If $f\in \mathbb C^\infty (M)$ and $(M,g)$ a Riemannian metric, does $g(fX,Y)=fg(X,Y)$?

Question

Let $(M,g)$ a riemanian manifold and $f\in \mathcal C^\infty (M)$. Does $$g(fX,Y)=fg(X,Y)\ \ ?$$

I know that $g$ is $\mathbb R-$bilinear, but is it also $\mathcal C^\infty (M)-$bilinear ?

In fact, in an exercise, I have $(M,g)$ and $(M,g')$ where $g'=e^{2u}g$. And I have to show that if $\{E_1,E_2\}$ is an othonormal basis of $T_pM$, then $\{\tilde E_1,\tilde E_2\}$ where $\tilde E_i=e^{-u}E_i$ is an orthonormal basisof $T_pM$ for the metric $g'$.

So if $g$ $\mathcal C^\infty (M)-$bilinear it's obvious, but since I don't have any result that says that, I have some doubt.

Answer 1

Yes, a Riemannian metric is an algebraic object (or tensor field): That is, $g(X, Y)$ at a point $p$ depends only on the values $X_{p}$ and $Y_{p}$ of the vector fields $X$ and $Y$. Bilinearity over smooth functions is another way to say the same thing.