x)$ is harmonic in $\R^{n}\del \curl{0}$ as it should be for $\la(T_{N})=\de$. We use “regularization” to make $N$ a $C^{\infty}$ function by “shifting” the “pole” at $x=0$ away into $\com^{n}$. Define

\begin{align*}N^{\e}(x)\define\frac{(\abs{x}^{2}+\e^{2})^{2-n}}{(2-n)\om_{n}}\end{align*}

Write $N^{\e}$ for $T_{N^\e}$ then $N^{\e}\to N$ in $\cali{D}'$. First $N^{\e}\to N$ pointwise and $N^{\e},N$ are dominated for $\e\leq 1$ by $\abs{N}\in C^{\infty}_{c}\suf L^{1}_{loc}$. By Dominated Convergence theorem

\begin{align*}\integral{}{}N^{\e}\pdx\to \integral{}{}N\p dx\end{align*}

for $\p\in\cali{D}=C_{c}^{\infty}$.

On radial functions

\begin{align*}\la=\der{^2}{r^2}+\frac{n-1}{n}\der{}{r}\end{align*}

hence

\begin{align*}\la N^{\e}(x)=\frac{n}{\om_{n}}\e^{2}(\abs{x}^{2}+\e^{2})^{-\frac{n+2}{2}}=\e^{-n}\psi\parens{\frac{x}{\e}}\end{align*}

where $\psi(x)=\frac{n}{\om_{n}}\parens{\abs{x}^2+1}^{-\frac{n+2}{2}}$. Observe that

\begin{align*}\la N^{\e}(-x)=\la N^{\e}(x)\end{align*}

\begin{align*}\inner{\la N^\e}{\p}=\integral{}{}(\la N^{\e})(x)\p(x)dx=\integral{}{}(\la N^{\e})(-x)\p(x)dx=(\p*\la N^{\e})(0)\end{align*}

By Lemma 12.4.10 with $g=\psi,f=\p$ then $\inner{\la N^\e}{\p}\to a\p(0)$ as $\e\to 0$. It remains to show that $a=1$.$\square$

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circleLemma 12.4.10: Approximate Identity Smoothing

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