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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?

1 Answers 1

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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 $\|I\| = 1$.