# Can two continuously differentiable functions be made $C^1$ close via a perturbation of the metric?

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.

• 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. Jun 14, 2021 at 9:05
• The difference I think is the requirement that the geodesic be unit speed with respect to $g$. Jun 14, 2021 at 9:08
• 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 $Dh(v)$. Jun 14, 2021 at 9:09
• This curve has $\gamma’(0) = \frac{v}{||v||_g}$ I believe.. Jun 14, 2021 at 9:14

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.