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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You made several mistakes. 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 $g$. Putting all these together, you get the correct statement claimed by Colding. 

