Smooth and integrable does not imply Schwartz

The denominator $1+x^2$ is strictly positive. Inductively $f^{(k)}(x)=Q_k(x)\mathrm{/}(1+x^2{)}^{k+1}$, where $Q_0=1$ and $Q_{k+1}=(1+x^2)Q_k^{\mathrm{\prime }}-2(k+1)x{Q}_k$; these derivatives are continuous, proving smoothness. On $[0,1]$, $f\le 1$, and for $x\ge 1$, $f(x)\le x^{-2}$. Thus for $R\ge 1$, ${\int }_0^{R}f\le 1+{\int }_1^R{x}^{-2}dx=2-1\mathrm{/}R\le 2$. The nonnegative improper integral exists and is finite; evenness gives the identical bound on the negative half-line. [F1] identifies the two improper integrals with the Lebesgue integrals, so $\mathrm{\parallel }f{\mathrm{\parallel }}_1\le 4$.

For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $x^{4}f(x)=x^4\mathrm{/}(1+x^2)\ge x^2\mathrm{/}2\to \mathrm{\infty }$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. Therefore $p_{4,0}(f)=\mathrm{\infty }$, violating the defining Schwartz condition despite step 1.1. Countable choice is used only through the improper-to-Lebesgue interface, not for the explicit smoothness or failed seminorm.