I am recently reading "Modern Fourier Analysis" by Grafakos and I try to solve one of his exercises:
Let $T$ be a bounded distribution and $P$ the Poisson-kernel in $\mathbb{R}^n$ defined by $$P(x) = \mathbf{c_n} \frac{1}{(1 + |x|^2)^{\frac{n + 1}{2}}},$$ where $\mathbf{c_n}$ is a constant, such that $\int P = 1$. Let $t > 0$ and $P_t$ denote the $L^1$-dilatation of $P$, namely $$P_t(x) = \frac{1}{t^n} \, P(\frac{x}{t}).$$ Prove that $T \ast P_t \to T$ in $\mathscr{S}'(\mathbb{R}^n)$ as $t \to 0$.
Grafakos made a hint to consider a Schwartzfunction $\psi$, which vanishes in a neighbourhood of $0$, and decompose the convolution, i.e. $$T \ast P_t = (T \ast \psi) \ast P_t + T \ast \big( (\delta_0 - \psi) \ast P_t \big).$$ I showed the convergence of the first term in $\mathscr{S'}(\mathbb{R}^n)$, but I am stuck concerning the second term. Grafakos hinted further, that one should show the convergence $\widehat{P_t} \, (1 - \widehat{\psi}) \, \widehat{T} \to (1 - \widehat{\psi}) \, \widehat{T}$ in $\mathscr{S}'$ and use the continuity of the Fouriertransform.
I know that $\widehat{P_t} = \exp\big(-2\pi \, t \, | \, \cdot \, | \big)$, thus $P_t \ast (\delta_0 - \psi)$ and $\widehat{P_t} \, (1 - \widehat{\psi})$ are both Schwartzfunctions. We know further, that $\widehat{P_t} \to 1$ pointwise and in $\mathscr{S}'$, but not uniform (and thus not in $\mathscr{S}$). Furthermore, we have $P_t \to \delta_0$ in $\mathscr{S}'$. But I don't know how to proceed, since I don't have any information about continuity of convolution when one factor is just integrable (not in Schwartzclass).
My question now is: How can I show the convergence $\widehat{P_t} \, (1 - \widehat{\psi}) \, \widehat{T} \to (1 - \widehat{\psi}) \, \widehat{T}$ in $\mathscr{S}'(\mathbb{R}^n)$ for $t \to 0$?
Thank you for your answers.
I think i solved it: One uses that $\widehat{P}_t \, (1 - \widehat{\psi}) \, \varphi \xrightarrow{t \to 0} (1 - \widehat{\psi}) \, \varphi$ in $\mathscr{S}$ for all $\varphi \in \mathscr{S}$. This can be seen by the decreasing-behaviour and the Lipschitz-continuity of $\widehat{P}_t$. Using the definition of the Schwartz-seminorms, this can be directly calculated. Then we have $$\langle \widehat{f} \, \widehat{P}_t \, (1 - \widehat{\psi}), \varphi \rangle = \langle \widehat{f} , \widehat{P}_t \, (1 - \widehat{\psi}) \, \varphi \rangle \xrightarrow{t \to 0}\langle \widehat{f}, (1 - \widehat{\psi}), \varphi \rangle = \langle \widehat{f} \, (1 - \widehat{\psi}), \varphi \rangle.$$ This concludes the proof.