# I don't understand this definition of upper semi-continuity

1. Oct 16, 2011

### moxy

Defn: $f: I → ℝ$ is upper semi-continuous at $x_0 \in I$ if $f(x_0) ≥ \limsup{f(x_0)}$.

The book goes on to say that "clearly" this is equivalent to saying,

For any $ε > 0$ there exists a neighborhood $U$ of $x_0$, relative to $I$ such that $f(x) < f(x_0) + ε , \forall x \in U$.

However, this isn't clear to me. Can someone please explain why these statements are equivalent?

Last edited: Oct 17, 2011
2. Oct 17, 2011

### lurflurf

It is quite clearly equivalent. What definition of lim sup are you using? If it is not a epsilon-delta definition that may be why it is not obvious. Try writing the lim sup using epsilon-delta.
Informally we might say
f is upper semi-continuous if f(x+h) is not more than a little bigger that f(x)
f is lower semi-continuous if f(x+h) is not more that a little less than f(x)
f is continuous if f(x+h) is not more than a little different than f(x)

As the name implies semi-continuous if like half continuous

3. Oct 17, 2011

### moxy

Gah. So in the first inequality, limsup f(x0) := limx-->x0 (sup f(x))?

limsupx-->x0 f(x) = L if for all ε>0 there exists δ>0 such that
f(x) < L + ε
whenever |x - x0| < δ

So, if we take limsup f(x) = L and f is usc at x0
==> f(x0) ≥ L

And by the defn of limsup, for any ε>0,
==> L > f(x) - ε

Then f(x0) ≥ L > f(x) - ε
==> f(x0) + ε > f(x)

Clearly I'm having a lot of trouble with neighborhoods and limsups/liminfs. They haven't quite sunk in yet, I guess.