Exercise 6.1.6

• (a)

  $\frac{d^{2}f}{d{x}^2}=\frac{1}{x}>0$ on the domain $x>0$ so the entropy is convex.

• (a)

  $\frac{d^{2}f}{d{x}^2}=\frac{e^{x+y}}{{(e^x+e^y)}^2}>0$, also the second derivative to $y$ is also greater than 0, so the function is convex on all $x,y$

• (a)

  We need compute the Hessian matrix $H$ for the $l^p$ norm and check if the matrix is poistive semi-definite. We check if the first leading determinant $D_1>0$, which is the element $H_{11}=\frac{{\partial }^2{l}^p}{\partial x_1^2}$

We have $\frac{\partial l^p}{\partial x_1}={\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-1}|x_1{\text{|}}^{p-1}\frac{d\left|x_1\right|}{d{x}_1}$, where $\frac{d\left|x_1\right|}{d{x}_1}=\mathit{sign}(x_1)$, take a second derivative, we have

$\frac{{\partial }^2{l}^p}{\partial x_1^2}=(1-p){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-2}{(\frac{d\left|x_1\right|}{d{x}_1})}^2|x_1{\text{|}}^{2p-2}+(p-1){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-1}{(\frac{d\left|x_1\right|}{d{x}_1})}^2|x_1{\text{|}}^{p-2}=(p-1){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-2}\left|x_1{\text{|}}^{p-2}\right|x_2{\text{|}}^p\ge 0$ when $p\ge 1$.

So we need compute other components as well. We have $\frac{{\partial }^2{l}^p}{\partial x_2^2}=(1-p){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-2}{(\frac{d\left|x_2\right|}{d{x}_2})}^2|x_2{\text{|}}^{2p-2}+(p-1){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-1}{(\frac{d\left|x_2\right|}{d{x}_2})}^2|x_2{\text{|}}^{p-2}=(p-1){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-2}\left|x_2{\text{|}}^{p-2}\right|x_1{\text{|}}^p$.

And $H_{12}=H_{21}=\frac{{\partial }^2{l}^p}{\partial x_1\partial x_2}=(1-p){\left(|x_1{\text{|}}^p+|x_2{\text{|}}^p\right)}^{\frac{1}{p}-2}\left|x_1{\text{|}}^{p-1}\right|x_2{\text{|}}^{p-1}\frac{d\left|x_1\right|}{d{x}_1}\frac{d\left|x_2\right|}{d{x}_2}$

To compute the determinant of $H$, we have

So $H$ is positive semi-definite, the norm $l^p$ is convex when $p\ge 1$.

• (a)

  If $v_1$ is the eigenvector corresponding to ${\lambda }_1$, then we have $S{v}_1={\lambda }_1{v}_1$, so $v_1^{T}S{v}_1={\lambda }_1{v}_1^T{v}_1$, and ${\lambda }_1=\frac{v_1^{T}S{v}_1}{v_1^T{v}_1}$, it’s clear the the 2nd derivative w.r.t. $S$ is 0, so ${\lambda }_1(S)$ is positive semi-definite. It is convex