Consider the 2D step function $f(x,y)$ which is 1 for $y>0$ and $0$ for $y\leq 0$.
I want to determine the $H^s$ order of the jump along the $x$ axis.
We know the singular support of $f$ is the whole $x$ axis and the wavefront set of $f$ is the $x$ axis together with its normal vectors. And I want to know the $H^s$ order of it. Take the point $x_0=(0,0)$ and direction $\xi_0=(0,1)$ for example. It does not belong to $H^s$ wavefront set if there exists cutoff function $\phi$ not zero at $x_0$ a conic neighbourhood $V$ of $\xi_0$ such that $$\int_V|\widehat{\phi f}(\xi)|^2(1+|\xi|^2)^s\,d\xi < \infty$$ Using integration by parts we know that $\widehat{\phi f}(\xi)$ decays faster than any order of $\xi_x$ for $\xi_x\neq0$ and in the direction $(0,1)$, I can only bound it by constant, then what is the largest $s$ for the above to hold. It seems by the reference any $s<\frac{1}{2}$ is ok, but I don't know why.
Indeed, $1/2$ is the critical exponent: the function is locally in $H^s$ when $s < 1/2$, but not for $s\ge 1/2$.
The Fourier transform of $\phi f$ is $$ F(u, v) = \int_{0}^\infty \int_{-\infty}^\infty e^{-i(ux+vy)} \phi(x, y)\,dx\,dy $$ Consider the inner integral (partial Fourier transform), $$ \Phi(u, y) = \int_{-\infty}^\infty e^{-iux} \phi(x, y)\,dx $$ Since $\phi$ is a test function, $\Phi$ is smooth and decays quickly at infinity. Then $$ F(u, v) = \int_{0}^\infty e^{-ivy} \Phi(u, y) \,dy = \frac{\Phi(u, 0)}{iv } + \frac{1}{iv} \int_{0}^\infty e^{-ivy} \frac{\partial}{\partial y}\Phi_y(u, y) \,dy \tag1 $$ Hence, $F(u, v) = O(|v|^{-1})$ as $v\to\infty$. Combined with the quick decay as $u\to\infty$, this is enough to conclude that $$ (u^2+v^2)^{s/2}F(u, v) \in L^2(\mathbb{R}^2) \quad \text{ when } s<1/2 \tag2 $$ because it's the integral over $v$ that we are concerned with, and the function in (2) is $O(|v|^{s-1})$ at infinity.
Returning to (1), another integration by parts shows the integral on the right is $O(|v|^{-2})$. So the principal term is $\Phi(u, 0)/v$. And here the key is that $\Phi(u,0)$, being the Fourier transform of $\phi(x, 0)$ (which is not identically zero), is also not identically zero. It being nonzero in some interval of $u$ shows we don't get any better integrability than what (2) says.