We gives a short proof of the inequality, which we call (*1) here, right after inequality (14) in [KT01], where they did not gives much details in the paper. We need the following Lemma. **Lemma 10:** In Euclidean space $\mathbb{R}^n$, we denote by $B(x,r)$ the ball centered at $x$ with radius $r$. If $$\sup_{x} \int_{B(x,1)} \abs{f(y)} dy < \infty, $$ then $$\int_{B(x_0,1)^c} \f{1}{\abs{x_0-y}^{n+1}}\abs{f(y)} dy \leq c \sup_{x} \int_{B(x,1)} \abs{f(y)} dy, $$ for all $x_0\in \mathbb{R}^n$ and some constant $c>0$ independent of $x_0$. **//Proof://** We prove the case when $n=1$, the general case can be proved similarly but a little bit sophisticated. WLOG, we assume that $x_0=0$. \begin{equation} \begin{split} \int_{B(0,1)^c} \f{1}{\abs{y}^2} \abs{f(y)} dy &= \sum_{k=1}^\infty \int_{B(\pm 2k ,1)} \f{1}{\abs{y}^2} \abs{f(y)} dy ~\\ &\leq \sum_{k=1}^\infty \f{1}{(2k-1)^2}\cdot 2\sup_{x} \int_{B(x,1)} \abs{f(y)}dy ~\\ &\lesssim \sup_{x} \int_{B(x,1)} \abs{f(y)}dy. \end{split} \end{equation} **Remark:** Since Lebesgue measure is doubling measure, we can prove Lemma 10 with cubes. We split the space $\mathbb{R}^n$ into many congruent cubes, and then apply similar arguments as for the case $n=1$. === Proof of (*1) By definition, we have \begin{equation} \begin{split} V\grad\Pi f = \int_0^t \grad\Pi S(s)f(t-s)ds ~\\ =\int_0^t \int k(y,s)f(x-y,t-s)dyds \end{split} \end{equation} === References [KT01] H. Koch and D. Tataru, Well-posedness for the Navier-Stokes equations, Adv. Math. 157, 22-35 (2001)