Skip to main content
MathOverflow

Return to Answer

added 205 characters in body
Source Link
Mateusz Kwaśnicki
Mateusz Kwaśnicki
  • 17.2k
  • 1
  • 33
  • 55

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} f(x)$$u_t(x) = e^{-t(-\Delta)^{\alpha/2}} f(x)$ and $v_t(x) = e^{t \Delta} f(x)$, then $$\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 $\alpha \in (1, 2)$.)

As suggested by Nate Eldredge, if $u_t(x) = e^{-t(-\Delta)^\alpha} f(x)$ 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 $\alpha \in (1, 2)$.)

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

Source Link
Mateusz Kwaśnicki
Mateusz Kwaśnicki
  • 17.2k
  • 1
  • 33
  • 55

As suggested by Nate Eldredge, if $u_t(x) = e^{-t(-\Delta)^\alpha} f(x)$ 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 $\alpha \in (1, 2)$.)