MHB Lower SemiContinuity Problem: Proofs & Step Funcs

[joypav]

Problem:
(a)
Show that $f: E \rightarrow R$ is lower semi-continuous at $x_0 \in E$ if and only if
$f(x_0) \leq \liminf_{x \rightarrow x_0}f(x)$

Proof:
"$\rightarrow$"
Assume f is l.s.c. at $x_0 \in E$.
$\implies \forall \epsilon > 0, \exists \delta >0$ such that if $\left|x-x_0 \right|<\delta$, then $f(x_0)-\epsilon \leq f(x)$

Consider $\liminf_{x \rightarrow x_0}f(x) = \liminf_{n \rightarrow \infty}\left\{f(x): \left| x-x_0 \right|<1/n\right\}$.

Let $y \in R, y = \liminf_{x \rightarrow x_0}f(x) \in R$. Then $y = f(x')$ for some $x' \in E$.

Because f is l.s.c. at $x_0$,
$\forall \epsilon > 0, \exists \delta >0$ such that if $\left|x'-x_0 \right|<\delta$, then $f(x') \geq f(x_0)- \epsilon$
$\implies \liminf_{x \rightarrow x_0}f(x) \geq f(x_0) - \epsilon$

Let $\epsilon \rightarrow 0$
$\implies \liminf_{x \rightarrow x_0}f(x) \geq f(x_0)$.

"$\leftarrow$"
Assume $f(x_0) \leq \liminf_{x \rightarrow x_0}f(x)$.
I am thinking I should proceed with proving by contradiction? Meaning, assume there is an $x_0$ where f is not l.s.c. (b)
A real-valued function $\phi$ defined on an interval $[a, b]$ is called a step function if there is a partition $a = x_0 < x_1 <...< x_n = b$ such that for each i the function $\phi$ assumes only one value in the interval $(x_{i−1}, x_i)$. Show that a step function $\phi$ is lower semi-continuous if and only if $\phi(x_i)$ is less than or equal to the smaller of the two values assumed in $(x_{i−1}, x_i)$ and $(x_i, x_{i+1})$.

For this part I think I am just confused by the definition given for step function. However, I am still thinking about the problem. So I will add onto the post if I make some headway.

 

[Euge]

Please state the definition you're using for l.s.c. functions. Also, are we assuming that $E$ is a subset of $\Bbb R$?

 

[joypav]

Please state the definition you're using for l.s.c. functions. Also, are we assuming that $E$ is a subset of $\Bbb R$?

Sorry! Yes, E is a subset of R.

The definition we were given is what I used in the proof:
A function f is l.s.c. on E if $\forall x_0 \in E$,
$\forall \epsilon > 0, \exists \delta >0$ such that if $\left|x-x_0 \right|<\delta$, then $f(x_0)-\epsilon \leq f(x)$

 

[Euge]

You do not seem to have the correct formulation of the inferior limit $\liminf\limits_{x\to x_0} f(x)$. If $x_0$ is a limit point of $E$, $\liminf\limits_{x\to x_0} f(x)$ is defined as $$\sup\limits_{\delta > 0}\, \inf\limits_{x\in E,\, 0 < |x - x_0|< \delta} f(x)$$ which is the same as $$\lim\limits_{\delta \to 0}\, \inf\limits_{x\in E,\, 0 < |x - x_0| < \delta} f(x)$$ Now suppose $f$ is l.s.c. at $x_0$. If $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x$, $|x - x_0| < \delta$ implies $f(x_0) - \epsilon \le f(x)$. So for $0 < \eta < \delta$, $f(x_0) - \epsilon$ is a lower bound for the of values $f(x)$ where $x\in E$ and $0 <|x - x_0| < \eta$. Hence $f(x_0) - \epsilon \le \inf\{f(x):x\in E,\, 0 < |x - x_0| < \eta\}$ for all $0 < \eta < \delta$. Therefore $f(x_0) - \epsilon \le \liminf\limits_{x\to x_0} f(x)$. As $\epsilon$ was arbitrary, $f(x_0) \le \liminf\limits_{x\to x_0} f(x)$.

Conversely, suppose $f(x_0) \le \liminf\limits_{x\to x_0} f(x)$. Set $L = \liminf\limits_{x\to x_0} f(x)$. Fix $\epsilon > 0$, and choose a positive number $\delta$ such that for all $x\in E$, $0 < |x - x_0| < \delta$ implies $f(x) > L - \epsilon$. By assumption $f(x_0) \ge L$, so $f(x) > f(x_0) - \epsilon$ whenever $|x - x_0| < \delta$. Hence, $f$ is l.s.c. at $x_0$.

 

[joypav]

Thank you!
Here is what I have for part b.

"$\leftarrow$"
Assume $\phi(x_i)$ is less than or equal to the smaller of the two values assumed in $(x_{i-1},x_i),(x_i,x_{i+1})$.

Case I: $x_0 \in (x_{i-1},x_i)$, some i
Then, choose $\delta = \frac{x_i-x_0}{2}$
If $\left| x-x_0 \right| < \delta = \frac{x_i-x_0}{2} \implies x \in (x_{i-1},x_i) \implies \phi(x)=\phi(x_0)$

Then, $\forall \epsilon > 0, \phi(x_0) - \epsilon \leq \phi(x_0) \implies \forall \epsilon > 0$ and for $\delta = \frac{x_i-x_0}{2}, \phi(x_0) - \epsilon \leq \phi(x)$
$\implies \phi$ is lower semi-continuousCase II: $x_0$ is an endpoint of an interval, WLOG say $x_0 = x_i$, some i
for $\forall x \in (x_{i-1}, x_i), \phi(x) = a$
for $\forall x \in (x_i, x_{i+1}), \phi(x) = b$
WLOG, assume a<b. Then, by assumption, $\phi(x_i) \leq a$.

Then, for $\delta = x_i - x_{i-1}$,
$\forall \epsilon>0$, if $\left| x-x_0 \right| = \left| x-x_i \right| < \delta \implies x \in (x_{i-1}, x_i)$ or $x \in (x_i, x_{i+1})$

if $x \in (x_{i-1}, x_i)$, then
$\phi(x_0) - \epsilon = \phi(x_i) - \epsilon \leq a - \epsilon < a = \phi(x) \implies \phi(x_0) - \epsilon \leq \phi(x)$

if $x \in (x_i, x_{i+1})$, then
$\phi(x_0) - \epsilon = \phi(x_i) - \epsilon \leq a - \epsilon < a < b = \phi(x) \implies \phi(x_0) - \epsilon \leq \phi(x)$

$\implies \phi$ is lower semi-continuous"$\rightarrow$"
Assume $\phi$ is lower semi-continuous
$\implies$ for every $x_0 \in [a, b], \forall \epsilon>0, \exists \delta>0, \left| x-x_0 \right|<\delta \implies \phi(x_0) - \epsilon \leq \phi(x)$
By way of contradiction, assume $\phi(x_i)$ is greater than the smaller of the two values assumed in $(x_{i-1},x_i),(x_i,x_{i+1})$.

for $\forall x \in (x_{i-1}, x_i), \phi(x) = a$
for $\forall x \in (x_i, x_{i+1}), \phi(x) = b$
WLOG, assume a<b. Then, by assumption, $\phi(x_i) > a$.

Choose $x_0=x_i$.
Then, if $x<x_0$,
$\phi(x_0) - \epsilon = \phi(x_i) - \epsilon > a - \epsilon = \phi(x) - \epsilon$
$\implies \phi(x_0) - \epsilon > \phi(x)$, a contradiction

$\implies \phi(x_i) \leq a$