In many books on pdes the expression "loss of derivatives" is used when some estimates on solution are proved. Can someone clarify to me (maybe with an example) the meaning of this expression? For instance: $$\Vert e^{tD^\alpha\partial_x}u_0\Vert_{L_t^4L_x^\infty}\leq\Vert D^{\frac{1-\alpha}{4}}u_0\Vert_{L^2}$$ has a lost of $\frac{1-\alpha}{4}$ derivatives.
Edit: Here is some more background and instances in which it appears.
- In the survey article by Ifrim and Tataru on local wellposedness for hyperbolic systems ("A key observation is that, whereas solving the linearized equation would cause a loss of derivatives, solving the paradifferential equation does not in general.")
- In the Encyclopedia of Math entry on nonlinear PDE ("The 'loss of one derivative' in the inversion of the second-order hyperbolic operator leads to principal obstacles in the study of non-linear hyperbolic equations")
- This paper on 2nd order hyperbolic PDE ("...the Cauchy problem for $L$ is well-posed in $H^\infty$ with no loss of derivatives.")
- I've heard it generally used to refer to when, in estimating some important quantity (usually an energy) $E$, the norm $\|{f}\|_{H^{k + 1}}$ appears, when somehow we only have a priori estimates for $\|{f}\|_{H^{k}}$.
Suppose we have some initial data $u_0$ where have control over $s$ derivatives (e.g. if we have control over $s$ derivatives in $L^2$, this is exactly $u_0 \in H^s$). Suppose also we have a solution operator $\mathcal U(t)$ that produces a solution $u(t) = \mathcal U(t)u_0$ to whatever problem you're working on, and suppose furthermore that $\mathcal U(t)$ commutes with derivatives (e.g. if it is a Fourier multiplier operator, since then on the Fourier side both $\mathcal U(t)$ and the derivative operators are multiplication hence commute).
Then suppose we have have a bound like the one in your post, where we can control $u = \mathcal U u_0$ in some space by $\beta$ many derivatives of $u_0$ (in the case of your example, $\beta = \frac{1-\alpha}4$. Then, the same bound will control $s-\beta$ derivatives of $u$, since if $t<s-\beta$, then the bound will bound $D^t u = D^t \mathcal U u_0 = \mathcal U D^t u_0$ by $D^\beta D^t u_0 = D^{\beta+t}u_0$, which we only have control over if $\beta+t<s \iff t<s-\beta$.
Thus, "loss of derivatives" refers to when a bound can only guarantee that the solution is less regular (more precisely: controlled $s-\beta$ derivatives) than the original initial data $u_0$ (for which we could control $s$ derivatives).