Operators $A$ such that $e^A$ is norm preserving

Question

Let $X$ be a Banach space. $A$ a bounded operator. We can define the exponential of $A$ by $$e^{A}=\sum_{n=0}^{+\infty}\frac{A^n}{n!},$$ which is also a bounded operator.

Is there any sufficient condition on the operator $A$ to ensure that $e^{A}$ is a norm preserving operator (isometry) i.e. $|e^Ax|=|x|$ for all $x\in X$ ?

If $X=\mathbb{C}$, we get this if $A=a\in i\mathbb{R}$. Maybe if my intuition is right, we get also this in the case $X=\mathbb{R^n}$ and $A$ has all its eigenvalues purely imaginary.

Answer 1

The natural generalization of what you noticed in $\Bbb C$ is the condition $A^* = -A$ in any Hilbert space.

Answer 2

It is sufficient to assume that $(A-\lambda I)$ is invertible for all non-zero real $\lambda$, and that $\|(A-\lambda I)^{-1}\| \le 1/|\lambda|$ for all such $\lambda$. This condition is equivalent to the condition that $e^{tA}$ is norm-preserving for all $t \ge 0$.

This condition is stated in a way that holds for real or complex Banach spaces, and I believe it holds for closed, densely-defined linear operators, too. If the Banach space $X$ is complex, then this condition forces the spectrum of $A$ to be a subset of the imaginary axis, and it forces $\|(A-\lambda I)^{-1}\|\le 1/|\Re\lambda|$ for all $\lambda$ for which $\Re\lambda \ne 0$.

Notice that the $0$ operator trivially satisfies the stated condition.

If $B$ is a densely-defined selfadjoint linear operator on a Complex Hilbert space $X$, then $A=iB$ satisfies this inequality, which is evident from the spectral theorem.

If $A$ is a complex $N\times N$ matrix, then the stated condition forces $A$ to have a basis of eigenvectors with purely imaginary or 0 eigenvalues. However, not all such $A$ satisfy the stated condition.

It is not sufficient for $A$ to have imaginary spectrum, even though the condition I stated implies that $A$ must have imaginary spectrum. For example, suppose $A^{2}=0$ for some $n > 1$ and $A\ne 0$. The spectrum of $A$ is $\{ 0\}$. If it were enough that the spectrum $\sigma(A)$ were on the imaginary axis, then $e^{zA}=I+zA$ would have to be norm preserving for all $z \in \mathbb{C}$. This $A$ does not satisfy the stated condition because $$ (\lambda I-A)^{-1} = \frac{1}{\lambda}I + \frac{1}{\lambda^{2}}A. $$

Reference: http://en.wikipedia.org/wiki/Hille%E2%80%93Yosida_theorem#Hille-Yosida_theorem_for_contraction_semigroups