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Let $T(t),t\geq 0$, be a $C_0$-semigroup on a Banach space $X$. If $A$ is the infinitesimal generator of $T(t),t\geq 0$, then $$T(t)x=\lim_{n\infty}(I-\frac{t}{n}A)^{-n}x$$ for every $x \in X, t\geq 0$, and the limit is uniform on any bounded interval $[a,b]\subset [0,\infty)$. This is the exponential formula.

I wonder if one could affirm that the sequence $(I-\frac{t}{n}A)^{-n}$ converges in norm to $T(t)$ and the convergence is uniform on any bounded interval $[a,b]\subset [0,\infty)$?

Thank you.

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up vote 2 down vote accepted

If you have an analytic semigroup (generated by a so-called sectorial operator), then the answer is well-known and yes, even convergence rates are possible. This and generalizations were proved by many, see for example the summary and references in

M. Crouzeix, S. Larsson, S. Piskarev, V. Thomée: The stability of rational approximations of analytic semigroups, BIT 33 (1993), 74-84, Theorem 3.

In general it is possible to obtain convergence rates for "nice" initial values. A nice summary of the existing results and vast generalizations are available in this paper by Alexander Gomilko and Yuri Tomilov.

ADDED: After discussion with Delio, let me mention that the result (in Hilbert spaces) is implicitely contained in

H. FUJITA & A. MIZUTANI, "On the finite element method for parabolic equations. I; Approximation of holomorphic semi-groups," J. Math. Soc. Japan, v. 28, 1976, pp. 749-771.

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In general, this is not true (think of the first derivative acting on $L^1(\mathbb R)$).

However, this might be true in some special cases: most notably, by the spectral theorem, if $A$ is a self-adjoint operator. Namely, in that case you can simultaneously diagonalize all resolvent operators, whose eigenvalues you explicitly know - so that you are essentially back to the scalar case. More generally, this is true for some (large) class of sectorial operators on Hilbert spaces, cf. Thm. 5.1 in Operator-Norm Approximation of Semigroups by Quasi-sectorial Contractions by Cachia-Zagrebnov, JFA 2001. Their answer has been extended to general Banach spaces in Some remarks on operator-norm convergence of the Trotter product formula on Banach spaces by Blunck, JFA 2002, see also Corollary 3.6 in Euler’s Exponential Formula for Semigroups by Cachia, Semigroup Forum 2004, if you are interested in an error estimate as well.

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Delio, it is always true for analytic semigroups. See my answer I posted earlier. – András Bátkai Feb 12 '13 at 10:33
Yes, András. I have not checked exactly, but I kind of remember that the results by Blunck and Zagrebnov even generalize those by Crouzeix & co. you have referred to. – Delio Mugnolo Feb 12 '13 at 10:55
sorry, I meant: by Blunck and Cachia. – Delio Mugnolo Feb 12 '13 at 10:55
They are about the Trotter product formula, not about Euler. – András Bátkai Feb 12 '13 at 11:07
uh? Andras, I do not understand. Have you checked those articles? E.g., Corollary 3.6 in Cachia's paper is clearly about Euler's formula and has nothing to do with Trotter. Btw, it is interesting to observe that for Trotter's product formula you can have even better convergence, e.g., trace norm convergence (this has been proved by Cachia and Zagrebnov) if you are in a Hilbert space, and the same holds for other relevant convergent sequences (e.g., the Dyson-Phillips formula), but I have never seen trace class convergence for the Euler formula. – Delio Mugnolo Feb 12 '13 at 12:07

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