I have been trying to find a good estimate on the constant in the inequality (in dimension $d=3$ to simplify) $$\label{0}\tag{0} \|(1-e^{t\Delta})f\|_{L^{5/3}(\Bbb R^3)} \leq C_0\,t^{2/5}\, \|\nabla \sqrt{f}\|_{L^2(\Bbb R^3)}. $$
- Here is a first method to estimate this constant. It uses the fact that by Hölder's and Sobolev's inequalities, $$ \|\nabla f\|_{L^{3/2}} = 2\,\|\sqrt{f}\nabla \sqrt{f}\|_{L^{3/2}} \\ \leq 2\,\|\sqrt{f}\|_{L^6}\,\|\nabla \sqrt{f}\|_{L^2} \leq 2\,\mathcal C_S^2 \,\|\nabla \sqrt{f}\|_{L^2}^2, $$ where $\mathcal C_S \simeq 0.427$ is the best constant in the Sobolev inequality $\dot H^1 \subset L^6$ in dimension $3$. Hence, I have been interested in finding the best possible constant in the inequality $$\label{1}\tag{1} \|(1-e^{t\Delta})f\|_{L^{5/3}(\Bbb R^3)} \leq C_1 \,t^{2/5}\,\|\nabla f\|_{L^{3/2}(\Bbb R^3)} $$ as it gives the following estimate $C_0 \leq 2\,\mathcal C_S^2\,C_1$.
Hence, main question: what are the methods to prove estimates like \eqref{1}? Or what keyword should I try to find results on this problem (Gaussian mollifier rate, Heat equation for small time, Fourier multiplier in $L^p$ spaces ...)? In particular, I was hoping that precise results exist in the spirit of the $L^2$ case, in which it is sufficient to write with $g = \nabla f$, $$ \|\tfrac{1-e^{t\Delta}}{(-t\Delta)^{1/2}}\,g\|_{L^2(\Bbb R^3)} \leq C \,\|g\|_{L^2} $$ thanks to the Fourier transform (or spectral theory). Here the constant is thus the supremum of the function $\tfrac{1-e^{-t}}{t^{1/2}}$, approximately $0.64$. Or would there be a direct method to prove Inequality \eqref{0}?
A method to estimate $C_1$
Recall the explicit formula of the heat kernel, $e^{t\Delta}f = g_t * f$ where $g_t(x) = (4\pi\,t)^{-3/2}\,e^{-|x|^2/(4t)}$. By the fundamental theorem of calculus, $$ (1-e^{t\Delta})f = \int_{\Bbb R^3} \left(f(x)-f(x-y)\right) g_t(y)\,\mathrm d y \\ = \int_{\Bbb R^3} \int_0^1 x\cdot\nabla\rho\!\left(x-\theta y\right) \,g_t(y) \,\mathrm d \theta \,\mathrm d y = \int_0^1 \theta^{-3} (\tfrac{x}{\theta}\,g_t(\tfrac{x}{\theta}))*\nabla f\,\mathrm d\theta. $$ Hence, it follows from Young's inequality with $p=5/3$, $q=3/2$ and $r = \frac{15}{14}$ so that $\frac{1}{r} = 1+\frac{1}{p}-\frac{1}{q}$, that $$ \|(1-e^{t\Delta})f\|_{L^p} \leq C_Y \int_0^1 \theta^{-3}\, \|\tfrac{x}{\theta}\,g_t(\tfrac{x}{\theta})\|_{L^r} \|\nabla\rho\|_{L^q} \,\mathrm d\theta \\ \leq C_Y\, t^\frac{2}{5} \int_0^1 \theta^{-3/r'} \|x\,g_1(x)\|_{L^r} \,\mathrm d\theta \,\|\nabla f\|_{L^q}, \\ \leq \frac{5}{4\,(4\pi)^{3/2}}\,C_Y\, t^\frac{2}{5} \|x\,e^{-|x|^2/4}\|_{L^r} \|\nabla f\|_{L^q}, $$ where $C_Y = \left(\frac{q^{1/q}\,r^{1/r}(p')^{1/p'}}{(q')^{1/q'}(r')^{1/r'}p^{1/p}}\right)^\frac{d}{2} \simeq 0.8$ is the optimal constant in Young's inequality. This gives $$ \|(1-e^{t\Delta})f\|_{L^p} \leq 4.4\, t^\frac{2}{5}\, \|\nabla f\|_{L^q}. $$
The reason I find this method not satisfying, is because applying the same method to $p=q=2$ gives $$ \|(1-e^{t\Delta})f\|_{L^2} \leq \frac{4}{\sqrt{\pi}}\, t^\frac{1}{2} \,\|\nabla f\|_{L^2} \simeq 2.26\, t^\frac{1}{2} \,\|\nabla f\|_{L^2}, $$ and $2.26$ is far from $0.64$ (more than 3 times). Moreover, wih a more careful analysis, one sees that the method giving $2.26$ would give something depending on the dimension (like $\sqrt d$), while the method giving $0.64$ is independent of the dimension!
Any interesting direction is welcome.
