metric scaling for an inequality

I read a lemma 1.12 in Tobias H.Colding's paper "Ricci curvature and volume convergence"."Suppose that $Ri{c_{{M^n}}} \ge \left( {n - 1} \right)\Lambda {R^{ - 2}}$,p and $q \in M$ with d(p,q)>8R,and ${b^ + } = d\left( {q, \cdot } \right) - d\left( {q,p} \right)$.Furthermore,let b be the harmonic function on ${B_{4R}}\left( p \right)$ with $b\left| {{S_{4R}}} \right.\left( p \right) = {b^ + }\left| {{S_{4R}}} \right.\left( p \right)$.Then$$\int_{{B_R}\left( p \right)} {{{\left| {Hess\left( b \right)} \right|}^2}} \le C\left( {\Lambda ,n} \right){R^{ - 2}}Vol\left( {{B_R}\left( p \right)} \right)$$.After proving when R=1,we use the metric Rg instead of g.Then $$Vo{l_{Rg}}\left( {{B_R}\left( p \right)} \right) = {R^n}Vo{l_g}\left( {{B_1}\left( p \right)} \right)$$.In metric Rg, $$\left| {Hess\left( b \right)} \right| = {R^{ - 2}}\left| {Hess\left( b \right)} \right|$$ in metric g.So$$\int_{{B_R}\left( p \right)} {{{\left| {Hess\left( b \right)} \right|}^2}} = {R^n}\int_{{B_1}\left( p \right)} {{R^{ - 4}}{{\left| {Hess\left( b \right)} \right|}^2}} \le C\left( {\Lambda ,n} \right){R^{ - 4}}Vo{l_{Rg}}\left( {{B_R}\left( p \right)} \right)$$.I get${R^{ - 4}}$ instead of ${R^{ - 2}}$.Where I am wrong?

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First, you should apply the result with $R=1$ to the manifold $(M,R^{-2}g)$, so that the unit ball in this scaled metric equals the $R$-size ball in the metric $g$.
Second, the correct scaling of the norm of the Hessian of a function $f$ is $|Hess(f)|^2_{g}=R^{-4}|Hess(f)|^2_{R^{-2}g}$.
Third, in Colding's paper, the function $f$ needs to be rescaled as well! In particular, the function $b$ for the metric $R^{-2}g$ is $R^{-2}$ times the corresponding function $b$ for the metric $g$.
@YangMills:Thank you!But I think ${R^{ - 1}}$ times instead. –  jiangsaiyin Jun 18 '13 at 3:14
Rescaling the metric tensor by $R^{-2}$ rescales distances like $R^{-1}$, which is what you want. –  YangMills Jun 18 '13 at 13:50