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!


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*}


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.

  • $\begingroup$ It is not clear to me how you did the first inequality. You would elaborate a bit on this? $\endgroup$ – Carl Christian May 31 '18 at 20:13
  • $\begingroup$ I do not think that the use of identity transforming $x^n-y^n$ is permitted in the case of matrices because we do not know if $A$ and $B$ commute. In particular, we have $$(A-B)(A+B)=A^2+AB-BA-B^2=A^2-B^2$$ if an only if $AB=BA$. $\endgroup$ – Carl Christian Jun 1 '18 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.