Assume that for all $k$, $\sup_{\omega \in \Omega}X_k(\omega) \leq 0$ with $\|X_k -X\|_ 2$ is converging in $L^2$. It seems obvious that $\sup(X) \leq 0$. It this true? What would I need to prove this, or is it a standard result?
I wanted first to use Hölder's inequality, but I cannot seem to make it work.
Edit: This probably does not hold? Could we set $X_k \equiv 0$ for all k, and define $X(\omega) = 0 $ for all $\omega\in \Omega_1$, $X(\omega') = 1$ for $\omega' \in \Omega_2$, with $P(\Omega) = 1, P(\Omega_1) = 1, P(\Omega_2) = 0$?
As a first thing, let's spell out what $\sup_{\omega \in \Omega} X_k(\omega) \leq 0$, $k \in \mathbb{N}$, is supposed to mean. Since functions in $L^2(\Omega, \mu)$ are identified whenever they differ only on a set of measure zero, it makes no sense to work with the usual supremum of a $L^2$-function. Instead, we have to consider the essential supremum, which is defined as $$ \mathrm{ess \; sup}(X_k) = \inf \{c \in \mathbb{R}: \mu(\{\omega \in \Omega: X_k(\omega) > c\}) = 0\}. $$
Then, your claim can be shown as follows: Suppose there is some $\epsilon > 0$ such that $S := \{ \omega \in \Omega: X(\omega) > \epsilon\}$ satisfies $\mu(S) > 0$, where $(\Omega, \mu)$ is the underlying measurable space. Then for any $k \in \mathbb{N}$
$$ \| X_k - X \|_{L^2}^2 = \int_\Omega (X_k-X)^2 d\mu \geq \int_S (X_k-X)^2 d\mu \geq \int_S \epsilon^2 d\mu = \epsilon \cdot \mu(S) > 0. $$