# How to get $\int_{\mathbb R^d} |\partial_i\partial_j(1-\Delta)^{-\frac{\delta}{2}}p_t(\cdot-y)(x)| \, \mathrm d x \lesssim t^{\frac{\delta}{2}-1}$?

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!

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.
• 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? Commented Oct 8, 2023 at 13:10
• 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. Commented Oct 8, 2023 at 14:39