4
$\begingroup$

Let $E$ be a Banach space and $T,S \in \mathcal L (E)$ be two commuting maps i.e., $TS = ST.$ Then, $$\exp (T + S) = \exp (T) \exp (S) = \exp (S) \exp (T).$$

I can able to prove it using the formula for exponential but the proof is very clumsy. Is there any easier way to prove the above equality? Any help in this regard will be appreciated.

Thanks in advance.

$\endgroup$
2

1 Answer 1

8
$\begingroup$

Here's a different proof than the one linked (it also works in any Banach algebra). First, you can check that the power series $$e^{tS}=\sum\limits_{n=0}^\infty\frac{t^n}{n!}S^n$$ makes sense for all $t\in\mathbb{R}, S\in\mathcal{L}(E).$ Differentiating in $t$ gives $$\frac{d}{dt}e^{tS}=Se^{tS}=e^{tS}S.$$

Next, compute that $$\frac{d}{dt}(e^{t(S+T)}e^{-tT}e^{-tS})=e^{t(S+T)}Se^{-tT}e^{-tS}-e^{t(S+T)}e^{-tT}Se^{-tS}.$$ We claim this equals zero. Indeed, the fact that $ST=TS$ implies that $$e^{-tT}S=\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^nS=S\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^n=Se^{-tT},$$ from which it follows that $Se^{-tT}=e^{-tT}S$.

Hence, $$e^{t(S+T)}e^{-tT}e^{-tS}$$ is constant in $t$. Evaluating at $t=0$ gives that $$e^{t(S+T)}e^{-tT}e^{-tS}=I.$$ All that we must show is that the inverse of $e^{tS}$ is $e^{-tS},$ as the result will follow from inversion (multiply on the right by $e^{tS}$, then by $e^{tT}$). Indeed, differentiate $e^{(s+t)S}e^{-tS}$ in $t$. You'll get that it's zero, so it's constant in $t$, and evaluation at $t=0$ gives $$e^{(s+t)S}e^{-tS}=e^{sS}.$$ Evaluate this at $s=0$ to get the inversion property.

This shows that $$e^{t(S+T)}=e^{tS}e^{tT}.$$ Evaluating at $t=1$ gives the result that you wanted (well, the first equality; to get the other, simply follow the same argument with $e^{t(S+T)}e^{-tT}e^{-tS}$).

Alternatively, this results directly from e.g. the holomorphic functional calculus. This also works in any Banach algebra.

$\endgroup$
5
  • $\begingroup$ Sorry I don't know Banach algebra yet. $\endgroup$ Commented Dec 11, 2020 at 16:19
  • 1
    $\begingroup$ Oh, they’re not necessary. I was just remarking that the proof generalizes to that situation. $\endgroup$
    – cmk
    Commented Dec 11, 2020 at 16:32
  • $\begingroup$ Sir I don't actually know how differentiation of an operator is done just like what you did for $e^{tS}.$ Also I don't know whether differentiation of an operator zero means the operator is a constant operator or not. $\endgroup$ Commented Dec 11, 2020 at 16:34
  • $\begingroup$ View $e^{tS}$ as a function $f(t)=e^{tS},$ being defined by that power series. For a fixed $S$, the power series converges for all $t$, so you can differentiate it term-by-term. $\endgroup$
    – cmk
    Commented Dec 11, 2020 at 20:16
  • $\begingroup$ For your second comment, it's just basic real analysis- we're showing that the derivative of a function of a real variable is zero. Thus, it must be constant. $\endgroup$
    – cmk
    Commented Dec 11, 2020 at 20:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .