5

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]...


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