Existance of the Lozinskii Measure In my ODEs class, we have the defined the Lozinskii Measure by
$\displaystyle\mu(A) = \lim_{h\to 0^{+}}\frac{\|I + hA\| - 1}{h}$, where $\|\cdot\|$ is the matrix norm defined  by
$\displaystyle\|A\| = \sup_{|x|\leq 1} |Ax|$ and $|\cdot|$ is any norm on $\mathbb{R}^{n}$.
I am asked to show several properties of this, the first of which is:
a) $\mu(A)$ exists for any $n\times n$ matrix $A$.
All I have worked out is this:
For fixed $h > 0$ (to avoid writing limit) I define $\mu_{h}(A) = \frac{|I + hA| - 1}{h}$.  Then if I apply the definition of matrix norm:
\begin{align*}
\mu_{h}(A) &= \frac{\sup_{|x|\leq 1}|(I + hA)x| - 1}h\\
&= \frac{\sup_{|x|\leq 1}|x + hAx| - 1}h.\end{align*}
But this gets me nowhere without any insight on the norm $|\cdot|$.
I'm stuck here and I was hoping someone might have a pointer for me?
 A: Fix $u, v$ in a normed vector space and consider the function $f: \mathbb{R} \to \mathbb{R}$ given by
$$
f(t) = \|u + tv\|, \qquad t \in \mathbb{R}.
$$
For any $x, y \in \mathbb{R}$ and $t \in [0,1]$ we have
$$
\begin{align}
f(tx + (1 - t)y) & = \|u + (tx + (1-t)y) v\| \\
& = \|tu + (1 - t) u + tx v + (1 - t) y v\| \\
& = \|t(u + xv) + (1 - t)(u + yv)\| \\
& \leq \|t (u + xv)\| + \|(1 - t)(u + yv)\| \\
& = t f(x) + (1 - t) f(y)
\end{align}
$$
showing that the function $f$ is convex.  
For any convex $f$ an elementary argument (which has probably been given somewhere on stackexchange before, although I can't find a link) shows that the function $g$ defined for $h  > 0$ by
$$
g(h) = \frac{f(h) - f(0)}{h}
$$
is a nondecreasing function of $h$.  It follows that $\lim_{h \to 0^{+}} g(h)$ exists (and is $\inf_{h > 0} g(h)$).
Apply this to the Banach space of all linear operators on $\mathbb{R}^n$ (with the norm induced on it by the norm you chose on $\mathbb{R}^n$), taking $u = I$ and $v = A$.  The only property of the operator norm you are really using here (besides the fact that it is a norm) is that $\|I\| = 1$.
