On the Kirchoff's formula to the 3D wave equation solution $u(x,t)$, it's written in some books that if the initial conditions are $u(x,0):=f\in C^k$ and $u_t(x,0):=g\in C^{k-1}$, then the solution $u\in C^{k-1}$, why? I understand why $f$ must be once more differentiable than $g$, but since the solution is given by a integral over the sphere, doesn't that mean that one can differentiate the solution at least $k$ times?
To simplify, let $h(y)$ be a $C^{k}$ function, then is it true that $u(x,t)=\int_{|y-x|=t} h(y) dS_y$ is a $C^{k}$ class function? Why not $C^{k+1}$?
What about the 2D case?
Let $h(y)$ be a $C^{k}$ function, then is it true that $u(x,t)=\int_{|y-x|\leq t} h(y) dy$ is a $C^{k}$ class function?
Thanks in advance.
To answer your first question, I would write $$ u(t, \mathbf x)=t^2\int_{\mathbb S^2} h(t \mathbf y +\mathbf x)\, dS(\mathbf y).$$ So, formally at least $$ \partial_{\mathbf x}^\alpha u(t, \mathbf x) =t^2\int_{\mathbb S^2}\partial^\alpha_{\mathbf x} h(t \mathbf y +\mathbf x)\, dS(\mathbf y),$$ meaning that to differentiate $u$ $k$ times in $\mathbf x$ one needs $k$ derivatives of $h$. Same goes with time derivatives.