Problem
We wish to prove the following limit:
For all $f \in L^1(\mathbb{R}^n)$:
$t^{n(1-1/p)} ||(P_{t} \ast f)(x) - P_t (x) \int_{\mathbb{R}^n} f(y) \text{d}y ||_{L^p_x(\mathbb{R}^n)} \rightarrow 0 $, as $t \rightarrow \infty$,
where $P_t$ is the poisson kernel: $P_t(x) = t^{-n}P(t^{-1}x) = t^{-n} \displaystyle \frac{\Gamma(\frac{n+1}{2})}{\pi^{\frac{n+1}{2}} (1+|t^{-1}x|^2)^{\frac{n+1}{2}} } $, for all $x\in \mathbb{R}^n$, $t>0$.
Proof
We start by assuming $f \in C^{\infty}_{0}(\mathbb{R}^n)$, with $R>0$ such that:
supp$f \subseteq \{ x \in \mathbb{R}^n \ | \ |x| \leq R \}$.
Then $t^{n(1-1/p)} ||(P_{t} \ast f)(x) - P_t (x) \int_{\mathbb{R}^n} f(y) \text{d}y ||_{L^p_x(\mathbb{R}^n)} \leq C t^{n(1-1/p)} ||t^{-n} \nabla P(t^{-1}x)||_{L^p_x(\mathbb{R}^n)} \int_{y \leq R} \frac{|y|}{t} f(y) \text{d}y $
$ \leq C t^{-1} R ||f||_{L^1(\mathbb{R}^n)} \rightarrow 0 $, as $t \rightarrow \infty$.
So we have shown that the statement holds for $f$ smooth with compact support. The next part is the part I don't understand.
For $f \in L^1(\mathbb{R}^n) $, we use the fact that $ C^\infty_0(\mathbb{R}^n) $ is dense in $L^1(\mathbb{R}^n)$, and use the following inequality which is proven earlier:
$ t^{n(1-1/p)} ||(P_{t} \ast f)(x) - P_t (x) \int_{\mathbb{R}^n} f(y) \text{d}y ||_{L^p_x(\mathbb{R}^n)} \leq C||f||_{L^1} $.
According to the paper I am reading, this is enough to get the desired convergence, by a density argument.
In other words, we should let $\{ f_k \} \subseteq C^\infty_0(\mathbb{R}^n)$ such that $f_k \rightarrow f$ in $L^1(\mathbb{R}^n)$. Then we know that the above inequality and convergence hold for each $f_k$, and we should be able to use convergence of $f_k$ into $f$ to get the final result.
I am struggling to see exactly how we can do this. If someone could please provide a hint/first step, that would be very much appreciated. Thank you very much.
This is a very standard argument. Say $\text{LHS}(f)$ is the stuff on the left side.
Say $f\in L^1$. Let $\epsilon>0$. Choose $g$ smooth with $$||f-g||_1<\epsilon/2.$$
$\newcommand\LHS[1]{\text{LHS}#1}$Choose $T$ so $$\LHS(g)\le\epsilon/2\quad(t>T).$$If $t>T$ then $$\LHS (f)\le\LHS (g)+\LHS(f-g)<\epsilon.$$
Exercise: Explain how this result is really just a special case of the fact that a uniform limit of continuous functions (on $[0,\infty]$) is continuous.