Lets say $f:\mathbb{R}\rightarrow \mathbb{R}$ is convex. Consider the following "divided difference" matrix: for a fixed $y,x_1,\ldots,x_n$, let $$ M_{ij}=\frac{1}{x_j-x_i}\cdot \Big(\frac{f'(x_j) -f'(y)}{x_j-y}-\frac{f'(x_i) -f'(y)}{x_i-y}\Big). $$ My question is: Is $M^i$ positive semidefinite?
My attempt: one can write the Hessian matrix $P_{i,j}=\nabla_i\nabla_j f$ which is known to be positive semi-definite. Now consider the following matrix, for a fixed $i\in [n]$, $$ Q^i_{jk}=\nabla_i\nabla_j\nabla_k f. $$ I believe $Q^i$ is positive semidefinite, but does that imply that the $M^i$ I wrote above is also PSD?