I read about this question about proving that the function of the optimal value of an LP is concave. Proof that $\xi^*(b)$ is concave for arbitrary b in a linear program
@Max who asked this question posted his proof as an answer underneath. But there's one step that doesn't make sense to me.
Perturb b1 to $tb_1+(1−t)b_2$. Then $y_1$ remains dual feasible, and $(tb_1+(1−t)b_2)^Ty_1$ provides a lower bound for the optimal objective value of this new problem.
My question is why does $(tb_1+(1−t)b_2)^Ty_1$ provides a lower bound to $\xi^*(tb_1+(1−t)b_2)$? Since $\xi^*(b)$ is a maximization problem, by weak duality it should provide an upper bound to $\xi^*(tb_1+(1−t)b_2)$?