It turns out that the answer is 'No, the fourth power of the geodesic distance from a point $p$ in a Finsler space $M$ whose norm, raised to the fourth power, is a smooth convex quartic form is not generally a smooth function on a neighborhood of $p$'.

While the desired smoothness does hold in the Minkowski case (i.e., when, in some local coordinate system, the functions $g_{\alpha\beta\gamma\delta}(X)$ are constant), it does not hold in general.

For example, let $M$ be the upper half-plane $y>0$ in the $xy$-plane, and let $$ ds^4 = \frac{\mathrm{d}x^4 + 2A\,\mathrm{d}x^2\mathrm{d}y^2 + \mathrm{d}y^4}{y^4}, $$ where $A$ is a constant satisfying $0<A<1$. It is easy to show that this does define a strictly convex Finsler metric on $M$ that is, in fact, complete. Moreover, it can also be shown that the geodesic distance $s_p:M\to [0,\infty)$ from any given point $p = (x_0,y_0)\in M$ has the property that $({s_p})^4$ is $C^5$ at $p$ but it is not $C^6$ there. (It is smooth everywhere else, of course.)

It does not seem to be so easy to explain why, however. Even in this example, which is homogeneous under translation in $x$ and simultaneous scaling in $x$ and $y$, simply computing the geodesic flow and using this explicitly to compute $s_p$ for some specific $p$ turns out to be surprisingly tricky. Ultimately, I had to use a different method.

Here is a sketch of the method I used; details can be supplied upon request:

In a Finsler space $(M,\mathrm{d}s)$ that is smooth (i.e., $\mathrm{d}s:TM\to[0,\infty)$ is strictly convex fiber wise and smooth away from the zero section in $TM$), the geodesic distance function from either a point or a submanifold satisfies the so-called eikonal equation wherever it is $C^1$. Here, the eikonal equation is a first order PDE for functions $f:M\to\mathbb{R}$ that is characterized by the condition that, for each $x\in M$, the $1$-form $\mathrm{d}f_x:T_xM\to \mathbb{R}$ lies in the Legendre transform $\Sigma^*\subset T^*M$ of the tangent indicatrix $\Sigma\subset TM$ consisting of 'unit vectors' in $TM$, i.e., $\Sigma = \{ u\in TM\ |\ \mathrm{d}s(u)=1 \}$. (For $u\in \Sigma_x = \Sigma\cap T_xM$, its Legendre transform is the element $u^*\in T^*_xM$ that satisfies $u^*(u)=1$ and $u^*(v)\le 1$ for all $v\in \Sigma_x$.)

When $\mathrm{d}s^4$ is a smooth function on $TM$ that is a homogeneous quartic on each fiber $T_pM$, one can write down the eikonal equation for $f:M\to\mathbb{R}$ as an explicit polynomial PDE on $F=f^4$. In the above example, the eikonal equation takes the form $P\bigl(F,\,\tfrac14yF_x,\,\tfrac14yF_y\bigr)=0$, where $P$ is an irreducible polynomial in its three arguments. In fact, one finds, after some calculation, $$ P(r_0,r_1,r_2) = \left(1-{A}^{2}\right)^{2}{r_{{0}}}^{9}+ \left(1-{A}^{2}\right) \left( \left( {A}^{2}-3 \right) {r_{{1}}}^{4}+12\,A{r_{{1}} }^{2}{r_{{2}}}^{2}+ \left( {A}^{2}-3 \right) {r_{{2}}}^{4} \right){r_ {{0}}}^{6} - \left( \left( -2\,{A}^{2}+3 \right) {r_{{1}}}^{8}+ \left( 10\,{A}^{3}-6\,A \right) {r_{{1}}}^{6}{r_{{2}}}^{2}+ \left( {A }^{4}+26\,{A}^{2}-21 \right) {r_{{1}}}^{4}{r_{{2}}}^{4}+ \left( 10\,{A }^{3}-6\,A \right) {r_{{1}}}^{2}{r_{{2}}}^{6}+ \left( -2\,{A}^{2}+3 \right) {r_{{2}}}^{8} \right) {r_{{0}}}^{3}- \left({r_{{1}}}^{4}+ 2\,A{r_{{1}}}^{2}{r_{{2}}}^{2}+{r_{{2}}}^{4} \right)^{3} $$

Now, if $f = s_{(0,1)}$ had the property that $F = f^4$ were smooth, then it is not hard to show that $$ F = x^4 + 2Ax^2(y{-}1)^2 + (y{-}1)^4 + R_5(x,y{-}1), $$ where $R_5(u,v)$ vanishes to order $5$ at $(u,v)=(0,0)$. Using the explicit form of the eikonal equation above, one can then show that $$ F = x^4 + 2A\,x^2(y{-}1)^2 + (y{-}1)^4 - 2\,x^4(y{-}1) - 2A\,x^2(y{-}1)^3 + R_6(x,y{-}1) $$ where $R_6(u,v)$ vanishes to order $6$ at $(u,v)=(0,0)$. However, now going back and substituting this into the eikonal equation and examining the lowest order terms (which have degree $38$), one finds that there is no solution to these equations for any term of order $6$. Hence $F=f^4$ cannot be $C^6$.

It turns out that the answer is 'No, the fourth power of the geodesic distance from a point $p$ in a Finsler space $M$ whose norm, raised to the fourth power, is a smooth convex quartic form is not generally a smooth function on a neighborhood of $p$'.

While the desired smoothness does hold in the Minkowski case (i.e., when, in some local coordinate system, the functions $g_{\alpha\beta\gamma\delta}(X)$ are constant) and (of course) in the case that the norm is the square root of a positive definite quadratic form, it does not hold in general. (In fact, at least in the $2$-dimensional case, it appears that the above two cases are the only cases in which, for all points $p$, the fourth power of the geodesic distance from $p$ is $C^6$ near $p$. — RLB)

For example, let $M$ be the upper half-plane $y>0$ in the $xy$-plane, and let $$ ds^4 = \frac{\mathrm{d}x^4 + 2A\,\mathrm{d}x^2\mathrm{d}y^2 + \mathrm{d}y^4}{y^4}, $$ where $A$ is a constant satisfying $0<A<1$. It is easy to show that this does define a strictly convex Finsler metric on $M$ that is, in fact, complete. Moreover, it can also be shown that the geodesic distance $s_p:M\to [0,\infty)$ from any given point $p = (x_0,y_0)\in M$ has the property that $({s_p})^4$ is $C^5$ at $p$ but it is not $C^6$ there. (It is smooth everywhere else, of course.)

It does not seem to be so easy to explain why, however. Even in this example, which is homogeneous under translation in $x$ and simultaneous scaling in $x$ and $y$, simply computing the geodesic flow and using this explicitly to compute $s_p$ for some specific $p$ turns out to be surprisingly tricky. Ultimately, I had to use a different method.

Here is a sketch of the method I used; details can be supplied upon request:

In a Finsler space $(M,\mathrm{d}s)$ that is smooth (i.e., $\mathrm{d}s:TM\to[0,\infty)$ is strictly convex fiber wise and smooth away from the zero section in $TM$), the geodesic distance function from either a point or a submanifold satisfies the so-called eikonal equation wherever it is $C^1$. Here, the eikonal equation is a first order PDE for functions $f:M\to\mathbb{R}$ that is characterized by the condition that, for each $x\in M$, the $1$-form $\mathrm{d}f_x:T_xM\to \mathbb{R}$ lies in the Legendre transform $\Sigma^*\subset T^*M$ of the tangent indicatrix $\Sigma\subset TM$ consisting of 'unit vectors' in $TM$, i.e., $\Sigma = \{ u\in TM\ |\ \mathrm{d}s(u)=1 \}$. (For $u\in \Sigma_x = \Sigma\cap T_xM$, its Legendre transform is the element $u^*\in T^*_xM$ that satisfies $u^*(u)=1$ and $u^*(v)\le 1$ for all $v\in \Sigma_x$.)

When $\mathrm{d}s^4$ is a smooth function on $TM$ that is a homogeneous quartic on each fiber $T_pM$, one can write down the eikonal equation for $f:M\to\mathbb{R}$ as an explicit polynomial PDE on $F=f^4$. In the above example, the eikonal equation takes the form $P\bigl(F,\,\tfrac14yF_x,\,\tfrac14yF_y\bigr)=0$, where $P$ is an irreducible polynomial in its three arguments. In fact, one finds, after some calculation, $$ P(r_0,r_1,r_2) = \left(1-{A}^{2}\right)^{2}{r_{{0}}}^{9}+ \left(1-{A}^{2}\right) \left( \left( {A}^{2}-3 \right) {r_{{1}}}^{4}+12\,A{r_{{1}} }^{2}{r_{{2}}}^{2}+ \left( {A}^{2}-3 \right) {r_{{2}}}^{4} \right){r_ {{0}}}^{6} - \left( \left( -2\,{A}^{2}+3 \right) {r_{{1}}}^{8}+ \left( 10\,{A}^{3}-6\,A \right) {r_{{1}}}^{6}{r_{{2}}}^{2}+ \left( {A }^{4}+26\,{A}^{2}-21 \right) {r_{{1}}}^{4}{r_{{2}}}^{4}+ \left( 10\,{A }^{3}-6\,A \right) {r_{{1}}}^{2}{r_{{2}}}^{6}+ \left( -2\,{A}^{2}+3 \right) {r_{{2}}}^{8} \right) {r_{{0}}}^{3}- \left({r_{{1}}}^{4}+ 2\,A{r_{{1}}}^{2}{r_{{2}}}^{2}+{r_{{2}}}^{4} \right)^{3} $$

Now, if $f = s_{(0,1)}$ had the property that $F = f^4$ were smooth, then it is not hard to show that $$ F = x^4 + 2Ax^2(y{-}1)^2 + (y{-}1)^4 + R_5(x,y{-}1), $$ where $R_5(u,v)$ vanishes to order $5$ at $(u,v)=(0,0)$. Using the explicit form of the eikonal equation above, one can then show that $$ F = x^4 + 2A\,x^2(y{-}1)^2 + (y{-}1)^4 - 2\,x^4(y{-}1) - 2A\,x^2(y{-}1)^3 + R_6(x,y{-}1) $$ where $R_6(u,v)$ vanishes to order $6$ at $(u,v)=(0,0)$. However, now going back and substituting this into the eikonal equation and examining the lowest order terms (which have degree $38$), one finds that there is no solution to these equations for any term of order $6$. Hence $F=f^4$ cannot be $C^6$.

It turns out that the answer is 'No, the fourth power of the geodesic distance from a point $p$ in a Finsler space $M$ whose norm, raised to the fourth power, is a smooth convex quartic form is not generally a smooth function on a neighborhood of $p$'.

While the desired smoothness does hold in the Minkowski case (i.e., when, in some local coordinate system, the functions $g_{\alpha\beta\gamma\delta}(X)$ are constant), it does not hold in general.

For example, when $M$ is the upper half-plane $y>0$ in the $xy$-plane and $$ ds^4 = \frac{\mathrm{d}x^4 + 2A\,\mathrm{d}x^2\mathrm{d}y^2 + \mathrm{d}y^4}{y^4}, $$ where $A$ is a constant satisfying $0<A<1$, then one can show that the geodesic distance $s_p:M\to [0,\infty)$ from any given point $p = (x_0,y_0)\in M$ has the property that $({s_p})^4$ is $C^5$ at $p$ but it is not $C^6$ there. (It is smooth everywhere else, of course.)

It turns out, though, that this is a delicate problem, and I don't have any simple way to explain it. Even in the above example, which is homogeneous under translation in $x$ and simultaneous scaling in $x$ and $y$, simply computing the geodesic flow and using this to derive $s_p$ for some specific $p$ turns out to be surprisingly difficult. Ultimately, I had to use a different method.

Here is a sketch of the method I used; details can be supplied upon request: In a Finsler space $(M,\mathrm{d}s)$ that is smooth (i.e., $\mathrm{d}s:TM\to[0,\infty)$ is smooth and strictly convex fiberwise away from the zero section in $TM$), the geodesic distance function from either a point or a submanifold satisfies the eikonal equation wherever it is $C^1$. The eikonal equation is a first order PDE for functions $f:M\to\mathbb{R}$ that is characterized by the condition that $\mathrm{d}f_x:T_xM\to \mathbb{R}$ lies in the Legendre transform $\lambda(\Sigma)\subset T^*M$ of the tangent indicatrix $\Sigma\subset TM$ consisting of 'unit vectors' in $TM$. When $\mathrm{d}s^4$ is a smooth function on $TM$ that is a homogeneous quartic on each fiber $T_pM$, one can write down the eikonal equation as an explicit polynomial PDE on $F=f^4$. Thus, in the above example, the eikonal equation takes the form $P(y,F,F_x,F_y)=0$ where $P$ is a polynomial in its arguments. ($P$ has degree 9 in $F$ and $12$ in $F_x$ and $F_y$.) Using this explicit expression, one can show that this equation has no nonzero solution that vanishes to order $3$ at a point $p$ (which $F=f^4$ would if $f$ were the geodesic distance from the point $p$) and yet is $C^6$ at $p$. (There is a solution that is $C^5$ at $p$, and it is unique.)

It turns out that the answer is 'No, the fourth power of the geodesic distance from a point $p$ in a Finsler space $M$ whose norm, raised to the fourth power, is a smooth convex quartic form is not generally a smooth function on a neighborhood of $p$'.

While the desired smoothness does hold in the Minkowski case (i.e., when, in some local coordinate system, the functions $g_{\alpha\beta\gamma\delta}(X)$ are constant), it does not hold in general.

For example, let $M$ be the upper half-plane $y>0$ in the $xy$-plane, and let $$ ds^4 = \frac{\mathrm{d}x^4 + 2A\,\mathrm{d}x^2\mathrm{d}y^2 + \mathrm{d}y^4}{y^4}, $$ where $A$ is a constant satisfying $0<A<1$. It is easy to show that this does define a strictly convex Finsler metric on $M$ that is, in fact, complete. Moreover, it can also be shown that the geodesic distance $s_p:M\to [0,\infty)$ from any given point $p = (x_0,y_0)\in M$ has the property that $({s_p})^4$ is $C^5$ at $p$ but it is not $C^6$ there. (It is smooth everywhere else, of course.)

It does not seem to be so easy to explain why, however. Even in this example, which is homogeneous under translation in $x$ and simultaneous scaling in $x$ and $y$, simply computing the geodesic flow and using this explicitly to compute $s_p$ for some specific $p$ turns out to be surprisingly tricky. Ultimately, I had to use a different method.

Here is a sketch of the method I used; details can be supplied upon request:

In a Finsler space $(M,\mathrm{d}s)$ that is smooth (i.e., $\mathrm{d}s:TM\to[0,\infty)$ is strictly convex fiber wise and smooth away from the zero section in $TM$), the geodesic distance function from either a point or a submanifold satisfies the so-called eikonal equation wherever it is $C^1$. Here, the eikonal equation is a first order PDE for functions $f:M\to\mathbb{R}$ that is characterized by the condition that, for each $x\in M$, the $1$-form $\mathrm{d}f_x:T_xM\to \mathbb{R}$ lies in the Legendre transform $\Sigma^*\subset T^*M$ of the tangent indicatrix $\Sigma\subset TM$ consisting of 'unit vectors' in $TM$, i.e., $\Sigma = \{ u\in TM\ |\ \mathrm{d}s(u)=1 \}$. (For $u\in \Sigma_x = \Sigma\cap T_xM$, its Legendre transform is the element $u^*\in T^*_xM$ that satisfies $u^*(u)=1$ and $u^*(v)\le 1$ for all $v\in \Sigma_x$.)

When $\mathrm{d}s^4$ is a smooth function on $TM$ that is a homogeneous quartic on each fiber $T_pM$, one can write down the eikonal equation for $f:M\to\mathbb{R}$ as an explicit polynomial PDE on $F=f^4$. In the above example, the eikonal equation takes the form $P\bigl(F,\,\tfrac14yF_x,\,\tfrac14yF_y\bigr)=0$, where $P$ is an irreducible polynomial in its three arguments. In fact, one finds, after some calculation, $$ P(r_0,r_1,r_2) = \left(1-{A}^{2}\right)^{2}{r_{{0}}}^{9}+ \left(1-{A}^{2}\right) \left( \left( {A}^{2}-3 \right) {r_{{1}}}^{4}+12\,A{r_{{1}} }^{2}{r_{{2}}}^{2}+ \left( {A}^{2}-3 \right) {r_{{2}}}^{4} \right){r_ {{0}}}^{6} - \left( \left( -2\,{A}^{2}+3 \right) {r_{{1}}}^{8}+ \left( 10\,{A}^{3}-6\,A \right) {r_{{1}}}^{6}{r_{{2}}}^{2}+ \left( {A }^{4}+26\,{A}^{2}-21 \right) {r_{{1}}}^{4}{r_{{2}}}^{4}+ \left( 10\,{A }^{3}-6\,A \right) {r_{{1}}}^{2}{r_{{2}}}^{6}+ \left( -2\,{A}^{2}+3 \right) {r_{{2}}}^{8} \right) {r_{{0}}}^{3}- \left({r_{{1}}}^{4}+ 2\,A{r_{{1}}}^{2}{r_{{2}}}^{2}+{r_{{2}}}^{4} \right)^{3} $$

Now, if $f = s_{(0,1)}$ had the property that $F = f^4$ were smooth, then it is not hard to show that $$ F = x^4 + 2Ax^2(y{-}1)^2 + (y{-}1)^4 + R_5(x,y{-}1), $$ where $R_5(u,v)$ vanishes to order $5$ at $(u,v)=(0,0)$. Using the explicit form of the eikonal equation above, one can then show that $$ F = x^4 + 2A\,x^2(y{-}1)^2 + (y{-}1)^4 - 2\,x^4(y{-}1) - 2A\,x^2(y{-}1)^3 + R_6(x,y{-}1) $$ where $R_6(u,v)$ vanishes to order $6$ at $(u,v)=(0,0)$. However, now going back and substituting this into the eikonal equation and examining the lowest order terms (which have degree $38$), one finds that there is no solution to these equations for any term of order $6$. Hence $F=f^4$ cannot be $C^6$.