[example: Jump Across A Hypersurface]
Let $u=\mathbb 1_{\{x_1>0\}}$ on $\mathbb R^n$, and write $x=(x_1,x')$. Away from the hyperplane $\{x_1=0\}$ the function is locally constant, hence smooth; at any point $(0,y_0)$ on the hyperplane every neighbourhood contains points where $u=0$ and points where $u=1$, so no smooth representative can agree with $u$ there. Thus
\begin{align*}
\operatorname{sing\,supp}u=\{x_1=0\}.
\end{align*}
Choose a cutoff of product form $\chi(x_1,x')=\alpha(x_1)\beta(x')$, with $\alpha,\beta\in C_c^\infty$, $\alpha(0)\ne 0$, and $\int \beta(x')\,dx'\ne 0$. Then
\begin{align*}
\widehat{\chi u}(\xi_1,\xi')=\left(\int_0^\infty \alpha(x_1)e^{-ix_1\xi_1}\,dx_1\right)\left(\int_{\mathbb R^{n-1}}\beta(x')e^{-ix'\cdot \xi'}\,dx'\right).
\end{align*}
Along the normal ray $\xi' =0$, set $t=\xi_1$. Since $\alpha$ is compactly supported, [integration by parts](/theorems/210) on $[0,\infty)$ gives
\begin{align*}
\int_0^\infty \alpha(s)e^{-its}\,ds=\frac{\alpha(0)}{it}+\frac{1}{it}\int_0^\infty \alpha'(s)e^{-its}\,ds.
\end{align*}
Applying the same [integration by parts](/theorems/2098) to the remaining integral gives
\begin{align*}
\int_0^\infty \alpha(s)e^{-its}\,ds=\frac{\alpha(0)}{it}+O(t^{-2}).
\end{align*}
Therefore
\begin{align*}
\widehat{\chi u}(t,0)=\left(\frac{\alpha(0)}{it}+O(t^{-2})\right)\int_{\mathbb R^{n-1}}\beta(x')\,dx',
\end{align*}
so $\widehat{\chi u}(t,0)$ decays like $|t|^{-1}$, not faster than every power of $|t|$. In tangential directions with $|\xi'|$ large and $\xi_1$ bounded, repeated integration by parts in the $x'$ variables gives rapid decay because derivatives fall only on the smooth compactly supported factor $\beta$. Thus the singularity is detected in the conormal directions, parallel to $dx_1$, rather than in directions tangent to the hyperplane.
[/example]