Grafakos, Modern Fourier Analysis, Third Edition, Exercise 2.1.4. Bounded Tempered Distributions In Grafakos book, Modern Fourier Analysis, the exercise 2.1.4 is as follows:
Let $P$ be the Poisson Kernel. Show that for any bounded tempered distribution $f$ we have $P_t \ast f \to f$ in $\mathcal{S}'(\mathbb{R}^n)$ as $t \to 0$.
[Hint: Fix a smooth function $\phi$ whose Fourier transform is equal to 1 in a neighborhood
of zero. Show that $P_t \ast (\phi \ast f ) \to \phi \ast f$ in $\mathcal{S}'(\mathbb{R}^n)$
and that $\hat{P_t}(1 - \hat{\phi})\hat{f} \to (1 - \hat{\phi})\hat{f}$ in $\mathcal{S}'(\mathbb{R}^n)$ as $t \to 0$.]
The definition of bounded tempered distribution can be found here and $P_t(x) = t^{-n}P(t^{-1}x)$. The existence of $\phi$ can be proved using the $C^{\infty}$ Urysohn Lemma, but the convergences above are a trouble for me. Rememeber that $P_t$ is not a Schwartz Function and so we can not apply a tempered distribution on $P_t$. Maybe density arguments using that $\mathcal{S}$ is dense in $L^1$ ($P_t \in L^1$) allied with the approximation identities theorems are a good way. Another possibility is try to reconstruct the theory of convolutions between tempered distributions and Schwartz functions for this case.
I appreciate any help or comment about this.
 A: The first is a consequence of the fact that $\phi*f\in L^\infty$ by definition of bounded distribution. Indeed, since the kernel of $P_t$ is even and real we have
$$
\langle P_t (\phi* f), g\rangle = \langle \phi*f, P_t g\rangle = \int_{\mathbb{R}^n} (\phi*f) P_t g\, dx,
$$
for every $g\in \mathcal{S}$. Since we know that $\|P_t g-g\|_{L^1}\to 0$, the result follows.
For the second one, we use that $\hat{\phi}$ is one in a neighborhood of the origin, so that multiplying $\hat{f}$ by $\hat{P}_t(1-\hat{\phi})\in \mathcal{S}$ is well-defined and
$$
\langle \hat{P}_t(1-\hat{\phi})\hat{f}, \hat{g}\rangle = \langle \hat{f}, \hat{P}_t(1-\hat{\phi})\hat{g}\rangle = \langle \hat{f}, e^{-2\pi t|\xi|}(1-\hat{\phi})\hat{g}\rangle=:  \langle \hat{f}, h_t\rangle. 
$$
Now it's a matter of proving that $h_t\to (1-\hat{\phi})\hat{g}=:h$ in $\mathcal{S}$. This is cumbersome to write down, but the idea is simple: You want to show that for all multiindices $\alpha, \beta$ we have
$$
\sup_{\xi\in \mathbb{R}^n} |\xi^\alpha \partial_\xi^\beta [h_t(\xi)- h(\xi)]| \to 0, \qquad t\to 0.
$$
Consider first the case $\beta=0$ (i.e. no derivatives). Since $\hat{\phi}$ is equal to one in, say, $B_r(0)$ we have $h_t=h$ in this same ball. Therefore
$$
\sup_{\xi\in \mathbb{R}^n}|\xi^\alpha[h_t(\xi)-h(\xi)]| = \sup_{\xi\notin B_r(0)} |\xi^\alpha (1-\hat{\phi})\hat{g}[e^{-2\pi t|\xi|}-1]| \lesssim_{\phi, g} |e^{-2\pi t r}-1| \to 0,
$$
since $(1-\hat{\phi})\hat{g}\in \mathcal{S}$. From this and the product rule you see that when derivatives are added, the only possible issues arise when the derivatives hit the exponential, however this only adds powers of $|\xi|$ which we can add to the $\alpha$ and control thanks to the fact that $(1-\hat{\phi})\hat{g}\in \mathcal{S}$.
