Problem #7.7, MA598R, Summer 2009, Weigel
Let $ K(x) = (4\pi)^{\frac{-n}{2}}e^{\frac{-|x|^2}{4}} $, $ x \in \mathbb{R}^n $, and let $ K_t(x) = t^{\frac{-n}{2}}K\left(\frac{x}{\sqrt{t}}\right) $, $ t>0 $.
Given $ f: \mathbb{R}^n \rightarrow \mathbb{R} $ measurable, define $ P_tf(x) = (K_t * f)(x) = \int_{\mathbb{R}^n} K_t(x-y)f(y)dy $
a) Prove that for every $ 1 \leq p < \infty $, one has $ \lim_{t \rightarrow 0^+}\|P_tf - f\|_p=0 \text{, for every } f \in L^p(\mathbb{R}^n) $ and moreover, for every $ 1 \leq p \leq \infty $, $ \|P_tf\|_p \leq \|f\|_p $.
Proof. First we find the integral by using substition and the result of 7.3:
$ \begin{align} \int_{\mathbb{R}^n}K(x)dx &= \int_{\mathbb{R}^n}(4\pi)^{\frac{-n}{2}}e^{\frac{-|x|^2}{4}}dx\\ &= \int_{\mathbb{R}^n}(4\pi)^{\frac{-n}{2}}e^{-|\frac{x}{2}|^2}dx\\ &= \int_{\mathbb{R}^n}(4\pi)^{\frac{-n}{2}}e^{-|y|^2}(2^n)dy\\ &= \int_{\mathbb{R}^n}(\pi)^{\frac{-n}{2}}e^{-|y|^2}dy\\&= (\pi)^{\frac{-n}{2}}(\pi)^{\frac{n}{2}} =1\end{align} $
Similarly using the substitution $ y = \frac{x}{\sqrt{t}} $ we find that $ \int_{\mathbb{R}^n}K_t(x)dx = \int_{\mathbb{R}^n}t^{\frac{-n}{2}}K\left(\frac{x}{\sqrt{t}}\right)dx = \int_{\mathbb{R}^n}K(y)dy $.
Using the exact same method as 7.10(c) we get the desired result.
The moreover result comes from $ \|P_tf\|_p = \|K_t * f\|_p \leq \|K_t\|_1\|f\|_p = \|K\|_1\|f\|_p = \|f\|_p $ where the inequality is a result of the convolution theorem.
b) Prove that if $ 1 \leq p < \infty $, one also has $ P_t: L^p(\mathbb{R}^n) \rightarrow L^{\infty}(\mathbb{R}^n) $ with $ \|P_tf\|_{\infty} \leq (p^{\prime})^{\frac{-n}{2p^{\prime}}}(4\pi t)^{\frac{-n}{2p}}\|f\|_p, $ where $ \frac{1}{p} + \frac{1}{p^{\prime}} = 1 $. Here, one must take $ (p^{\prime})^{\frac{-n}{2p^{\prime}}} = 1 $ when $ p=1 $.
Proof. Using Young's Convolution Theorem with $ r = \infty $ we find:
$ \|P_tf\|_{\infty} \leq \|f\|_p\|K_t\|_{p^{\prime}} $, where $ \frac{1}{p} + \frac{1}{p^{\prime}} = 1 $
$ \begin{align} \|K_t\|_{p^{\prime}} &= \left(\int_{\mathbb{R}^n} (K_t(x))^{p^{\prime}}dx \right)^{\frac{1}{p^{\prime}}}\\ &= \left(\int_{\mathbb{R}^n} \left(t^{\frac{-n}{2}}K\left(\frac{x}{\sqrt{t}}\right)\right)^{p^{\prime}}dx \right)^{\frac{1}{p^{\prime}}}\\ &= \left(\int_{\mathbb{R}^n} t^{\frac{-np^{\prime}}{2}}\left(K\left(\frac{x}{\sqrt{t}}\right)\right)^{p^{\prime}}dx \right)^{\frac{1}{p^{\prime}}}\\ &= \left(\int_{\mathbb{R}^n} t^{\frac{-np^{\prime}}{2}}\left(4\pi\right)^{\frac{-np^{\prime}}{2}} e^{\frac{-|x|^2p^{\prime}}{4t}} dx \right)^{\frac{1}{p^{\prime}}}\\ &= \left(\int_{\mathbb{R}^n} t^{\frac{-np^{\prime}}{2}}\left(4\pi\right)^{\frac{-np^{\prime}}{2}} e^{-\left|\frac{x\sqrt{p^{\prime}}}{2\sqrt{t}}\right|^2} dx \right)^{\frac{1}{p^{\prime}}} \\ &= \left(\int_{\mathbb{R}^n} t^{\frac{-np^{\prime}}{2}}\left(4\pi\right)^{\frac{-np^{\prime}}{2}} e^{-|y|^2} \left(\frac{4t}{p^{\prime}}\right)^{\frac{n}{2}}dy \right)^{\frac{1}{p^{\prime}}}\\ &= \left(\left(p^{\prime}\right)^{\frac{-n}{2}}\left(4t\right)^{\frac{-np^{\prime}}{2} + \frac{n}{2}}\left(\pi\right)^{\frac{-np^{\prime}}{2}}\right)^{\frac{1}{p^{\prime}}}\left(\int_{\mathbb{R}^n}e^{-|y|^2}dy\right)^{\frac{1}{p^{\prime}}}\\ &= (p^{\prime})^{\frac{-n}{2p^{\prime}}}(4\pi t)^{\frac{-n}{2p}} \end{align} $
Thus we again get the desired result.
