Spring 2016 - Problem 1

Solution.

1. First, observe that $\frac{1}{3} + \frac{1}{3/2} = 1$. Then by Hölder's inequality,

   $$\begin{aligned} \abs{t^{1/2} \p{K_t * f}\p{x}} &\leq t^{1/2} \int_{\R^3} \abs{K_t\p{x-y}}\abs{f\p{y}} \,\diff{y} \\ &\leq t^{1/2} \norm{K_t}_{L^{3/2}} \norm{f}_{L^3}. \end{aligned}$$

   Next, we have

   $$\begin{aligned} \norm{K_t}_{L^{3/2}}^{3/2} &= \int_{\R^3} \p{4\pi t}^{-9/4} e^{-3\abs{x}^2/2t} \,\diff{x} \\ &= \int_{\R^3} \p{4\pi t}^{-9/4} e^{-3\abs{x}^2/2} \,\diff{x} && \p{x \mapsto \frac{x}{t^{1/2}}} \\ &= \frac{t^{3/2}}{\p{4\pi t}^{9/4}} \int_{\R^3} e^{-3\abs{x}^2/2} \,\diff{x} \\ &\eqqcolon C^{3/2}t^{-{3/4}} \\ \implies \norm{K_t}_{L^{3/2}} &= Ct^{-1/2}. \end{aligned}$$

   Putting these together,

   $$\abs{t^{1/2} \p{K_t * f}\p{x}} \leq C\norm{f}_{L^3},$$

   so the map is bounded.

2. Let $\epsilon > 0$. By density, there exists $g \in C_c\p{\R^3}$ such that $\norm{f - g}_{L^3} < \epsilon$. Thus, by the triangle inequality and the estimate above,

   $$\begin{aligned} \abs{t^{1/2} \p{K_t * f}\p{x}} &\leq \abs{t^{1/2} \p{K_t * \p{f - g}}\p{x}} + \abs{t^{1/2} \p{K_t * g}\p{x}} \\ &\leq C\norm{f - g}_{L^3} + C\abs{t^{1/2} \p{K_t * g}\p{x}} \\ &\leq C\epsilon + C\abs{t^{1/2} \p{K_t * g}\p{x}} \end{aligned}$$

   Since $g$ is compactly supported, it is bounded by some $M > 0$ and so

   $$\begin{aligned} \abs{\int_{\R^3} K_t\p{x - y} g\p{y} \,\diff{y}} &\leq \p{4\pi t}^{-3/2} \int_{\R^3} e^{-\abs{x-y}^2/4t} \abs{g\p{y}} \,\diff{y} \\ &= \p{4\pi t}^{-3/2} \int_{\R^3} e^{-\abs{x-y}^2/4t} \abs{g\p{y}} \,\diff{y} && \p{y \mapsto \frac{x - y}{\sqrt{t}}} \\ &= \p{4\pi t}^{-3/2} t^{3/2} \int_{\R^3} e^{-\abs{y}^2/4} \abs{g\p{x - y\sqrt{t}}} \,\diff{y} \\ &\leq \p{4\pi}^{-3/2} M \int_{\R^3} e^{-\abs{y}^2/4} \,\diff{y} \\ &\eqqcolon A. \end{aligned}$$

   Thus,

   $$\abs{t^{1/2} \p{K_t * f}\p{x}} \leq C\epsilon + CAt^{1/2} \implies \limsup_{t\to0}\,\abs{t^{1/2} \p{K_t * f}\p{x}} \leq C\epsilon,$$

   which was what we wanted to show.