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.
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.