I am having trouble with exercise 26 in chapter 2 of Peter Petersen's text "Riemannian Geometry." The exercise is stated:
"Using Polarization show that the norm of the curvature operator on $\Lambda^2 T_pM$ is bounded by $|\mathcal{R}|_p \leq c(n)|\text{sec}|_p$ for some constant $c(n)$ depending on dimension and where $|\text{sec}|_p$ denotes the largest absolute value for any sectional curvature of a plane in $T_pM$."
I understand how to write the norm of the curvature tensor as $|R|^2 = R_{ijk}^l R^{ijk}_l$. And I see that \begin{eqnarray*} R_{ljj}^l &=& g(R(\frac{\partial}{\partial x^l},\frac{\partial}{\partial x^j})\frac{\partial}{\partial x^j},\frac{\partial}{\partial x^l})\\ &=& \sum^{n}_1 c \sec(\frac{\partial}{\partial x^j},\frac{\partial}{\partial x^l})\\ &\leq& c(n)|\text{sec}|_p, \end{eqnarray*} where $c$ is some constant, but I'm not sure how to proceed or in what way the author intends us to use polarization...
EDIT (Attempt at a Solution): Since $\mathcal{R}$ is self adjoint, there exists an orthonormal basis for $\Lambda^2 T_pM$ consisting of eigenvectors of $\mathcal{R}$. Let $v_1,...,v_n$ be such a basis, i.e. $\mathcal{R}v_i = \lambda_i v_i, \; g(v_i,v_j) = 0 \; \text{and} \; ||v_i|| = 1$. We prove this using the operator norm defined in the text $|\mathcal{R}|_p = \max\{|\lambda_j|\}$ where $\lambda_j$ is an eigenvalue of $\mathcal{R}.$ Let $|\lambda_k| = \max\{|\lambda_j|\}$ \begin{eqnarray*} |\mathcal{R}|^2_p &=& (\max\{|\lambda_j|\})^2\\ &=& |\lambda_{k}|^2\\ &=& |g(\mathcal{R}(v_k),\mathcal{R}(v_k)|\\ &=& |g(\mathcal{R}(v_k),\lambda_k v_k|\\ &=& |\lambda_{k}||g(\mathcal{R}(v_k), v_k)|\\ &=& |\lambda_{k}| |\sec{(v_k)}|\\ &\leq& |\lambda_{k}| |\sec|_p.\\ \end{eqnarray*} Thus, $|\mathcal{R}|_p = |\lambda_{k}| \leq |\sec|_p$. Then, by equivalence of norms we obtain $|\mathcal{R}|_p| \leq c(n)|\sec|_p$ for the Euclidean norm, where $c(n)$ is some constant $c(n)$ depending on dimension.
The problem with the above is that the $v_i$ might not be simple ($v_i$ might not equal $x_1 \wedge x_2$ where $x_1,x_2 \in T_PM$) so that $g(\mathcal{R}(v_k), v_k)$ isn't a sectional curvature....