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Definitions:

We say a smooth Riemannian metric on $\mathbb R^n$ is smoothly equivalent to Lebesgue measure if the Radon Nikodym derivative of the associated Riemannian volume measure with respect to Lebesgue measure is smooth.

Given a smooth Riemannian metric $g$ on $\mathbb R^n$, and a point $x \in \mathbb R^n$, denote by $\gamma_v$ the unit speed geodesic starting at $x$ in the direction of $v$.

For a scalar function $F: \mathbb R^n \to \mathbb R$, denote by $\partial_v {F|}_g$ the scalar function given by $$ \partial_v {F|}_g (x) := \lim_{t \to 0} \frac{F(\gamma_v (t)) - F(x)}{t} $$

Question: Let $f: \mathbb R^n \to \mathbb R$, $h: \mathbb R^n \to \mathbb R$ be $C^1$ functions. For any $\varepsilon > 0$, does there always exist a smooth Riemannian metric $g$ that is smoothly equivalent to Lebesgue measure such that $$ \sup_{v \in S^{n-1}} \left\|\partial_v {h|}_g - \frac{\partial f}{\partial v}\right\|_{C^0} < \varepsilon $$ for each $1 \leq i \leq n$?

Note: Here $\|\cdot\|_{C^0}$ denotes the sup norm, and $S^{n-1}$ the unit sphere in $\mathbb R^n$ under the Euclidean metric.

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  • $\begingroup$ Defining the derivative of a function does not require a metric; recall for example the definition of tangent vectors to a manifold. The scalar function $\partial_v h$ that you define is just $Dh(v)$, it seems to me. $\endgroup$
    – Leo Moos
    Jun 14, 2021 at 9:05
  • $\begingroup$ The difference I think is the requirement that the geodesic be unit speed with respect to $g$. $\endgroup$
    – Nate River
    Jun 14, 2021 at 9:08
  • $\begingroup$ I must confess I don't follow your point. Any curve $\gamma$ with $\gamma(0) = x$ and $\gamma'(0) = v$ will produce the same derivative, namely $Dh(v)$. $\endgroup$
    – Leo Moos
    Jun 14, 2021 at 9:09
  • $\begingroup$ This curve has $\gamma’(0) = \frac{v}{||v||_g}$ I believe.. $\endgroup$
    – Nate River
    Jun 14, 2021 at 9:14

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Edit. For an even simpler example, let $f \in C^1(\mathbf{R}^n)$ be non-constant and $h = 0$. Such a metric cannot exist when the constant is so small that $\epsilon < \lvert Df \rvert_\infty$.

Let $f \in C^1(\mathbf{R}^n)$ be an arbitrary, non-constant function, and $h = -f$. Let moreover the metric $g$ be arbitrary. Then for all euclidean unit vectors $v \in \mathbf{R}^n$, that is vectors with $\sum_i (v^i)^2 = 1$, one has $D_v h/g(v,v)^{1/2} - D_v f = (1/g(v,v)^{1/2} + 1) D_v h$. Taking absolute values, the supremum of this taken over points $x \in \mathbf{R}^n$ is at least $\lvert D_v h \rvert_{\infty} = \lvert D_v f \rvert_{\infty}$. In particular if $\epsilon < \sup_{v \in \mathbf{S}^{n-1}} \lvert D_v f \rvert_{\infty}$ then the inequality cannot hold, regardless of the metric.

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  • $\begingroup$ Ah.. I guess changing the metric can only alter the magnitude of the derivative, not the direction, so if they happen to be of opposite signs then we can’t do anything. $\endgroup$
    – Nate River
    Jun 14, 2021 at 10:11

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