# 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.

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 $$(1-\hat{\phi})\hat{g}\in \mathcal{S}$$.