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Let us consider the fractional heat semigroup $\left(e^{-t(-\Delta)^\alpha}\right)_{t\ge 0}$ for $\alpha\in (0,1)$ (the fractional power is taken in whole $\mathbb R^d$). Is there any result about the limit of $e^{-t(-\Delta)^\alpha} f$ when $\alpha \to 1^-$ with respect to an appropriate norm, e.g., $L^2(0,T;L^2(D))$ for a fixed domain $D\subset \mathbb R^d$ ? In particular, I am interested to some kind of explicit convergence rates with respect to $\alpha$. Any reference on similar topics would be helpful.

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    $\begingroup$ For starters, the spectral theorem tells you that up to unitary transformation, you are working with $e^{-th^\alpha}$ for some nonnegative measurable function $h$. Then dominated convergence should get you convergence in $L^2$. If you know the spectral gap of $\Delta$ on your domain, then $h$ is bounded away from 0 and this may tell you more. $\endgroup$ May 21, 2021 at 19:04
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    $\begingroup$ What do you mean by $(-\Delta)^\alpha$: the fractional power of $\Delta$ in full space, its restriction to $D$ with (say) zero exterior condition ("Dirichlet fractional Laplacian"), or the fractional power of (say) DIrichlet Laplacian in $D$ ("fractional Dirichlet Laplacian" or "spectral fractional Laplacian")? In each case some estimates are available, I think, but of course they are of different type. $\endgroup$ May 22, 2021 at 8:14
  • $\begingroup$ Thanks for your comments. @Mateusz I meant the fractional power in whole $\mathbb R^d$. I would also be interested to the techniques used in other cases. $\endgroup$
    – Migalobe
    May 22, 2021 at 11:34
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    $\begingroup$ For the full space, and a global estimate, see my answer below. Alternative way would be to use Bochner's subordination formula — this might also work for fractional powers of other operators, in particular for the fractional Dirichlet Laplacian. If the domain is bounded, estimates of the least eigenvalue (or for the spectral gap in the Neumann case) enter into the game, as Nate Eldredge suggested. The Dirichlet fractional Laplacian seems to be least accessible, but also here some $\alpha$-continuity results should be available — I would have to search references. $\endgroup$ May 22, 2021 at 22:00

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Edit: I was thinking about $(-\Delta)^{\alpha/2}$ rather than $(-\Delta)^\alpha$, so the $\alpha$ in the following answer is equal to $2\alpha$ with the notation of the statement of the question.

As suggested by Nate Eldredge, if $u_t(x) = e^{-t(-\Delta)^{\alpha/2}} f(x)$ and $v_t(x) = e^{t \Delta} f(x)$, then $$\begin{aligned} \int |u_t(x) - v_t(x)|^2 dx & = \frac{1}{(2 \pi)^d} \int |\hat u_t(\xi) - \hat v_t(\xi)|^2 d\xi \\ & = \frac{1}{(2\pi)^d} \int |e^{-t |\xi|^\alpha} - e^{-t |\xi|^2}|^2 |\hat{f}(\xi)|^2 d\xi . \end{aligned}$$ It follows that $$ \int |u_t(x) - v_t(x)|^2 dx \leqslant \frac{M_t^2}{(2\pi)^d} \int |\hat{f}(\xi)|^2 d\xi = M_t^2 \|f\|_2^2 , $$ where $M_t$ is the supremum of $|e^{-t|\xi|^\alpha} - e^{-t|\xi|^2}|$ over all $\xi$. Integrating this with respect to $t \in [0, T]$, we obtain $$ \int_0^T \int |u_t(x) - v_t(x)|^2 dx dt \le N_T \|f\|_2^2 , $$ where $$ N_T = \int_0^T M_t^2 dt . $$ Evaluation of $M_t$ (or rather finding an appropriate upper bound for $M_t$) is standard, but rather involved. Below I give a very rough estimate; one can probably do much better using more refined tools.


Clearly, $|e^{-p} - e^{-q}| \leqslant |p - q|$ and $|e^{-p} - e^{-q}| \leqslant \max\{e^{-p}, e^{-q}\}$. We distinguish two cases:

  • If $|\xi| < 1$, then $|\xi|^\alpha > |\xi|^2$ and $$||\xi|^\alpha - |\xi|^2| \leqslant (2 - \alpha) |\xi|^\alpha |\log |\xi|| \leqslant \frac{2-\alpha}{e \alpha} ,$$ so that $$ |e^{-t |\xi|^\alpha} - e^{-t |\xi|^2}| \leqslant \frac{2-\alpha}{e \alpha} t . $$

  • On the other hand, if $|\xi| > 1$, then $|\xi|^\alpha < |\xi|^2$ and $$||\xi|^\alpha - |\xi|^2| \leqslant (2 - \alpha) |\xi|^2 \log |\xi| ,$$ so that $$ |e^{-t |\xi|^\alpha} - e^{-t |\xi|^2}| \leqslant \min\bigl\{(2 - \alpha) t |\xi|^2 \log |\xi|, e^{-t |\xi|^\alpha} \bigr\} . $$ Using rather crude bounds $e^{-p} \leqslant p^{-1}$ and $\log |\xi| \leqslant |\xi|$, we find that $$ |e^{-t |\xi|^\alpha} - e^{-t |\xi|^2}| \leqslant \min\bigl\{(2 - \alpha) t |\xi|^3, t^{-1} |\xi|^{-\alpha} \bigr\} \leqslant (2 - \alpha)^{\frac{\alpha}{3 + \alpha}} t^{-\frac{3 - \alpha}{3 + \alpha}} . $$

It follows that $$ M_t \leqslant \max\biggl\{\frac{2 - \alpha}{\alpha e} \, t, (2 - \alpha)^{\frac{\alpha}{3 + \alpha}} t^{-\frac{3 - \alpha}{3 + \alpha}}\biggr\} . $$ Thus, $$ N_T = \int_0^T M_t^2 dt \leqslant C_1 (2 - \alpha)^2 T^3 + C_2 (2 - \alpha)^{\frac{2 \alpha}{3 + \alpha}} T^{\frac{3 (\alpha - 1)}{3 + \alpha}} $$ for some constants $C_1$ and $C_2$ (here $\alpha \in (1, 2)$.)

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    $\begingroup$ Ah, sorry: I am used to having $\alpha/2$ as the exponent: $(-\Delta)^{\alpha/2}$ rather than $(-\Delta)^\alpha$. Everything works just fine, I think, it is just a different notation. I will edit momentarily. $\endgroup$ May 26, 2021 at 16:23

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