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For $f(t,x,v)=f_0(x-ta(v),v)$, apply the area formula to the velocity map and bound the initial data by their velocity essential supremum. This gives the displayed mixed Lebesgue norm estimate whenever the weighted inverse multiplicity is essentially bounded. If $a$ is injective and $|\det Da|\ge \alpha >0$, the constant is at most $1/\alpha$. In dimension one a derivative bounded away from zero gives a global diffeomorphism; in higher dimensions determinant bounds alone do not.

For the free transport equation, substitute $y=x-tv$ into $\int |f_0(x-tv,v)|dv$. The resulting factor $|t{|}^{-d}$ and the bound by ${\sup }_v|f_0(y,v)|$ prove this mixed Lebesgue norm estimate. The right side contains initial data. The same-time mixed norm is not a conserved replacement: prescribing a product $\psi (x)\chi (v)$ at a later time and dilating $\chi$ disproves such a claim.

A mixed Lebesgue norm first takes the Lp norm in one variable and then the norm in another. For finite $p,q$, $\Vert f{\Vert }_{L_x^p{L}_v^q}=(\int (\int |f(x,v){|}^{q}dv{)}^{p/q}dx{)}^{1/p}$; an infinite exponent replaces that integral norm by an essential supremum. The order matters: generally $L_x^p{L}_v^q$ and $L_v^q{L}_x^p$ are different spaces. This distinction is crucial in free transport dispersion.

For the free transport equation, the mixed Lebesgue norm dispersion estimate isIndeed, use the method of characteristics and then the change of variables $y=x-tv$:Taking the essential supremum over $x$ proves the result. This decay measures the spreading of a velocity average; the total phase space Lp norm is conserved rather than decaying. The datum on the right is the initial datum, not the solution at the same time.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.