$M^n \subset \mathbb R^{n+1}$ is smooth hypersurface. $\nabla $ is Euclidean connection. $g$ is induced metric. $f$ is a function on $M$. How to show $$ \mathcal L_{\nabla f} g= \nabla ^2 f $$
This question is from the conformal tangent vector ($\mathcal L_X g=\lambda g$). I see a another definition that $\nabla f$ is conformal if $\nabla ^2 f =\lambda g$. So , I try to prove $\mathcal L_{\nabla f} g= \nabla ^2 f$.
But as wiki, $$ \mathcal L_Xg= (X^c \partial_c g_{ab}+g_{cb}\partial_a X^c + g_{ca} \partial _b X^c) dx^a\otimes dx^b $$ I can't turn it to $\nabla ^2 f$.
Besides, in Euclidean space, how to show $\nabla^2 |x|=\frac{1}{|x|}(g+\nabla|x| \otimes \nabla |x|)$ ?
$\def\L{\mathcal L}$ From the product rule for the Lie derivative we have $$\L_X g (Y,Z) = X(g(Y,Z)) - g(\L_X Y,Z) - g(Y, \L_X Z).$$
Expanding the first term using the metric-compability of $\nabla$ and the second two terms using the fact that $\nabla$ is torsion-free we get
$$ \L_X g(Y,Z) = g(\nabla_X Y,Z) + g(Y, \nabla_X Z) - g(\nabla_X Y - \nabla_Y X,Z) - g(Y, \nabla_X Z - \nabla_Z X) \\ =g(\nabla_Y X,Z)+g(Y,\nabla_Z X).$$
Thus $\L_X g$ is twice the symmetrization of the tensor obtained by lowering the index of $\nabla X$; so $$\L_{\mathrm{grad}f} g=2\mathrm{Sym}(\nabla\nabla f)=2\nabla^2 f$$ since the Hessian is already symmetric.
Your second question should be a simple coordinate calculation - just show $$\partial_i \partial_j |x| = \frac 1 {|x|} (\delta_{ij} + \partial_i |x| \partial_j |x|).$$ The product rule applied to $|x|^2$ should help.