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The exponential series of a bounded operator

Statement

Let $X$ be a real or complex Banach space and let $A \in B (X)$ (A bounded linear operator between normed spaces). For $t \in R$ define $E (t) : = \sum_{n = 0}^{\infty} \frac{t^{n}}{n!} A^{n}$ . Then the series converges absolutely in the operator norm $∥ \cdot ∥$ of $B (X)$ , uniformly for $t$ in compact subsets of $R$ ; $E (t) \in B (X)$ with $∥ E (t) ∥ \leq e^{∣ t ∣ ∥ A ∥}$ ; $E (0) = I$ and $E (t + s) = E (t) E (s)$ for all $s, t \in R$ ; $t \mapsto E (t)$ is of class $C^{\infty}$ in the operator norm with $E^{'} (t) = A E (t) = E (t) A,$ and $∥ E (t) - I - t A ∥ \leq \frac{t^{2}}{2} ∥ A ∥^{2} e^{∣ t ∣ ∥ A ∥}$ , so that $\frac{E (t) - I}{t} \to A$ in operator norm as $t \to 0$ . In particular $E$ is a uniformly continuous (hence strongly continuous) group of bounded operators on $X$ whose generator is the bounded operator $A$ .

Facts & Assumptions

Given: A real or complex Banach space $X$ , an operator $A \in B (X)$ , and for $t \in R$ and $N \in N$ the partial sums $s_{N} (t) : = \sum_{n = 0}^{N} \frac{t^{n}}{n!} A^{n}$ and the series $E (t) : = \sum_{n = 0}^{\infty} \frac{t^{n}}{n!} A^{n}$ .

[F1]

The operator norm $∥ \cdot ∥$ is a norm on $B (X)$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The operator norm is a norm on the space of bounded linear operators, A bounded linear operator between normed spaces), and $B (X)$ is complete for it because $X$ is a Banach space (If (Y) is Banach then (\mathcal B(X,Y)) is Banach).

[F2]

Composition in $B (X)$ is associative and bilinear, $I$ is its identity, and $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ for all $S, T \in B (X)$ (Composition satisfies |ST|\le|S|,|T|); consequently $∥ A^{n} ∥ \leq ∥ A ∥^{n}$ for every $n \geq 0$ . Completeness of $B (X)$ for the operator norm [F1] together with these facts is all the structure used below; no separate Banach-algebra packaging is needed, and the estimates are identical over $R$ and $C$ .

[F3]

In a Banach space a series converges whenever it converges absolutely, i.e. whenever the series of norms converges; its partial sums are then Cauchy (Series criterion for Banach spaces, An absolutely convergent series has Cauchy partial sums, Series and absolute convergence in a normed space).

[F4]

For real $u \geq 0$ the exponential series satisfies $\sum_{n \geq 0} u^{n} / n! = e^{u} < \infty$ , and $e^{u} > 0$ (The real exponential function and the number $e$ by a power series, The exponential is a continuous bijection from $R$ onto $(0, \infty)$ ); by $n! \geq 2 (n - 2)!$ for $n \geq 2$ , its tail obeys $\sum_{n \geq 2} u^{n} / n! \leq \frac{u^{2}}{2} e^{u}$ .

Proof

technique · direct, by norm estimates on the exponential series and an elementary Cauchy-product lemma

1.1F2F4

The series converges absolutely for every real $t$ : by [F2] the general term obeys $∥ t^{n} A^{n} / n! ∥ \leq ∣ t ∣^{n} ∥ A ∥^{n} / n!$ , so with $u : = ∣ t ∣ ∥ A ∥$ the comparison series $\sum_{n} u^{n} / n!$ is the scalar exponential of [F4] and converges.

1.2F1F2F3

Cauchy-product step. If $\sum_{k} x_{k}$ and $\sum_{j} y_{j}$ converge absolutely in $B (X)$ and $z_{n} : = \sum_{k + j = n} x_{k} y_{j}$ , then $\sum_{n} z_{n}$ converges absolutely with sum $(\sum_{k} x_{k}) (\sum_{j} y_{j})$ . Indeed $\sum_{n} ∥ z_{n} ∥ \leq \sum_{n} \sum_{k + j = n} ∥ x_{k} ∥ ∥ y_{j} ∥ = (\sum_{k} ∥ x_{k} ∥) (\sum_{j} ∥ y_{j} ∥) < \infty$ by [F2], so $\sum_{n} z_{n}$ converges by [F3]. Writing $P_{N} : = \sum_{k \leq N} x_{k}$ , $Q_{N} : = \sum_{j \leq N} y_{j}$ , $T_{N} : = \sum_{n \leq N} z_{n}$ , the product $P_{N} Q_{N} = \sum_{n \leq 2 N} \sum_{k + j = n, k, j \leq N} x_{k} y_{j}$ differs from $T_{2 N}$ only by the terms with $k > N$ or $j > N$ , so $∥ P_{N} Q_{N} - T_{2 N} ∥ \leq (\sum_{k > N} ∥ x_{k} ∥) (\sum_{j} ∥ y_{j} ∥) + (\sum_{k} ∥ x_{k} ∥) (\sum_{j > N} ∥ y_{j} ∥) \to 0$ by [F2] and absolute convergence; since $P_{N} Q_{N} \to (\sum x_{k}) (\sum y_{j})$ by continuity of the product, the subsequence $T_{2 N}$ converges to that product, and a subsequence of a convergent sequence has the same limit, so $\sum_{n} z_{n} = (\sum_{k} x_{k}) (\sum_{j} y_{j})$ .

2.1F3F4step 1.1

Hence $E (t) : = \sum_{n \geq 0} t^{n} A^{n} / n! \in B (X)$ is defined for every $t$ by [F3], the family of series is dominated by the convergent scalar series $\sum_{n} R^{n} ∥ A ∥^{n} / n!$ on every compact interval $∣ t ∣ \leq R$ , so the convergence is uniform there and in particular $t \mapsto E (t)$ is continuous in operator norm; and $∥ E (t) ∥ \leq \sum_{n} ∣ t ∣^{n} ∥ A ∥^{n} / n! = e^{∣ t ∣ ∥ A ∥} < \infty$ .

2.2step 1.2step 1.1algebra

Applying [step 1.2] to $x_{k} = \frac{t^{k}}{k!} A^{k}$ and $y_{j} = \frac{s^{j}}{j!} A^{j}$ , whose series converge absolutely by [step 1.1], gives $E (t) E (s) = \sum_{n} (\sum_{k + j = n} \frac{t^{k} s^{j}}{k! j!}) A^{n} = \sum_{n} \frac{(t + s)^{n}}{n!} A^{n} = E (t + s)$ , where $A^{k} A^{j} = A^{k + j}$ and the binomial theorem in the commutative subalgebra generated by $A$ were used.

2.3F2F4step 1.1algebra

For $h \neq 0$ and $n \geq 1$ , the binomial expansion gives $\frac{(t + h)^{n} - t^{n}}{h} - n t^{n - 1} = \sum_{j = 2}^{n} (\binom{n}{j}) h^{j - 1} t^{n - j}$ ; subtracting the two absolutely convergent series and using $\sum_{n \geq 1} \frac{n t^{n - 1}}{n!} A^{n} = A E (t)$ , the difference quotient obeys $∥ \frac{E (t + h) - E (t)}{h} - A E (t) ∥ \leq \sum_{n \geq 2} \sum_{j = 2}^{n} \frac{(\binom{n}{j}) ∣ h ∣^{j - 1} ∣ t ∣^{n - j} ∥ A ∥^{n}}{n!} = (\sum_{j \geq 2} \frac{∣ h ∣^{j - 1} ∥ A ∥^{j}}{j!}) (\sum_{m \geq 0} \frac{∣ t ∣^{m} ∥ A ∥^{m}}{m!}) \leq \frac{∣ h ∣ ∥ A ∥^{2}}{2} e^{∣ h ∣ ∥ A ∥} e^{∣ t ∣ ∥ A ∥}$ , which tends to $0$ as $h \to 0$ ; the rearrangement of the nonnegative double series is legitimate and the tail estimate is [F4].

3.1F1F2step 2.1

At $t = 0$ all terms with $n \geq 1$ vanish, so $E (0) = I$ ; and since multiplication is continuous in the operator norm by [F1] and [F2], the product of the partial sums converges, which is what the next steps quantify.

3.2F2F4step 2.1

The quadratic remainder is $∥ E (t) - I - t A ∥ = ∥ \sum_{n \geq 2} \frac{t^{n}}{n!} A^{n} ∥ \leq \sum_{n \geq 2} \frac{∣ t ∣^{n} ∥ A ∥^{n}}{n!} \leq \frac{t^{2} ∥ A ∥^{2}}{2} e^{∣ t ∣ ∥ A ∥}$ by [F2] and the tail bound of [F4], since $n! \geq 2 (n - 2)!$ for $n \geq 2$ .

3.3step 2.3F2algebra

Therefore $E$ is differentiable on $R$ with $E^{'} (t) = A E (t)$ ; since $A$ commutes with every power $A^{n}$ , continuity of multiplication and [step 2.1] give $A E (t) = \lim_{N} A s_{N} (t) = \lim_{N} s_{N} (t) A = E (t) A$ as well. Iterating, if $E$ is $m$ times differentiable with $E^{(m)} = A^{m} E$ , then $t \mapsto A^{m} E (t)$ is differentiable with derivative $A^{m} E^{'} (t) = A^{m + 1} E (t)$ because $A^{m}$ is bounded; hence $E^{(m)} = A^{m} E$ for all $m$ , that is $E$ is $C^{\infty}$ with $E^{(m)} (t) = A^{m} E (t)$ .

4.1step 2.1step 2.2step 3.2step 3.3∎

Taking $t \to 0$ in [step 3.2] gives $∥ \frac{E (t) - I}{t} - A ∥ \leq \frac{∣ t ∣ ∥ A ∥^{2}}{2} e^{∣ t ∣ ∥ A ∥} \to 0$ , so the difference quotients of $E$ at $0$ converge to $A$ in operator norm; consequently $E$ is a uniformly continuous group: by [step 2.2] $E (0) = I$ and $E (t) E (- t) = E (0) = I = E (- t) E (t)$ , so each $E (t)$ is invertible with inverse $E (- t)$ , and $t \mapsto E (t)$ is norm continuous on $R$ by [step 2.1], hence strongly continuous, and the difference-quotient limit at $0$ identifies its generator with the bounded operator $A$ .

Notes. The same estimates give $∥ E (t) ∥ \leq e^{∣ t ∣ ∥ A ∥}$ and show the series converges in operator norm uniformly on compact $t$ -intervals; nothing here uses a choice principle, and the zero space $X = {0}$ is included through the estimates $∥ A ∥ = 0$ , $E (t) = I = 0$ .

Depends on

Used by

A first resolvent estimate does not ensure the prescribed semigroup bound Counterexample
The sector changes under the sign convention Counterexample
The analytic semigroup generated by a bounded operator Example
The exponential of a bounded operator is a uniformly continuous semigroup Example
Bounded Yosida semigroups converge to the generated semigroup Theorem

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Sources

Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)
Klaus-Jochen Engel and Rainer Nagel, One-Parameter Semigroups for Linear Evolution Equations, Graduate Texts in Mathematics 194 (complete author-hosted monograph) (standard reference, not scraped)