# Why does $\frac{d}{dt}e^{X+tY} |_{t=0}$ depend linearly on $Y$ with $X$ fixed?

I'm studying the proof of Baker-Campbell-Hausdorff formula from Brian Hall's book Lie Groups, Lie Algebras and Representations. I am stuck at this part:

I don't get why continuity of exp implies that the directional derivative depends linearly on Y with X fixed. Any help will be much appreciated.

• It seems to me that it is a simple consequence of the fact that, by definition, the derivative is the linear approximation of the function at a point. – Emilio Novati Mar 30 '16 at 21:23
• @EmilioNovati : you meant the differential more than the derivative, and don't you think continuous differentiability implies that the differential at some point is a bounded linear operator (in case of $X,Y$ belong to some Hilbert or Banach space) ? – reuns Mar 30 '16 at 23:23

## 1 Answer

I'd say that $$\frac{d}{dt}e^{X+tY} = \frac{d}{dt}\sum_{k \ge 0} \frac{(X+tY)^k}{k!} = \sum_{k \ge 1} \frac{\frac{d}{dt}(X+tY)^k}{k!}$$ with (because of non-commutativity) $$\frac{d}{dt}(X+tY)^k = \sum_{m=0}^{k-1} (X+tY)^{m} Y (X+tY)^{k-1-m}$$ and at $t=0$ : $$\frac{d}{dt}{(X+tY)^k}_{|t=0} = \sum_{m=0}^{k-1} X^{m} Y X^{k-1-m}$$ and $$\frac{d}{dt} {e^{X+tY}}_{|t=0} =\sum_{k=1}^\infty \frac1{k!}\sum_{m=0}^{k-1} X^{m} Y X^{k-1-m}$$

now the general case : if $f$ is continuously differentiable at $X$, then $$f(X+tY) = f(X) + g(t Y) + \mathcal{O}(\|t Y\|^{1 +\epsilon})$$ where $g$, the differential at $X$, is some (bounded, because of the continuity) $\mathbb{R}$-linear operator, i.e. $g(tY) = t g(Y)$ and $$\frac{d}{dt}{f(X+tY)}_{|t=0} = g(Y)$$ hence the directional derivative (at some point) of any $C^1$ function is bounded $\mathbb{R}$-linear in the direction.

• What if $f$? An entire function? – Friedrich Philipp Mar 30 '16 at 21:15
• no in the neighborhood of $X$ it is any locally analytic function plus a residual term $\mathcal{O}(\|t Y\|^{1+\epsilon})$ : i.e. a $C^1$ function, it is not correct ? – reuns Mar 30 '16 at 21:17
• @FriedrichPhilipp : I precised $C^1 \implies$ the directional derivative is $\mathbb{R}$-linear and bounded in the direction. is it why you asked if $f$ was entire ? – reuns Mar 30 '16 at 21:52
• Thanks. For the general part, your argument acted as a hint and I think I have a somewhat clearer explanation. Let $g: \mathbb{R} \to M_n(\mathbb{C})$ be $g(t) = X+tY$. Then $\frac{d}{dt} f(X+tY) |_{t=0} = \frac{d}{dt}f(g(t))|_{t=0}$. By the chain rule, this is $f'(g(0)).g'(0) = f'(X).Y$. I still don't see where the continuous differentiability of $f$ is used though. – Seven Mar 30 '16 at 22:42
• I don't think I can do this. I am confused again. – Seven Mar 30 '16 at 22:59