I've been trying to do this exercise for a while now and I'm getting nowhere, I'd really appreciate some help:
Let $A: \mathcal{D}(A) \to C([0, 1])$ be the linear operator given by $(Ax)(t) = x'(t) + a(t) x(t)$ for all $x \in C([0, 1])$, where $a \in C([0, 1])$ is fixed and $\mathcal{D}(A) = \{x \in C^1([0, 1]) \ \vert \ x(0) = 0 \}$ . Prove that $A^{-1}$ exists in $C([0, 1])$ and is continuous.
Observation: the norm on $\mathcal{D}(A)$ is given by $\|x\| = \displaystyle\max_{x \in [0, 1]} |x(t)| + \displaystyle\max_{x \in [0, 1]} |x'(t)|$, whereas the norm in $C([0, 1])$ is the usual one, $\|f\| = \displaystyle\max_{x \in [0,1]} |f(x)|$.
I know that a linear operator $T: \mathcal{D}(T) \subset E \to F$ between two normed vector spaces has a continuous inverse iff there exists a constant $m > 0$ such that $\|T(x)\| \geq m \|x \|$ for all $x \in \mathcal{D}(T)$. I tried to find such an estimate for $\|T\|$ but to no avail, I can't seem to find any way to control $\|T(x)\|$ in terms of $\|x\|$ like that. I also tried to find the explicit inverse but it's way too complicated to show that is continuous. I also tried the usual $\varepsilon$-$\delta$ characterization but that didn't help either. I'm stuck. Would really like some help! Thanks beforehand.
You are making it too complicated. Let $g\in C[0,1]$. Consider the initial value problem $$ x'+ax=g,\ \ \ x(0)=0. $$ This is a first order linear differential equation. Integrating factor is $e^{a(t)}$, and we get $$ x(t)=e^{-a(t)}\,\int_0^t e^{a(s)}\,g(s)\,ds. $$ And that's $A^{-1}$: $$ (A^{-1}g)(t)=e^{-a(t)}\,\int_0^t e^{a(s)}\,g(s)\,ds. $$ Note that $$ \|A^{-1}g\|\leq\|e^{-a}\|\,\|e^{a}\|\,\|g\|, $$ so $A^{-1}$ is bounded and $\|A^{-1}\|\leq\|e^{-a}\|\,\|e^a\|$.