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This is a follow-up to this question.

Let $T$ be a tempered distribution on $\mathbb{R}^d$. Then there is a multiindex $\alpha \in \mathbb{N}_0^d$, an $n \in \mathbb{N}_0$ and a bounded continuous function $f$ on $\mathbb{R}^d$ such that

\begin{equation} T = \partial^\alpha \left( 1 + \left \Vert x \right \Vert^2 \right)^n f \end{equation}

Suppose now, that for all $\phi \in C_c^\infty \left( \mathbb{R}^d \right)$

\begin{equation} \phi \ast T \in L^\infty \left( \mathbb{R}^d \right) \end{equation}

can we then constrain $\alpha, n$ or $f$? Intuitively, I would expect to conclude that we can choose $n = 0$ which would be in line with

\begin{equation} \begin{aligned} \left \vert \phi \ast T \left( x \right) \right \vert &\le \int \left \vert \left( \partial^\alpha \phi \right) \left( x - y \right) \right \vert \left( 1 + \left \Vert y \right \Vert^2 \right)^n \left \vert f \left( y \right) \right \vert \mathrm{d} y \\\\ &\le A^n \left \Vert f \right \Vert_{L^\infty} \left( 1 + \left \Vert x \right \Vert^2 \right)^n \int \left \vert \left( \partial^\alpha \phi \right) \left( x - y \right) \right \vert \left( 1 + \left \Vert x - y \right \Vert^2 \right)^n \mathrm{d} y \\\\ &\le A^n C_{\partial^\alpha \phi,n} \left \Vert f \right \Vert_{L^\infty} \left( 1 + \left \Vert x \right \Vert^2 \right)^n \end{aligned} \end{equation}

where $0 < C_{\partial^\alpha \phi,n} < \infty$ and $A > 0$ is a constant chosen such that

\begin{equation} 1 + \left \Vert y \right \Vert^2 \le A \left( 1 + \left \Vert x \right \Vert^2 \right) \left( 1 + \left \Vert x - y \right \Vert^2 \right) \end{equation}

for all $x, y \in \mathbb{R}^d$.

For e.g $f = 1$, it is easy tow show that we can find a $\phi$ producing the above asymptotics, that is "saturating" the inequality - up to a constant factor. But generally $f$ could be wildly oscillating and do all sorts of weird things. Hence, using any specific $\phi$ seems out of the question - but then I really struggle with ideas to move further... (I would really like to be able to show $n = 0$, so my perspective is probably biased. On the other hand it seems extremely natural.)

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