We consider the Gaussian heat kernel $p_t$ on $\mathbb R^d$, i.e., $$ p_t (x) := (4\pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{4t}}, \quad t>0, x \in \mathbb R^d, $$ and define the operator $P_t$ by $$ P_t f (x) := \int_{\mathbb R^d} p_t ( x- y) f(y) \, \mathrm d y. $$

We fix $t>0,\delta \in (0, 1), y \in \mathbb R^d$ and $i,j \in \{1, 2, \ldots, d\}$. Let $$ I := \int_{\mathbb R^d} | \partial_i \partial_j (1- \Delta)^{- \frac{\delta}{2}} p_t ( \cdot - y) (x) | \, \mathrm d x. $$

At page $584$ of this paper, the author obtains an upper bound $$ I \lesssim t^{\frac{\delta}{2}-1} \tag{$*$} \label{*} $$ by using an auxiliary result (in the same paper)

Lemma 5.2(2) For any $\alpha, \beta, k \geq 0$, there exists a constant $c>0$ such that $$ \left\|(1-\Delta)^{-k} P_t f\right\|_{C_b^{\alpha+\beta}} \leq c t^{-\left(\frac{\alpha}{2}-k\right)^{+}}\|f\|_{C_b^\beta}, \quad t>0 . $$

The related function space is defined as

For any $n \in \mathbb{Z}^{+}$and $\alpha \in(0,1), C_b^{n+\alpha}\left(\mathbb{R}^d\right)$ is the space of functions $f \in C_b^n\left(\mathbb{R}^d\right)$ such that $$ \|f\|_{C_b^{n+\alpha}}:=\|f\|_{C_b^n}+\sup _{x \neq y} \frac{\left|\nabla^n f(x)-\nabla^n f(y)\right|}{|x-y|^\alpha}<\infty . $$

Unfortunately, I could not see how to get (\ref{*}). We have $$ \begin{align*} (1- \Delta)^{-k} P_t f &:= (1- \Delta)^{-k} (P_t f) \\ & = (1- \Delta)^{-k} \left ( \int_{\mathbb R^d} p_t ( \cdot - y) f(y) \, \mathrm d y \right ). \end{align*} $$

So the operator $\partial_i \partial_j (1- \Delta)^{- \frac{\delta}{2}}$ is inside the integral in $I$ whereas $(1- \Delta)^{-k}$ is outside the integral in $(1- \Delta)^{-k} P_t f$. I don't know how to put $I$ in a form that Lemma 5.2(2) is applicable.

Could you explain how to get (\ref{*})?

Thank you so much for your help!


1 Answer 1


This is a duality argument (the author is really invoking the adjoint of Lemma 5.2(2), rather than Lemma 5.2(2) directly). We can write $$ I = \sup_g \left|\int_{{\bf R}^d} \partial_i \partial_j (1-\Delta)^{-\delta/2} p_t(\cdot-y)(x) g(x)\ dx\right|$$ where $g$ ranges over test functions in $C^\infty_c({\bf R}^d)$ of supremum norm one. The expression inside the absolute value can be rearranged (after an "integration by parts") as $$ |\partial_i \partial_j (1-\Delta)^{-\delta/2} P_t g(y)|.$$ By Lemma 5.2 (with $\alpha=2$, $\beta=0$, $k = \delta/2$), this expression is $\lesssim t^{\frac{\delta}{2}-1} \|g\|_{C^0} = t^{\frac{\delta}{2}-1}$, as claimed.

  • 1
    $\begingroup$ Dear professor Tao, thank you so much for your help! Unfortunately, I could not see how to use integration by parts to get $$ \begin{align*} &\int_{\mathbb R^d} \partial_i \partial_j (1- \Delta)^{- \frac{\delta}{2}} p_t ( \cdot - y) (x) g(x) \, \mathrm d x \\ = &\partial_i \partial_j (1- \Delta)^{- \frac{\delta}{2}} \left ( \int_{\mathbb R^d} p_t ( x - \cdot) (x) g(x) \, \mathrm d x \right ) (y). \end{align*} $$ Could you elaborate more on this point? $\endgroup$
    – Akira
    Commented Oct 8, 2023 at 13:10
  • 1
    $\begingroup$ One can use the integral kernel of $(1-\Delta)^{-\delta/2}$ (which is even, as is $p_t$) to move them over to $g$, and move the derivatives over by integration by parts. (one can also restrict attention to smooth $g$ if desired to make it easier to justify the calculations). Alternatively, you can express everything in Fourier space where the self adjointness of the relevant operators becomes transparent. $\endgroup$
    – Terry Tao
    Commented Oct 8, 2023 at 14:39
  • $\begingroup$ Thank you so much for your explanation! I will check it out. Have a nice weekend! $\endgroup$
    – Akira
    Commented Oct 8, 2023 at 14:44

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