Consider $E:=C^0([a,b])\times\mathbb{R}^n$ and $F:=C^n([a,b])$ equipped with the product norms. Consider $$ u^{(n)}+\sum_{i=0}^{n-1}a_i(t)u^{(i)}=f $$ with $$u(t_0)=w_1,\dots,u^{(n-1)}(t_0)=w_n \\ a_i\in C^0([a,b]),w_i\in\mathbb{R},t_0\in [a,b]$$ Then, let $T:E\to F$, defined by $$T(f,w)=u $$ where $u$ is the unique solution of the Cauchy problem.
My goal is to prove that $T$ is linear and bounded.
For what concerns the boundedness, I know that I have to apply the Closed Graph Theorem, since $E$ and $F$ are Banach. However, I can't prove any of the two thesis. Any idea?
Consider the map $D\colon F\to E$ given by
$$D(u) = \left(u^{(n)} + \sum_{i=0}^{n-1} a_i\cdot u^{(i)}, (u^{(i)}(t_0))\right).$$
It is elementary to verify that $D$ is continuous, and then you can note that $D = T^{-1}$ to conclude.
Alternatively, if you want to directly use the closed graph theorem, consider sequences $\bigl((f_k,w_k)\bigr)$ and $(u_k)$ with $u_k = T(f_k,w_k)$ such that $(f_k,w_k) \to (f,w)$, and $u_k\to u$. Then
$$u^{(i)}(t_0) = \lim_{k\to\infty} u_k^{(i)}(t_0) = \lim_{k\to\infty} w_k^i = w^i$$
for $0 \leqslant i < n$ and
$$u^{(n)} + \sum_{i=0}^{n-1} a_i\cdot u^{(i)} = \lim_{k\to\infty} u_k^{(n)} + \sum_{i=0}^{n-1} a_i\cdot u_k^{(i)} = \lim_{k\to\infty} f_k = f,$$
so $u = T(f,w)$, which shows that the graph of $T$ is closed.