Let $X$ be a Banach space and $\mathcal{L}(X)$ the Banach algebra of all bounded linear operators $L:X\to X$, where the norm is given by


and the product is the composition of operators.

If $T,F:(0,\infty)\to\mathcal{L}(X)$ are differentiable functions, can we apply the product rule to derive $T(t)F(t)$?

Particularly, I'm interested in the case that $\{T(t)\}_{t\geq 0}$ and $\{F(t)\}_{t\geq 0}$ are both uniformly continuous semigroups of bounded linear operators.


  • 1
    $\begingroup$ For that, the product rule is relevant. You don't have a composition $g(f(t))$, but a product (which is a composition of linear maps, true, but that's not the sort of composition you apply the chain rule to). $\endgroup$ – Daniel Fischer Jan 15 '14 at 15:41
  • $\begingroup$ @DanielFischer I should have written "product rule" instead of "chain rule". I will correct it. $\endgroup$ – Pedro Jan 15 '14 at 15:46
  • $\begingroup$ I am pretty sure that you can copy and paste the standard proof for the product rule in $\mathbb{R}$ to this situation - all that you really use is that the difference quotient converges to the differential and the continuity of the multiplication. $\endgroup$ – Sebastian Jan 15 '14 at 15:49

Yes, you can. For $h \ne 0$, $$ \begin{align} & \frac{1}{h}\left[T(t+h)F(t+h)-T(t)F(t)\right]\\ & =\frac{1}{h}\{T(t+h)-T(t)\}F(t+h)+T(t)\frac{1}{h}\{F(t+h)-F(t)\}\\ & =\frac{1}{h}\{T(t+h)-T(t)\}F(t) + T(t)\frac{1}{h}\{F(t+h)-F(t)\} \\ & + \frac{1}{h}\{T(t+h)-T(t)\}\{F(t+h)-F(t)\} \end{align} $$ Therefore, $$ \begin{align} & \left\|\frac{1}{h}\left[T(t+h)F(t+h)-T(t)F(t)\right]-\left[T'(t)F(t)+T(t)F'(t)\right]\right\| \\ & \le \left\|\frac{1}{h}[T(t+h)-T(t)]-T'(t)\right\|\|F(t)\| +\|T(t)\|\left\|\frac{1}{h}[F(t+h)-F(t)]-F'(t)\right\| \\ & + \left\|\frac{1}{h}[T(t+h)-T(t)]\right\|\left\|\frac{1}{h}[F(t+h)-F(t)]\right\||h|. \end{align} $$ The right side tends to 0 as $h\rightarrow 0$ assuming that $T$ and $F$ are differentiable at $t$.

  • $\begingroup$ Could you explain me how do we get $$\left\|\frac{1}{h}[T(t+h)-T(t)]-T'(t)\right\|\left\|\frac{1}{h}[F(t+h)-F(t)]-F'(t)\right\||h|$$ instead of $$\left\|\frac{1}{h}[T(t+h)-T(t)]\right\|\left\|\frac{1}{h}[F(t+h)-F(t)]\right\|?$$ $\endgroup$ – Pedro Jan 17 '14 at 0:00
  • $\begingroup$ @Pedro: I accidentally put in the $-T'(t)$ and $-F'(t)$. I corrected that now. The end result is the same because there is a term $|h|$ multiplying terms which remain bounded as $h\rightarrow 0$. $\endgroup$ – DisintegratingByParts Jan 17 '14 at 3:36
  • $\begingroup$ @Pedro: Please note that the $|h|$ multiplying the last two terms is correct because the original product had only one $1/h$; so I added another to go with $F(t+h)-F(t)$. $\endgroup$ – DisintegratingByParts Jan 17 '14 at 3:52

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