Norm of difference in exponential of matrices

Question

I would like to prove that $\|e^A-e^B\| \leq \|A-B\|e^{max\{\|A\|,\|B\|\}}$, where $A,B \in \mathbb{R}^{n \times n}$.

So far I was able to create the first difference term, but I have no idea how to incorporate the max norm. I've read this post, where the Fréchet calculus was mentioned, but I'm still stuck.

Any help would be appreciated. Thank you in advance!

Answer 1

You can first get $\|B^k-A^k\|\leq k\|B-A\|(\max(\|A\|,\|B\|))^{k-1}$ as follows

\begin{align*} \|B^k-A^k\| &= \left\lVert\sum_{l=0}^{k-1}B^{l}(B-A)A^{k-1-l}\right\rVert \leq \sum_{l=0}^{k-1}\left\lVert B^{l}(B-A)A^{k-1-l}\right\rVert\\ &\leq \sum_{l=0}^{k-1}\|B\|^l\|B-A\|\|A\|^{k-1-l} \leq k\|B-A\|\left(\max(\|A\|,\|B\|)\right)^{k-1} \end{align*}

Then\begin{align*}\|e^B-e^A\|=\left\lVert \sum_{k=0}^\infty \frac{B^k-A^k}{k!}\right\rVert\leq \sum_{k=0}^\infty \frac{k\left(\max(\|A\|,\|B\|)\right)^{k-1}\|B-A\|}{k!}=\left|B-A\right|e^{\max(\|A\|,\|B\|)}\end{align*}

Answer 2

So after some sleepless nights I came up with, what I hope is the answer. First let's use the Taylor series expansion: $\|e^A-e^B\| = \|\displaystyle\sum_{k=0}^\infty\frac{A^k-B^k}{k!}\|$

We can then use the property of the binomial polynoms $(x-y)^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})$ as $\|e^A-e^B\| = \|\displaystyle\sum_{k=0}^\infty\frac{(A-B)(A^{k-1}+A^{k-2}B+\cdots+AB^{k-2}+B^{k-1})}{k!}\|$.

The norm of product is the product of norm, and the first term is independent of $k$, so I can bring it to the front $\|e^A-e^B\| = \|A-B\|\|\displaystyle\sum_{k=0}^\infty\frac{(A^{k-1}+A^{k-2}B+\cdots+AB^{k-2}+B^{k-1})}{k!}\|$, and after that we can apply the inequality of the norm of the sums: $\|x+y\| \leq \|x\|+\|y\|$ as

$\|e^A-e^B\|\leq \|A-B\|\displaystyle\sum_{k=0}^\infty\frac{\|A\|^{k-1}+\|A\|^{k-2}\|B\|+\cdots+\|A\|\|B\|^{k-2}+\|B\|^{k-1}}{k!}=\|A-B\|S.$ If $\|A\| \geq \|B\|$, then $S\leq\displaystyle\sum_{k=0}^\infty\frac{k\|A\|^{k-1}}{k!}=\displaystyle\sum_{k=0}^\infty\frac{\|A\|^{k-1}}{(k-1)!}=e^{\|A\|}$. Similarly if $\|A\| \leq \|B\|$ then $S \leq e^{\|B\|}$, from which we can state that $S \leq e^{max\{\|A\|,\|B\|\}}$, for every $A,B\in \mathbb{R}^{n\times n}$.

This proves my initial problem.

Answer 3

Here's a slightly different proof using integration instead of the Taylor expansion. The key to what follows is the fact that${}^1$ for all $A,B\in\mathbb C^{m\times m}$ and all $t\geq0$ $$ e^{tA}-e^{tB}=\int_0^t e^{(t-s)B}(A-B)e^{sA}\,ds\,. $$ An easy way to prove this is to re-write this as $e^{-tB}e^{tA}-{\bf1}=\int_0^t e^{-sB}(A-B)e^{sA}\,ds$ and to 1. note that this is true for $t=0$, 2. differentiate after $t$ and see that the two sides are the same.

With this let $\|\cdot\|$ be any norm on $\mathbb C^{m\times m}$ which is submultiplicative (which is a necessary assumption that was also made by the other answers). We combine the above with the standard inequalities $\|\int f\,d\mu\|\leq\int\|f\|\,d\mu$ and $\|e^A\|\leq e^{\|A\|}$ to obtain \begin{align*} \|e^A-e^B\|&\leq \int_0^1 \big\|e^{(1-s)B}(A-B)e^{sA} \big\|\,ds\\ &\leq \int_0^1 \big\|e^{(1-s)B}\big\|\,\|A-B\|\,\big\|e^{sA} \big\|\,ds\tag{1}\\ &\leq \int_0^1 e^{(1-s)\|B\|}\|A-B\|e^{s\|A\|}\,ds\\ &\leq \int_0^1 e^{(1-s)\max\{\|A\|,\|B\|\}}\|A-B\|e^{s\max\{\|A\|,\|B\|\}}\,ds\\ &=e^{\max\{\|A\|,\|B\|\}}\|A-B\|\int_0^1 1\,ds=\|A-B\|e^{\max\{\|A\|,\|B\|\}} \end{align*} in the fourth step we used monotonicity of the exponential.

The advantage of this proof is that if you---for whatever reason---have a better grip on $\|e^{(t-s)A}\|,\|e^{sB}\|$ (as opposed to $e^{(t-s)\|A\|},e^{s\|B\|}$), then you can stop at (1) and upper bound the norms of the exponentials to become a better overall estimate. As an example, if $A,B$ are from the unitary algebra $\mathfrak u(n)$ and $\|\cdot\|$ is the norm which outputs the largest singular value, then $e^{(t-s)A},e^{sB}$ are unitary for all $s,t$ and their norm is always $1$. This shows the substantially better bound $\|e^A-e^B\|\leq\|A-B\|$ for this special case.

---

_{1: This is a special case of more general identity that is sometimes called "Duhamel's Formula", cf. Ch. 1, Thm. 5.1 in Dollard and Friedman, Product Integration with Application to Differential Equations, Cambridge University Press (1984)}