I am reading Hale's Oscillations in Nonlinear Systems, and I want to generalize some results to Banach spaces. However, I cannot find a way to generalize the following part (which is in the demonstration of Lemma 6-1):
Let $g:\mathbb{R}\rightarrow \mathbb{R}^n$ be a continuous, T-periodic function with mean value zero, and let $x^*(t)$ be the unique periodic solution of
$$\dot{x}=g(t)$$
Then there exists a constant $K$ independent of $g(t)$ such that:
$$\lvert x^*(t) \rvert \le \frac{K}{T}\int^T_0 \lvert g(u)\rvert du$$
The demonstration in the book uses coordinates to reduce the problem to the one-dimensional case, but I cannot do that in a Banach space.
If one is interested in the demonstration in the one-dimensional case, it can be solved by showing that there is a $0 \le \xi_1 \le T$ such that $x^*(t) = \int_{\xi_1}^t g(u)du$, so that $\lvert x^*(t) \rvert \le \int_{\xi_1}^t \lvert g(u) \rvert du \le \int_0^T \lvert g(u) \rvert du$. This is done by observing that, for any $0 \le \xi \le T$, the periodic solution with zero mean value $x^*(t)$ is given by integrating $g(t)$ from $\xi$ to $t$ and removing the mean value:
$$x^*(t) = \int_\xi^tg(u)du - \frac{1}{T}\int_0^T\left( \int_\xi^tg(u)du \right)dt$$
Then since it is a real function with mean value zero, there must be a $\xi_1$ for which $x^*(\xi_1) = 0$. Fiddling with the expression for $x^*(t)$ yields the desired equality.
I have the feeling I am forgetting something very basic here, but I cannot figure it out.
Let $(E, | \cdot |)$ be a Banach space and let $g : \mathbb{R} \to E$ be a continuous $T$-periodic function with vanishing mean value. Let $x : \mathbb{R} \to E$ be the unique $T$-periodic function with vanishing mean value satisfying $\dot{x}(t) = g(t)$ (which exists and is continuous by the Picard-Lindelöf theorem). Then $|x(t)| \le \int_0^T |g(u)|du$ for all $t \in \mathbb{R}$.
Since I do not quite understand the proof you sketched when $E = \mathbb{R}$, please allow me to give here my own proof. Since the estimate we want to prove is uniform in $t$, it suffices to prove that $|x_s(t)| \le \int_0^T |g(u)|du$ for all $t$ and for our favorite $s$, where we set $x_s(t) := x(t+s)$. Since $x$ has vanishing mean value, there are $t_1$ and $t_2$ such that $x(t_1) \le 0$ and $x(t_2) \ge 0$; since $x$ is continuous, the intermediate value theorem implies that there is $s$ such that $x(s) = 0$, thus $x_s(0) = 0$. Then we have $x_s(t) = x_s(t) - x_s(0) = \int_0^t g_s(u) du$, hence $$|x_s(t)| \le \int_0^t |g_s(u)| du \le \int_0^T |g_s(u)|du = \int_0^T |g(u)|du$$ where we used the $T$-periodicity of $g$ in the last equality.
Let now $E$ be arbitrary and let $E'$ denotes its topological dual. For $l \in E'$, the function $l \circ g : \mathbb{R} \to \mathbb{R}$ is continuous, $T$-periodic and has vanishing mean value, whereas the function $l \circ x : \mathbb{R} \to \mathbb{R}$ is differentiable, verifies $\frac{d(l(x(t)))}{dt} = l(g(t))$ and has vanishing mean value. Hence it follows that $$ |l(x(t))| \le \int_0^T |l(g(u))|du \, . $$
Now, given any element $y \in E$, it is well-known that $|y| = \max\{ |l(y)| : l \in E', |l| = 1 \}$. Hence for each $t$, we can choose $l_t \in E'$ with $|l_t| =1$ such that $|l_t(x(t))| = |x(t)|$, in which case we get $$ |x(t)| \le \int_0^T |l_t(g(u))|du \le \int_0^T |g(u)| du .$$ This concludes the proof.