Monotone matrix norms [Ciarlet 2.2-10]


*

*Let $\mathscr{S}_n$ be the set of symmetric matrices and $\mathscr{S}_n^+$ the subset of non-negative definite symmetric matrices. A matrix norm $\|\cdot\|$ to be monotone if
$$A\in\mathscr{S}_n^+\; \wedge\; B-A\in\mathscr{S}_n^+\ \Rightarrow\ \|A\|\ \leq\ \|B\|.$$
Show that the norms $\|\cdot\|_2$ and $\|\cdot\|_F$ (Frobenus norm) are monotone.

*More generally, show that if a matrix norm $\|\cdot\|$ is invariant under unitary transformations, that is, if $\|A\| = \|AU\| = \|UA\|$ for every unitary matrix $U$, then it is monotone.

*Let $\|\cdot\|$ be a monotone norm and $\mbox{cond}(\cdot)$ the condition number function associated with it. Prove that
$$A,B\in\mathscr{S}_n^*\ \Rightarrow\ \mbox{cond}(A+B)\ \leq\ \max\left\{\mbox{cond}(A),\; \mbox{cond}(B)\right\}$$
where $\mathscr{S}_n^*$ denotes the subset of positive definite symmetric matrices.



I already have proved (1), and I proved that $\lambda_i(A) \leq \lambda_i(B)$, $\forall\ i=1,2,\ldots,n$ and $\forall A,B-A\in\mathscr{S}_n^+$. But I have had problems in order to prove (2) and (3). For (2), i proved that 
\begin{eqnarray*}
\|A\| & = & \|U^*AU\|\ =\ \|\mbox{diag}(\lambda_i(A))\|,\\[0.3cm]
\|B\| & = & \|V^*BV\|\ =\ \|\mbox{diag}(\lambda_i(B))\|.
\end{eqnarray*}
but I don't know what I should do next. Please help me and thanks in advance.
 A: For part 2:
To simplify notation, define a function $g:\mathbb R^n\to\mathbb R$ by  $g(x_1,\ldots,x_n)=\|\text{diag}(x_j)\|$. Since $\|\cdot\|$ is a unitarily invariant matrix norm, we have that

*

*$g$ is a norm on $\mathbb R^n$


*$g(x_1,\ldots,x_n)=g(|x_1|,\ldots,|x_n|)$ (this comes from the unitary invariance)


*$g(x_1,\ldots,x_n)=g(x_{\sigma(1)},\ldots,x_{\sigma(n)})$ for any permutation $\sigma$.
Such a $g$ is called a gauge function.
Now, if $t\in[0,1]$, then (writing $x=(x_1,\ldots,x_n)$)
\begin{align}
g(tx_1,x_2,\ldots,x_n)&=g\left(\frac{1+t}2\,x+\frac{1-t}2(-x_1,x_2,\ldots,x_n)\right)\\ \ \\
&\leq\frac{1+t}2\,g(x)+\frac{1-t}2g(-x_1,x_2,\ldots,x_n)\\ \ \\
&=\frac{1+t}2g(x)+\frac{1-t}2g(x)=g(x).
\end{align}
Applying the above inductively, we get
$$
g(t_1x_1,\ldots,t_nx_n)\leq g(x)
$$
whenever $t_1,\ldots,t_n\in [0,1]$.
Since $0\leq\lambda_j(A)\leq\lambda_j(B)$ for all $j$, we have $\lambda_j(A)=t_j\lambda_j(B)$ for appropriate $t_1,\ldots,t_n\in[0,1]$.
Then
\begin{align}
\|A\|&=\|\text{diag}(\lambda_j(A))\|=\|\text{diag}(t_j\,\lambda_j(B))\|\\ \ \\
&\leq \|\text{diag}(\lambda_j(B))\|=\|B\|
\end{align}
For part 3, I know of the original proof by Marshall and Olkin (1973). Assume $\text{cond}(A)\leq\text{cond}(B)$. Let $A'=A/\|A\|$, $B'=B/\|B\|$, and $t=\frac{\|A\|}{\|A\|+\|B\|}$.
We have, in the new notation, that $$\tag{0}\|(A')^{-1}\|\leq\|(B')^{-1}\|.$$ And
$$\tag{1}
\|tA'+(1-t)B'\|\leq t\|A'\|+(1-t)\|B'\|=1.
$$
Also, as the inverse is convex on the set of positive-definite matrices,
$$\tag{2}
(tA'+(1-t)B')^{-1}\leq t(A')^{-1}+(1-t)(B')^{-1}. 
$$
Thus, using $(2)$ and $(0)$,
\begin{align}\tag{3}
\|(tA'+(1-t)B')^{-1}\|&\leq\|t(A')^{-1}+(1-t)(B')^{-1}\|\\ 
&\leq t\|(A')^{-1}\|+(1-t)\|(B')^{-1}\|\\ &\leq\|(B')^{-1}\|
\end{align}
Now, combining $(1)$,  and $(3)$,
\begin{align}\tag{4}
\|tA'+(1-t)B'\|\,\|(tA'+(1-t)B')^{-1}\|&\leq \|(tA'+(1-t)B')^{-1}\|\leq\|(B')^{-1}\|.
\end{align}
If we now use the definitions of $A'$ and $B'$, we get
$$
tA'+(1-t)B'=\frac1{\|A\|+\|B\|}\,\left(A+B\right),\ \ \ \ (B')^{-1}=\|B\|\,B^{-1}.
$$
We may thus rewrite $(4)$ as
$$
\|A+B\|\,\|(A+B)^{-1}\|\leq \|B\|\,\|B^{-1}\|,
$$
as desired.
