Consider the wave equation
$$\begin{cases} \partial_t^2 u &= \Delta u, t \in \mathbb R \\ u(0) &= u_0 \in S'(\mathbb R^n) \\ \partial_t u(0) &= u_1 \in S'(\mathbb R^n) \end{cases}$$
where $n \geq 2$ and $S'(\mathbb R^n)$ denotes the space of tempered distributions.
We denote by $\hat{u}_0$, $\hat{u}_1$ the Fourier transform of $u_0$, $u_1$ with respect to the space variable $x \in \mathbb R^n$. Here, we assume that $u_0, u_1 \in L^1(\mathbb R^n)$ and $\hat{u}_0$, $\hat{u}_1$ are compactly supported in some annulus $C$.
Applying the Fourier transform and solving the ODE, we find the solution
$$u(t) = K^{+}(t, \cdot) *_x \gamma^+(\cdot) + K^{-}(t, \cdot) *_x \gamma^-(\cdot)$$
where $$\hat{K}^{\pm}(t,\xi) = e^{\pm i t |\xi|}, \quad \hat{\gamma}^{\pm}(\xi) = \dfrac{1}{2}( \hat{u}_0(\xi) \pm \dfrac{1}{i|\xi|} \hat{u}_1(\xi))$$
We can multiply $\hat{K}^{\pm}$ and $\hat{\gamma}^{\pm}$ by a smooth and compactly supported function $\phi(\xi)$ such that $\phi = 1$ on $C$. This won't change $u(t)$, so let's assume we have done this.
Now, using Young inequality, we have a decay
$$||u(t)||_{L^{\infty}} \leq C ||K^{\pm}(t,x)||_{L^{\infty}(\mathbb R^n)}(||u_0||_{L^1} + ||u_1||_{L^1})$$
and $$||K^{\pm}(t,x)||_{L^{\infty}(\mathbb R^n)} = ||K^{\pm}(t,tx)||_{L^{\infty}(\mathbb R^n)}$$
for $t \neq 0$.
Here, $K^{\pm}(t,tx)$ is an oscillatory integral of the first kind with phase $\Phi_x(\xi) = \pm |\xi| + \langle x, \xi \rangle$ and amplitude $\phi(\xi)$. The method of stationary phase should get us an upper bound $|K^{\pm}(t,tx)| \leq C |t|^{-n/2}$.
However, in my book, it is only proven that $|K^{\pm}(t,tx)| \leq C |t|^{-(n-1)/2}$.
Why do we have this loss of regularity ?