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Let $X$ and $Y$ denote symmetric real matrices of the same fixed size. Let $\|\cdot\|$ denote a submultiplicative matrix norm. (For concreteness, let $\|\cdot\|$ be the Frobenius norm, but I am interested in other norms too.)

Define $$f(X) = \sup_Y \|e^{X+Y}e^{-Y}-I\|$$ and $$g(X)=\inf_Y \|e^{X+Y}e^{-Y}-I\|.$$

Trivially, $0 \le g(X) \le \|e^X-I\| \le f(X)$. The question is to prove nontrivial bounds. Specific questions:

  1. Is $f(X)$ finite?
  2. Is $g(X)$ nonzero?
  3. Is $f(X) = \Theta(\|X\|)$ as $\|X\|\to0$?
  4. Is $g(X) = \Theta(\|X\|)$ as $\|X\|\to0$?
  5. Is there a simple closed form expression for $f(X)$ or $g(X)$?

Note that $\|e^X-I\|=\Theta(\|X\|)$ as $\|X\|\to0$.

The reason I'm asking this question is that I want to have some understanding of how "smooth" the matrix exponential is.

A related bound is $\|e^{X+Y}-e^Y\| \le \|X\|e^{\|X\|+\|Y\|}$ and, more generally, $\|e^X-e^Y\| \le \|X-Y\| e^{\max\{\|X\|,\|Y\|\}}$.

Thomas
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  • If you expand $e^{X+Y}e^{-Y}-I$ you get $X + (X^2 + XY - YX)/2 + \dots$. That seems to at least suggest the quantity can be taken to 0 for fixed $Y$. For non-fixed $Y$ I would guess $f(X)=\infty$, but I don't know. – Thomas Ahle May 14 '19 at 19:11

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