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I am currently reading through Groups and Symmetries by Yvette Kosmann-Schwarzbach. I am working through a lemma with proof in the section titled One-Parameter subgroups of $GL(n,\mathbb{K})$, and I am having trouble understanding certain steps.

I understand the first few steps until I get to "To prove $f$ is differentiable, it is sufficient to show that there is a real number $a>0$ such that $\int_0^a f(t)dt$ is invertible."

My question is why is this a sufficient condition for differentiability of $f$ and how do the following steps imply invertibility of $\int_0^a f(t)dt$?

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1 Answer 1

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Because $f$ is matrix valued, $\int_0^a f(t)dt$ is a matrix. So by invertible, the author means invertible as a matrix.

If there exists an $a > 0$ such that $\int_0^a f(t)dt$ is invertible, then we can solve to obtain $$ f(s) = \left(\int_0^a f(t)dt\right)^{-1}\int_s^{s+a}f(t)dt. $$ But we know that the function on the right is differentiable, so the one on the left is too.

Now to see why the condition established above shows that $\int_0^a f(t)dt$ is invertible we can use the Neumann series. Specifically, if $\left\lVert T \right\rVert < 1$, then $$(I - T)^{-1} = \sum_{k=0}^\infty T^k,$$ where you can show that the series converges using essentially the same proof that is used to prove the formula for the sum of a convergent geometric series. Now we just notice that $\frac{1}{a}\int_0^a f(t)dt = I - (I - \frac{1}{a}\int_0^a f(t)dt)$ is invertible since $$\left\lVert \frac{1}{a}\int_0^a f(t)dt - I \right\rVert < 1.$$

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