Is the operator norm of the derivative the same as the Riemannian norm

derivativesdifferential-geometrynormed-spacesoperator-theory

On a Riemannian manifold, $T_p M$ is the set of all derivative operators at that point. The norm is given by $$\left\|\frac{d}{dx}\right\|=\sqrt{g\left(\frac{d}{dx}, \frac{d}{dx}\right)}=\sqrt{g_{i,j}\frac{dx^i}{dx} \frac{dx^j}{dx}}$$

where $g$ is the Riemannian metric. The operator norm of an operator is given by $$\left \|\frac{d}{dx}\right\|= \sup_{\|f\|=1} \frac{df}{dx}$$

My question is this:
Given the two different ways to look at the derivative operator; do the align in any meaningful way? Are these two norms possibly the same or have some other interesting connections to each other?

Best Answer

Not an answer, too long for a comment:

J.V. Gaiter's comment is spot-on. If you are curious, however, about possible links between functional analysis and Riemannian geometry, I encourage you to take a look at Hodge theory.

One of the reasons why your intuition does not give rise to a nicely set-up functional analysis theory is because you are not acting on a Hilbert space of functions ($C^\infty$ germs at points). If you complete $C^\infty$ functions to $L^2$ (here the metric $g$ will be used, $\langle f,g\rangle_{L^2(M)} = \int_M f g \text{vol}_g$ then you end up with a Hilbert space.

You also want, as in the comment, to consider "functions" whose derivatives yield "functions" of the same type. Then you are headed towards Sobolev spaces rather than $L^2$, where you can consider vector fields as operators.

All in all, I started reading about this theory and wanted to let you know about possible interaction points between Riemannian geometry and analysis on function spaces.

Apologies if you were already aware of this.