Diagonal approximation of the inverse Hessian matrix While reading chapter 5 of Data Networks [1] by Bertsekas and Gallager, I came across the following statement (p. 467):

A simple choice that often works well is to take $B^k$ as a diagonal approximation to the inverse Hessian, that is
  $$B^k=\begin{pmatrix}\left(\frac{\partial^2f(x^k)}{\partial x_1^2}\right)^{-1} & 0 & \cdots & 0 \\ 0 & \left(\frac{\partial^2f(x^k)}{\partial x_2^2}\right)^{-1} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \left(\frac{\partial^2f(x^k)}{\partial x_n^2}\right)^{-1}\end{pmatrix}.$$

How can such a matrix be a good approximation of the Hessian matrix of $f$ at point $x^k$, assuming that for all $x$, $\nabla^2f(x)$ is a positive semidefinite matrix that depends continuously on $x$ ?
 A: The statement should probably be viewed in the context of the book. When e.g. the matrix is diagonally dominated you may use the diagonal inverse to calculate the inverse using a von Neumann series. So it is 'sort' of an approximation. Better if the diagonal elements dominate with some large factor the other elements. But in general, you are right that the formula need not make sense.
A: It can be derived as a consequence for Kolmogorov-Frechet Theorems. Take some U-statistic and derive it as a symmetric, measurable (invertible without expansion of the domain), which has a bounded quadratic integrated value.(Consider Cauchy-Schwarz Theorem and some measure theoretic results)- (Mann-Whitney U Test- Rank sum test will be a good approx.) The resulting functions are differentiable with respect to Arzelo Ascoli Theorems. Derive the Hessian matrix and by Young´s Theorem. Notice, that any partial differentiation of this sort will yield ordinal results of the same domain. (Young´s theorem implies the neighborhood of size 1 of a variable of differentiation will alone imply the strong continuity with reference to the the Integral of the derivative.) 
By the approximation theorems, see, that any differentiation with reference to a variable of your choice won´t effect its ordinality in their domain; hence, the invertibility with respect to some squared integrated mean or median. This obeys the invertibility conditions and the arithmatic mean linearity at the same time. You have a diagonal, hence a normal matrix (Consider solvebility of a characteristic function with regards to its eigenvalues.)       
