Differentiation in Banach spaces

Question

Let $E$ be a Banach space, and $F:=L(E,E)$, with $L(E,E)$ the set of continuous linear funtions in $E$.

Prove that the function $\exp: F → F$, defined by $$\exp(A)=\sum\limits_{n=0}^\infty\frac{A^n}{n!}$$ is of class $C^{\infty}$.

Any help please?

Answer 1

The claim is proven if

$$\lim_{h\rightarrow 0} \frac{||\exp(A+h) - \exp(A) - \exp(A) h||}{||h||} = 0.$$

Now

$$\exp(A+h) = \sum_{n=0}^{\infty} \frac{(A+h)^n}{n!} = I + \sum_{n=1}^{\infty} \frac{A^n}{n!} + \sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{A^{m} h A^{n-m-1}}{n!} + o(||h||)$$

so

$$ \exp(A + h) = \exp(A) + \sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{A^{m} h A^{n-m-1}}{n!} + o(||h||). $$

Thus

$$\lim_{h\rightarrow 0} \frac{||\exp(A+h) - \exp(A) - \exp(A) h||}{||h||} = \lim_{h\rightarrow 0} \frac{||\sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{A^{m} h A^{n-m-1}}{n!} + o(||h||) - \exp(A) h||}{||h||} \leq \lim_{h\rightarrow 0} \frac{||(|| \sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{A^{m} h A^{n-m-1}}{n!} ||) - (||\exp(A)||||h||)||}{||h||} \leq \lim_{h\rightarrow 0} \frac{||(|| \sum_{n=1}^{\infty} \frac{||h|| ||A||^{n-1}}{(n-1)!} ||) - (||\exp(A)||||h||)||}{||h||} =0 .$$

Answer 2

My suggestion is to consider the formal differential $$ \sum_{n=1}^\infty \frac{1}{(n-1)!}A^{n-1}h, $$ prove that it is (totally) convergent, and conclude that it must coincide with the derivative of $\exp (A)$ evaluated at $h$. Then iterate.

Answer 3

You don't want to prove this by writing down the formula for all the derivatives. The formula would be an extension of the Campbell-Baker-Hausdorff formula to infinite dimensions and more non-commutative terms. It would have $p+1$ non-commuting terms for the $p$th derivative where C-B-H only has $2$ for all derivatives.

Even the differentiability of a polynomial function is not completely trivial. For the function $f(A) = A^2$, the first derivative is $Df(A)(H) = AH+HA$. Here I abuse the notation slightly. The second derivative is $D^2f(A)(H,K) = HK + KH$. This doesn't look much like the value $2$ for the $1$-dimensional case! It is constant, so $D^3f = 0$. These formulas are easy to prove by expanding $(A+H+K)^2$ as in the $1$-dimensional case except for being more careful with non-commutativity and starting with $H+K$ instead of $H$ (or start with H for $D^1$ and introduce $K$ for $D^2$). Similarly for any derivative of any monomial function $A^n$, except the notation would be messier. This also follows by induction if you know that the product of $C^\infty$ functions is $C^\infty$. Similarly but much more easily for addition of functions. Polynomials are $C^\infty$ since they are sums of monomials.

Addition for series is almost as easy to handle as addition for polynomials. For exp(), the $1$-dimensional estimates apply with minor changes. Only $o(n)$ estimates are needed on the derivatives. From the monomial case, we have estimates like $$||D^p(A^n)(H,K,...)|| \le (p+1)^nA^{n-p}(||H||+||K||...)^p.$$ Here $(p+1)^n$ is the sum of the coefficients in $(A+H+K+...)^n$. This is not a very precise estimate, but it is good enough since for fixed $p$, $n!$ grows much more rapidly than $(p+1)^nA^{n-p}$.

This uses the lemma that sufficient uniform convergence of derivatives of a sequence implies sufficient differentiability of the result. I tried to avoid tacit dependence on simple lemmas like this partly because most books on calculus in Banach spaces don't have many for higher derivatives.