Lower semicontinuous function
let $E$ be a topological space and $\phi:E\to\mathbb{R}\cup\{\infty\}$. then the following are equivalent:
1) $\liminf_{x\to x_0}\phi(x)\geq\phi(x_0)$
2) $epi(\phi):=\{(x,\lambda)\in E\times\mathbb{R}:\phi(x)\leq\lambda\}$ is closed in $E\times\mathbb{R}$
3) $C_{\lambda}:=\{x\in E:\phi(x)\leq\lambda\}$ is closed in $E$ $\forall\lambda\in\mathbb{R}$
have fun!
1) $\liminf_{x\to x_0}\phi(x)\geq\phi(x_0)$
2) $epi(\phi):=\{(x,\lambda)\in E\times\mathbb{R}:\phi(x)\leq\lambda\}$ is closed in $E\times\mathbb{R}$
3) $C_{\lambda}:=\{x\in E:\phi(x)\leq\lambda\}$ is closed in $E$ $\forall\lambda\in\mathbb{R}$
have fun!
Risposte
i'm still here...
all i wanted to say is that i wrote $epi(\phi)$ because "epi" stands for "epigraph"...
however your proof seem to be correct...
all i wanted to say is that i wrote $epi(\phi)$ because "epi" stands for "epigraph"...
however your proof seem to be correct...
If someone is still listening... I write my proof of implication $2\Rightarrow 1$:
We'll call $\omega:=(E\times \RR) -\pi(\phi)$: this is open by hypotesis $2$.
Let be $x\in E$ and $\epsilon > 0$.
We define $\lambda :=\phi(x)-\epsilon$: obviously $(x,\lambda)\in\omega$.
Since $\omega$ is open, there is a set of the base containing $(x,\lambda)$, and contained in $\omega$.
Such set will be $A\times B$, with $A$ open in $E$ containing $x$, and $B$ open in $\RR$ containing $\lambda$.
But since $A\times B\subseteq\omega$, we have $\phi(x) > \lambda$ for every $y\in A$.
We'll call $\omega:=(E\times \RR) -\pi(\phi)$: this is open by hypotesis $2$.
Let be $x\in E$ and $\epsilon > 0$.
We define $\lambda :=\phi(x)-\epsilon$: obviously $(x,\lambda)\in\omega$.
Since $\omega$ is open, there is a set of the base containing $(x,\lambda)$, and contained in $\omega$.
Such set will be $A\times B$, with $A$ open in $E$ containing $x$, and $B$ open in $\RR$ containing $\lambda$.
But since $A\times B\subseteq\omega$, we have $\phi(x) > \lambda$ for every $y\in A$.
For the next days I will be away from home, but I will continue thinking about this interesting problem, and I hope the next week I will have an answer...
For the implication $3\Rightarrow 2$, I was thinking of something like this:
$(E\times \RR) -\pi(\phi)=\bigcup_{\lambda\in\RR}(E-C_{\lambda})\times (-\infty,\lambda)$
Hope you forgive me if I made any mistakes, but I'm quite new to these things...
For the implication $3\Rightarrow 2$, I was thinking of something like this:
$(E\times \RR) -\pi(\phi)=\bigcup_{\lambda\in\RR}(E-C_{\lambda})\times (-\infty,\lambda)$
Hope you forgive me if I made any mistakes, but I'm quite new to these things...
(I think) I proved the implication $1\Rightarrow 3$:
Let be $\lambda\in \RR$: let's show that $E-C_{\lambda}$ is open.
If $E-C_{\lambda}$ is empty there is no problem.
In the other case, we have $x\in E$ such that $\phi(x)>\lambda$.
We define $\epsilon :=\phi(x)-\lambda$.
So $\epsilon>0$, and by hypotesis, there is an open set $G_x$ containing $x$ such that
for every $y\in G_x$, we have $\phi(y)>\phi(x)-\epsilon=\lambda$
$\Rightarrow G_x\subseteq E-C_{\lambda}$.
So we have that $E-C_{\lambda}=\bigcup_{x\in E-C_{\lambda}}G_x$
$\Rightarrow E-C_{\lambda}$ is open (it's the union of open sets).
Let be $\lambda\in \RR$: let's show that $E-C_{\lambda}$ is open.
If $E-C_{\lambda}$ is empty there is no problem.
In the other case, we have $x\in E$ such that $\phi(x)>\lambda$.
We define $\epsilon :=\phi(x)-\lambda$.
So $\epsilon>0$, and by hypotesis, there is an open set $G_x$ containing $x$ such that
for every $y\in G_x$, we have $\phi(y)>\phi(x)-\epsilon=\lambda$
$\Rightarrow G_x\subseteq E-C_{\lambda}$.
So we have that $E-C_{\lambda}=\bigcup_{x\in E-C_{\lambda}}G_x$
$\Rightarrow E-C_{\lambda}$ is open (it's the union of open sets).
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