mean valued theorem in Banach space

Question

Let $X$ be a Banach space and $U$ be a nonempty,closed,convex subset of $X$. $I$ is a subset of $\mathbb{R}$ with the Lebesgue measure $0<|I|<\infty$. $y(\cdot): [0,\infty) \to X$ with $\dot{y}(s) \in U$ a.e. $s\in I$. Do we have the following $$\frac{1}{|I|}\int_{I}{\dot{y}(s)ds} \in U$$

Answer 1

Sure! Let's write $\dot y = v$ for short.

Suppose $v : [0,\infty) \to X$ is norm-continuous. Then $\int$ can be interpreted as the norm limit of Riemann sums. It's not hard to check that $$ \frac{1}{b - a} \int_a^b v(s) \, ds = \lim_{n \to \infty} \frac1n \sum_{i = 1}^n v(a + i \frac{b-a}{n}) \, . $$ By convexity, each of the finite Riemann sums belongs to $U$. By closedness, the norm limit belongs to $U$.