Smoothing-Strichartz estimates for the heat-Schrodinger evolution Consider the heat-Schrodinger evolution $e^{(1+i)t\Delta}$, $t\geq 0$. For simplicity, suppose to work in dimension three. Due to the Strichartz estimates for the Schrodinger equation, and the smoothing effect of the heat flow, I expect that the following estimate holds true:
\begin{equation}\label{tre}
(1)\qquad \left\|\int_0^te^{(1+i)(t-s)\Delta}F(s)ds\right\|_{L^2((0,t),W^{1,6}(\mathbb{R}^3))}\lesssim \|F\|_{L^2((0,t),L^2(\mathbb{R}^3))}
\end{equation}
Nevertheless, I have not been able to prove it.
Is estimate (1) actually true? In case, is it already known in the literature?
Thank you for your suggestions.
 A: I think the estimate you want does not really require Strichartz. 
First, your estimate is equivalent to the following, which is a bit easier for me to think about: let $u$ be the solution to the equation
$$ \partial_t u - (1 + i) \triangle u = F \tag{*}$$
with initial data
$$ u(0,x) \equiv 0 $$
Then the desired estimate is 
$$ \| u \|_{L^2_t W^{1,6}_x} \leq \|F\|_{L^2_t L^2_x}$$
We can derive this estimate by multiplying the equation (*) by $\triangle \bar{u}$ and taking the real part, which gives
$$ \partial_t u \triangle \bar{u} + \partial_t \bar{u} \triangle u - (1+i) \triangle u \triangle \bar{u} - (1-i) \triangle u \triangle\bar{u} = F \triangle \bar{u} + \bar{F} \triangle u $$
Integrate by parts in space we get
$$ \partial_t \| \nabla u\|^2_{L^2_x} + 2 \|\triangle u\|_{L^2_x}^2 = - \int_{\mathbb{R}^3} F\triangle \bar{u} + \bar{F} \triangle u ~dx$$
Integrate between time 0 and time T you get (using the triviality of the data) 
$$ \|\nabla u(T)\|_{L^2_x}^2 + 2 \int_0^T \| \triangle u\|^2_{L^2_x} ~dt \leq \int_{[0,T]\times \mathbb{R}^3} 2 |F|\cdot |\triangle u| ~dx~dt (**)$$
Applying Young's inequality on the right, you can pull out $\epsilon \|\triangle u\|_{L^2_t L^2_x}$ which can be absorbed on the left hand side, and this reduces to 
$$  \| \triangle u\|_{L^2_t L^2_x} \lesssim \| F\|_{L^2_t L^2_x} $$
Finally use elliptic estimates to bound $\|\nabla^2 u\|$ by $\|\triangle u\|$, and then Sobolev inequality in space gets you 
$$ \| \nabla u\|_{L^2_t L^6_x} \lesssim \|F\|_{L^2_t L^2_x} $$
This takes care the homogeneous part of the Sobolev norm. 
I don't think the inhomogeneous part holds. It doesn't have the correct scaling. 
If you rescale $u(t,x) \mapsto u(\lambda^2 t, \lambda x)$ by the parabolic scaling, then $F(t,x) \mapsto \lambda^2 F(\lambda^2 t, \lambda x)$. The $L^2_t L^2_x$ norm of $F$ scales like $\lambda^{-1/2}$, but the $L^2_t L^6_x$ norm of $u$ scales like $\lambda^{-3/2}$. The term with the correct scaling should be $\|F\|_{L^1_t L^2_x}$. And the estimate 
$$ \| u\|_{L^2_t L^6_x} \lesssim \|\nabla u\|_{L^2_t L^2_x} \lesssim \|F\|_{L^1_t L^2_x} $$ 
can be proven by an analogous manner as above but instead of multiplying by $\triangle u$, multiply by $u$. 
Incidentally: both estimates also hold for the heat equation, without the Schroedinger part. 

Addendum: I just noticed that your inequality is only stated for $L^2((0,T),X)$ for some Banach space $X$. Do you only care about the local estimate where your constant can depend on the length of the interval? Or do you actually want uniform estimates for all $T$? I ask because quite obviously on a fixed time interval you have that $\|F\|_{L^1_t L^2_x} \lesssim \|F\|_{L^2_t L^2_x}$, and in fact you can get the same result by noting that (**) also implies $$\|u\|_{L^\infty_t L^6_x} \lesssim \|F\|_{L^2_t L^2_x}$$ and can be easily localized to intervals. 
