TITLE: show $x$ is not tempered distribution
QUESTION [1 upvotes]: Wrong statement: $x\notin\mathcal{S}'(\mathbb{R})$,i.e. $x$ is not a tempered distribution.

However, we know polynomials($P$) or even $Pf$ for $f\in L^p,1\leq p\leq \infty$ , are tempered distribution(am I right?).
But here's my proof for this wrong statement.
take any $\varphi_n\rightarrow0$ in $\mathcal{S}$,the Schwartz space,  then $(x,\varphi_n)\rightarrow 0$ if $x\in \mathcal{S}'(\mathbb{R})$, now let's consider function 
$$
\varphi(x)=\begin{cases}
e^{-\frac{1}{1-x^2}},|x|<1\\
0,|x|\geq 1
\end{cases}
$$,
and consider $\varphi_n=\epsilon_n\varphi(\frac{2}{n}(x-n))$, then $supp \,\,\varphi_n=[\frac{n}{2},\frac{3n}{2}]$, and they are all in Schwartz space for sure. Now consider the following integral:
$$
\begin{split}
\int_{\mathbb{R}}f\varphi_ndx&=\epsilon_n\int_{\frac{n}{2}}^\frac{3n}{2}x\varphi(\frac{2}{n}(x-n))dx\\
&\geq\epsilon_n\frac{n}{2}\frac{n}{2}\int_{-1}^1\varphi(x)dx
\end{split}\tag{1}
$$
then if we take $\epsilon_n=\frac{1}{n}$, equation (1) will go to infinity as $n$ goes to infinity. which is a contradiction.
What's wrong with this proof? Thanks.

REPLY [0 votes]: Actually, x is tempered distribution.
details: using (c), take p = 1 and take N=2
Source : Rudin Functional Analysis 7.12 example (c) and (d) (book p190 -191)
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