Estimate of norm of tangent map of a function on a Riemannian manifold

Question

Let $f: X\to \mathbb{R}$ be a function on a complete Riemannian manifold. If the gradient of $f$ satisfies $||\nabla f||\leq 2$, then

$$||f_* v||\leq 2,\qquad \text{for any unit tangent vector}\ v\in T_xX$$

where $f_*$ is the differential(or say, tangent map) of the map $f$.

I'm a little confused with the question, are the gradient and differential of $f$ the same in this special case that the image of the map is $\mathbb{R}$? If not, how do we make use of the estimate of the gradient to imply the estimate of the differential of the map $f$?

Could you please give me some help with the details? Thanks

Answer 1

If $||\nabla f||\le 2$ then $||f_*v||\le 2$ for every $v$ such that $||v||\le 1$ - the statement in your question is incorrect.

Alternatively, $||f_*v||\le 2||v||\, \forall v\in T_xM$, or better and more general, $||f_*v||\le ||f_*||\,||v||\, \forall v\in T_xM$.

$\nabla f $ is the vector field which satisfies $f_{*,x}v = g(x)(\nabla f(x), v)$ for every tangent vector $v\in T_xM$, where $g$ is the metric of the manifold (that is, actually, the very definition of the gradient of a function in a Riemannian manifold).

It should be clear from this equation how the norm of $\nabla f$ relates to that of $f_*$.

Answer 2

If $(M,g)$ is a Riemannian manifold and $f \colon M\to \Bbb R$ is smooth, the definition of the gradient of $f$ is $$ \forall v \in TM, \quad g(\nabla f,v) = f_*v $$ and Cauchy-Schwarz inequality yields $ |f_*v|\leqslant \|\nabla f \|\|v\| $. Since $\|f_*\| = \sup_{\|v\|=1}|f_*v|$, it follows that $\|f_*\|\leqslant \|\nabla f\|$. The reverse inequality is true since either $\nabla f = 0$ and there is nothing to do, or $\nabla f \neq 0$ and we have $|f_* (\frac{\nabla f}{\|\nabla f\|})| = \|\nabla f \|$.