Loss of Hölder continuity by solutions of the Euler equations

[Coelum]

Homework Statement

Show that if the initial conditions $u(0,x), x\in\mathbb R^3$ for a three-dimensional shear flow are such that $u_x(0,y,z)$ is $\gamma$-Holder continuous and $u_y(z)$ is $\beta$-Holder continuous, then the solution $u(t,x)$ for $t\neq 0$ need not be any smoother than $\alpha=\beta\gamma<\min\{\beta,\gamma\}$.
[This is part of project 8.5 from Craig's "A Course on Partial Differential Equations", AMS 2018]

Relevant Equations

A function $v(y):\mathbb R^d\rightarrow\mathbb R^d$ is a Holder continuous function of $y$ with Holder exponent $0<\beta<1$ if $\exists C_\beta\in\mathbb R: \forall y,y'\in\mathbb R^d$
$$\begin{equation*} \lVert v(y)-v(y')\rVert\leq C_\beta\lVert y-y'\rVert^\beta. \end{equation*}$$
A shear flow in 3D takes the form
$$\begin{align*} u_x(t,x,y,z)&=u_x(y - t u_y(z)) \\ u_y(t,x,y,z)&=u_y(z) \\ u_z(t,x,y,z)&=c \end{align*}$$
where c is a constant.

Our thesis can be restated as follows: $\exists C_\alpha\in\mathbb R_+$ s.t. $\forall w\in\mathbb R^2_+$
$$\begin{align*} \lVert u(t,w)-u(t,w')\rVert^2 \leq C_\alpha^2\lVert w-w'\rVert^{2\alpha} \end{align*}$$
where $w=(y,z)$ and $\alpha=\beta\gamma$.
We get an upper bound for each (squared) component of $u(w)$ by applying the Holder continuity conditions:
$$\begin{align*} |u_x(t,w)-u_x(t,w')|^2&=|u_x(y-tu_y(z))-u_x(y'-tu_y(z'))|^2 \\ &\leq C_\gamma^2|(y-tu_y(z))-(y'-tu_y(z'))|^{2\gamma} \\ &\leq C_\gamma^2|(y-y')-t(u_y(z)-u_y(z'))|^{2\gamma} \\ &\leq C_\gamma^2|(y-y')-tC_\beta|z-z'|^\beta|^{2\gamma} \\ |u_y(t,w)-u_y(t,w')|^2&=|u_y(z)-u_y(z')|^2 \\ &\leq C_\beta^2|z-z'|^{2\beta} \\ |u_z(t,w)-u_z(t,w')|^2&=|c-c|^2 \\ &=0. \end{align*}$$
The RHS of the inequality we want to prove is bound by the sum of the bounds on each component:
$$\begin{equation*} \lVert u(t,w)-u(t,w')\rVert^2 \leq C_\gamma^2[(y-y')-tC_\beta|z-z'|^\beta]^{2\gamma} + C_\beta^2|z-z'|^{2\beta} . \end{equation*}$$
Now, in order to prove our thesis, we need to show that $\exists C_\alpha\in\mathbb R_+$ such that
$$\begin{equation*} C_\gamma^2[Y-tC_\beta|Z|^\beta]^{2\gamma} + C_\beta^2|Z|^{2\beta} \leq C_\alpha^2[Y^2+Z^2]^{\beta\gamma} \end{equation*}$$
where we let $Y=y-y', \; Z=|z-z'|$.
Unfortunately, the last inequality is wrong - as it is easy to spot by letting $Z=0$.