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Your first argument is correct, for as you noted when $X\succ 0$, $$\log\det(X^{-1})=\log((\det(X))^{-1})=-\log\det(X),$$ which is the negative of a concave function, hence convex. Your issue with the second argument is you seem to be using the "claim" that if $f:C\to \mathbb{R}$ is concave, where $C$ is a convex set, then for any invertible map $\phi:C\... 3 A differentiable function$f$is convex on an interval$(a,b)$if and only if its derivative$f'$is increasing. Therefore, your assumptions imply that$f'$is increasing on$(0,z)$and on$(z,1)$. Now your assumption that$f'$is continuous immediately gives you that$f'$is increasing on$(0,1)$and therefore convex. 3 A differentiable function is convex in an interval$I$if and only if$f'$is increasing in$I$. Now we have that $$f'(x)\leq f'(z)\leq f'(y)$$ for any$0\leq x<z<y\leq 1$. Can you take it from here and show that the proposition is true? 2 A crude but simple way to bound the quantity you describe can be found if$f$is twice differentiable; in this case we have $$|pf(x)+(1-p)f(y)-f(px+(1-p)y)|\leq p(1-p)|x-y|^2\sup_{z\in[a,b]}|f^{\prime\prime}(z)|$$ for all$x,y\in[a,b]$and$p\in[0,1]$. This can be proven by thrice applying the mean value theorem: WLOG assume that$x<y$, then there exist ... 2 That “therefore” makes no sense. See what happens with the real numbers greater than$0$:$\log(x)$is concave and$\log\left( x^{-1}\right)=\bigl(-\log(x)\bigr)$is convex, not concave. 2 It is convex. Since$E$admits an orthogonal diagonalisation$Q\operatorname{diag}(n-1,\,-1,\ldots,-1)Q^T$, if we put$c=Q^Tb$, the function in question can be rewritten as $$x\mapsto\sum_{i=1}^n|c_i|^2\exp(-\lambda_i\langle a,x\rangle),$$ where$\lambda_1=n-1$and$\lambda_2=\cdots=\lambda_n=-1$. This is a non-negatively weighted sum of convex functions. ... 1 No, because$\sigma$"might be alternating". However each$\sigma$has such representation up to an arbitrary alternating tensor. Consider the maps$\operatorname{Alt}, \operatorname{Sym} \colon V \otimes V \rightarrow V \otimes V$given (on elementary tensors and extended linearly) by $$\operatorname{Alt}(v \otimes w) = \frac{v \otimes w - w \otimes v}{2},... 1 Yes it's true. f is convex iff f' is monotonically non-decreasing, but it works on [0, z) and (z, 1]. But f' is continuous at z so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex. 1 I think it is precisely the opposite, as it must follow from the reasoning below. I'm using the Wikipedia definition of strictly convex function, that is, f : X \to \mathbb R is strictly convex if, whenever x_1, x_2 \in X are such that x_1 \neq x_2, and 0 < t < 1, then$$f(tx_1 + (1-t)x_2) < tf(x_1) + (1-t) f(x_2).$$Now suppose$f,g : [0,1]...