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.



Limit of $e^{-t(-\Delta)^\alpha}$ when $\alpha \to 1$
*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.




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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 $\alpha \in (1, 2)$.)


