# Constrain representation of tempered distribution

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

$$$$T = \partial^\alpha \left( 1 + \left \Vert x \right \Vert^2 \right)^n f$$$$

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

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

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{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}

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

$$$$1 + \left \Vert y \right \Vert^2 \le A \left( 1 + \left \Vert x \right \Vert^2 \right) \left( 1 + \left \Vert x - y \right \Vert^2 \right)$$$$

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.)