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I was reading Carleman Inequalities: An Introduction and More Book by Nicolas Lerner. In that Section 8.2 gives strong unique continuation for $L^{n/2}_{loc}$ proof. I understood whole proof except I have doubt in interpretation of Heaviside function of operator given in Lemma 8.36 and Projection operator on that function given in Lemma 8.37.

Details of problem:

Let $\mathbb P_l$ stands for the orthogonal projection in $L^{2}(S^{n-1})$ on the space of $\mathcal H_{n,l}$ of harmonic homogenous polynomial with degree $l$ in $n$ variable. By spectral decomposition theorem of Laplacian on sphere we have $$\mathrm{Id}_{L^{2}\left(\mathbb{S}^{n-1}\right)}=\sum_{l \geq 0} \mathbb{P}_{l}, \quad-\Delta_{\mathbb{S}^{n-1}}=\sum_{l \geq 0} l(l+n-2) \mathbb{P}_{l}.$$

By writing Laplacian in polar coordinate we get $$ r^{2} \Delta_{\mathbb{R}^{n}}=\left(r \partial_{r}\right)^{2}+(n-2)\left(r \partial_{r}\right)+\Delta_{\mathbb{S}^{n-1}}$$

Define $\Lambda=\sum_{k \geq 1} k \mathbb{P}_{k} $. By orthogonality we have $$|x|^{2} \Delta_{\mathbb{R}^{n}}=\left(r \partial_{r}\right)^{2}+(n-2)\left(r \partial_{r}\right)-\Lambda(\Lambda+n-2).$$

Now take $r=e^t$ we get $$|x|^{2} \Delta_{\mathbb{R}^{n}}=\partial_{t}^{2}+(n-2) \partial_{t}-\Lambda(\Lambda+n-2)=\left(\partial_{t}+\Lambda+n-2\right)\left(\partial_{t}-\Lambda\right)=\mathcal{L}$$ We note that for $\lambda \geq 1$ \begin{equation} \mathcal{L}=|x|^{2} \Delta_{\mathbb{R}^{n}}, \quad \mathcal{L}_{\lambda}=e^{-\lambda t} \mathcal{L} e^{\lambda t}=\mathcal{L}_{-, \lambda} \mathcal{L}_{+, \lambda}=\mathcal{L}_{+, \lambda} \mathcal{L}_{-, \lambda} \end{equation} Where \begin{equation} \mathcal{L}_{+, \lambda}=\partial_{t}+\lambda+\Lambda+n-2, \quad \mathcal{L}_{-, \lambda}=\partial_{t}+\lambda-\Lambda. \end{equation}

Fundamental solution of differential operator $\mathcal{L}_{+, \lambda} \mathcal{L}_{-, \lambda}$ is given by \begin{align} E=& H(\lambda-\Lambda)(2 \Lambda+n-2)^{-1}\left(e^{-(\lambda-\Lambda) t}-e^{-(\Lambda+\lambda+n-2) t}\right) H(t)\notag \\ &-H(\Lambda-\lambda)(2 \Lambda+n-2)^{-1}\left(e^{-(\lambda-\Lambda) t} H(-t)+e^{-(\Lambda+\lambda+n-2) t} H(t)\right) \end{align} where $H$ is Heaviside function. i.e. \begin{equation} v=E *\left(\mathcal{L}_{+, \lambda} \mathcal{L}_{-, \lambda} v\right) \end{equation}

I understood that exponential of operator can be thought of as series in terms of operator. But How to think about Heaviside function of operator and $(2\Lambda+n-2)^{-1}$?

Now we wanted to estimate $\left\|\mathbb{P}_{k} v(t)\right\|_{L^{p^{\prime}}\left(\mathbb{S}^{n-1}\right)}$.

For that purpose we write $F=\mathcal{L}_{+, \lambda} \mathcal{L}_{-, \lambda} v$.

By property of fundamental solution we have \begin{align*} v(t)&=\frac{H(\lambda-\Lambda)}{2 \Lambda+n-2} \int_{-\infty}^{t}\left(e^{-(\lambda-\Lambda)(t-s)}-e^{-(\lambda+\Lambda+n-2)(t-s)}\right) F(s) d s\\ &\quad-\frac{H(\Lambda-\lambda)}{2 \Lambda+n-2}\left(\int_{t}^{+\infty} e^{-(\Lambda-\lambda)(s-t)} F(s) d s\right.\\ &\quad\left.\quad+\int_{-\infty}^{t} e^{-(\lambda+\Lambda+n-2)(t-s)} F(s) d s\right) \end{align*} Now we apply $\mathbb P_k$ to both side.

I can think that $e^{-(\lambda-\Lambda)(t-s)}$ can be think as series of exponential in $\Lambda$. so after applying $P_k$ to that we get $e^{-(\lambda-k)(t-s)}\mathbb P_k$ because of orthogonality.

But I have doubt how to apply $\mathbb P_k$ on $\frac{H(\lambda-\Lambda)}{2 \Lambda+n-2}$.

I thought about Fourier series of Heaviside function but there is problem about convergence and Heaviside on whole real line is not periodic function as such so I can not get Fourier series. Also how to apply on its denominator $2 \Lambda+n-2$. Final expression is given as \begin{align*} \mathbb{P}_{k} v(t)&=\frac{H(\lambda-k)}{2 k+n-2} \int_{-\infty}^{t}\left(e^{-(\lambda-k)(t-s)}-e^{-(\lambda+k+n-2)(t-s)}\right) \mathbb{P}_{k} F(s) d s\\ &\quad-\frac{H(k-\lambda)}{2 k+n-2}\left(\int_{t}^{+\infty} e^{-(k-\lambda)(s-t)} \mathbb{P}_{k} F(s) d s\right.\\ &\quad\left.\quad+\int_{-\infty}^{t} e^{-(\lambda+k+n-2)(t-s)} \mathbb{P}_{k} F(s) d s\right) \end{align*}

Any help or hint will be appreciated.

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  • $\begingroup$ Here it is applied to a function, not an operator (the image via operators of H(t) and H(-t)...)? $\endgroup$ – username 2 days ago
  • $\begingroup$ @username I could not get your comment. Can you please explain me little bit. Thanks for replying. $\endgroup$ – MSE_Lover 2 days ago
  • $\begingroup$ Both $a$ and $b$ given by \begin{align} a=& (\lambda-\Lambda)(2 \Lambda+n-2)^{-1}\left(e^{-(\lambda-\Lambda) t}-e^{-(\Lambda+\lambda+n-2) t}\right) H(t)\notag \\ b=&(\Lambda-\lambda)(2 \Lambda+n-2)^{-1}\left(e^{-(\lambda-\Lambda) t} H(-t)+e^{-(\Lambda+\lambda+n-2) t} H(t)\right) \end{align} are functions, so $$E=H(a(t))-H(b(t)),$$ just a composition? I am probably missing something deeper in your question. $\endgroup$ – username 19 hours ago
  • $\begingroup$ @username I think there is one problem with this interpretation. If we consider H is composition with all remaining, then H is obviously function of t then during convolution it still remain inside the integral. But here some how $H(\lambda-\Lambda)$ is constant wrt to $t$. So one can take it outside. $\endgroup$ – MSE_Lover 5 hours ago

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