TITLE: Semi-continuity of a function
QUESTION [0 upvotes]: We say that $f$ is lower semi-continuous at $x_0$ if $$\liminf_{x\to x_0}f(x)\geq f(x_0).$$
Is this property equivalent at $$\liminf_{n\to \infty } f(x_n)\geq f(x_0)$$
for all $(x_n)$ that converge to $x_0$ ? Or if we are in a Hilbert space, $(x_n)$ converge weakly to $x_0$ is enough ?

REPLY [1 votes]: Yes, the sequence characterisation is the same. It's perhaps easier to consider the contrapositive: if $f$ is not LSC at $x_0$, we must have $\liminf_{x \to x_0} f(x) < f(x_0)$, and to show this it is sufficient to find one sequence so that $\liminf_{n\to \infty} f(x_n) < f(x_0)$. Hence for LSC to hold, the opposite has to happen: every subsequence must have its limit inferior at least $f(x_0)$.
In a Hilbert space, we have two sorts of continuity, weak and strong. Since these have different topologies, we have two different notions of continuity (strong continuity being a harder condition to satisfy. Changing the function at a point implies that the notions of semicontinuity must be different as well, so we have two distinct classes of functions: strongly lower semi-continuous and weakly lower semi-continuous.
{"set_name": "stack_exchange", "score": 0, "question_id": 2257624}