Let $A$ be a commutative Banach algebra.

Consider the exponential function $$\exp(\lambda x) = \sum_{n=1}^\infty\frac{(\lambda x)^n}{n!},$$ where $x \in A$ and $\lambda \in \mathbb C$. We can easily show that $\exp$ is entire (and hence continuous) in $A$ by showing that it converges for every $x \in A$.

Now, in a problem that I am working on, I require to show that $$\exp(-\lambda x) \cdot\exp(\lambda x)=1$$ using the above power series definition.

How can I go about showing this using the power series definition of $\exp$?

  • $\begingroup$ you meant that for any $x \in A$, $\lambda \to \exp(\lambda x)$ is an "entire function $\mathbb{C} \to A$", i.e. that for any linear functional $u \in A^*$, $ \lambda \to u(\exp(\lambda x))$ is an entire function $\mathbb{C} \to \mathbb{C}$ ? $\endgroup$ – reuns Mar 18 '16 at 20:49
  • $\begingroup$ The summation for the exponential function starts at $n=0$, not $n=1$. $\endgroup$ – user940 Mar 18 '16 at 20:50
  • $\begingroup$ or does it really mean something that $x \to \exp(\lambda x)$ is entire ? $\endgroup$ – reuns Mar 18 '16 at 20:55


  1. Simplify - set $z = \lambda x$.
  2. Remember that $\left(\sum a_k x^k\right) \left(\sum b_k x^k \right) = \sum \sum a_k b_{n-k} x^n$ and compute coefficient of $x^j$ for any $j$


For example you can use the low end of the series to check yourself. $$ (1 + z + z^2/2 + \ldots)(1 - z + z^2/2 \pm \ldots) = 1 + z(1-1) + z^2(1/2-1+1/2) + \ldots = 1 $$


I don't want to give it away completely. Note that $$ \sum_{k=0}^n \frac{(-1)^k}{k! (n-k)!} = \frac{1}{n!} \sum_{k=0}^n (-1)^k \binom{n}{k} $$ and look into the Binomial Formula for $(a+b)^n$ to see a clever interpretation of the sum...

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  • $\begingroup$ I tried doing this just now. I then get $$\exp(-z)\cdot \exp(z) = \sum_{n=0}^\infty \left( \sum_{k=0}^\infty \frac{(-1)^k}{k! (n-k)!} \right)z^n.$$ The problem is, when $n=0$ and $k=1$, then $a_kb_{n-k}$ is undefinded, since we have $$\frac{(-1)^1}{1! (0-1)!}$$ which is undefined? $\endgroup$ – user229672 Mar 18 '16 at 20:27
  • $\begingroup$ Ohhh!!! I see my mistake now! Thank you! :D $\endgroup$ – user229672 Mar 18 '16 at 20:34
  • 2
    $\begingroup$ @user229672 The inner series goes to $k=n$, not to infinity... $\endgroup$ – bartgol Mar 18 '16 at 20:34
  • $\begingroup$ @bartgol that's the mistake I just picked up :) $\endgroup$ – user229672 Mar 18 '16 at 20:35
  • $\begingroup$ you have to show that $\sum_{k=0}^n \frac{1}{k!} \frac{(-1)^k}{(n-k)!} = 0$ if $n > 0$. when $n$ is odd it is easier since $2 \sum_{k=0}^n \frac{1}{k!} \frac{(-1)^k}{(n-k)!} = \sum_{k=0}^n (\frac{1}{k!} \frac{(-1)^k}{(n-k)!} + \frac{1}{(n-k)!} \frac{(-1)^{n-k}}{k!} ) = \sum_{k=0}^n 0$ $\endgroup$ – reuns Mar 18 '16 at 20:45


use the Cauchy rule for the product of two power series

$$ \left(\sum_{i=0}^{\infty}a_ix^i\right) \left(\sum_{j=0}^{\infty}b_jx^j\right)=\sum_{k=0}^{\infty} \sum_{l=0}^{k}a_lb_{k-l}x^k $$ to prove that, if $a,b$ commutes, than $e^a e^b=e^b e^a=e^{a+b}$.

$$ e^ae^b=\sum_{i=0}^{\infty}\frac{a^i}{i!}\sum_{j=0}^{\infty}\frac{b^j}{j!}=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{a^lb^{k-l}}{l!(k-l)!}= $$ $$ =\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}a^lb^{k-l}= \sum_{k=0}^{\infty}\frac{1}{k!}(a+b)^k= e^{a+b} $$

Note that the last step use the binomial formula that is true only if $a,b$ commutes.

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