Mixing
Show that lower semi-continuous function attains it's minimum. (Proof verification) (By contradiction)
$begingroup$
Let $f: [0,1]to mathbb{R}$ be a lower semi-continuous function, then
$$ liminf_{xto a} f(x) geq f(a), forall a in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$exists x_0 in [0,1]$ such that $f(x_0) le f(x)$, $forall x in [0,1]$.
My proof:
Assume that $f$ is lower semi-continuous, but $f$ doesn't attain it's minimum on $[0,1]$,
Since $f$ is lower semicontinuous at $x_0$,
$$forall epsilon > 0, exists delta > 0 mbox{ such that } |x - x_0| < delta, Rightarrow f(x) > f(x_0) - epsilon, forall x in (x_0 - delta,x_0 + delta)$$
Now since $f$ doesn't attain it's minimum on $[0,1]$, then
$$forall x_0 in [0,1], exists x in [0,1] mbox{ s.t } f(x_0) > f(x)$$
Let $epsilon = f(x_0) - f(x) > 0, exists delta > 0$, such that $|x-x_0| < delta$ and $f(x) > f(x_0) - epsilon Rightarrow f(x_0) - f(x) > epsilon = f(x_0) - f(x)$ which is a contradiction.
I am right? Thanks.
real-analysis proof-verification upper-lower-bounds semicontinuous-functions
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
$endgroup$
|
show 2 more comments
$begingroup$
Let $f: [0,1]to mathbb{R}$ be a lower semi-continuous function, then
$$ liminf_{xto a} f(x) geq f(a), forall a in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$exists x_0 in [0,1]$ such that $f(x_0) le f(x)$, $forall x in [0,1]$.
My proof:
Assume that $f$ is lower semi-continuous, but $f$ doesn't attain it's minimum on $[0,1]$,
Since $f$ is lower semicontinuous at $x_0$,
$$forall epsilon > 0, exists delta > 0 mbox{ such that } |x - x_0| < delta, Rightarrow f(x) > f(x_0) - epsilon, forall x in (x_0 - delta,x_0 + delta)$$
Now since $f$ doesn't attain it's minimum on $[0,1]$, then
$$forall x_0 in [0,1], exists x in [0,1] mbox{ s.t } f(x_0) > f(x)$$
Let $epsilon = f(x_0) - f(x) > 0, exists delta > 0$, such that $|x-x_0| < delta$ and $f(x) > f(x_0) - epsilon Rightarrow f(x_0) - f(x) > epsilon = f(x_0) - f(x)$ which is a contradiction.
I am right? Thanks.
real-analysis proof-verification upper-lower-bounds semicontinuous-functions
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
$endgroup$
1
$begingroup$
It's wrong. You should have said "$forallepsilon>0forall x_0in [0,1] $... such that $|x-x_0|<delta implies f(x)>f(x_0)color{red}- epsilon$".
$endgroup$
– Song
Jan 21 at 17:04
1
$begingroup$
It is $-epsilon$ in the definition, and the $forall x$ has to be after the “such that”.
$endgroup$
– Mindlack
Jan 21 at 17:04
$begingroup$
Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now.
$endgroup$
– Richard Clare
Jan 21 at 17:06
1
$begingroup$
No. The $delta$ is allowed to depend on $epsilon$ and $x_0$, not $x$.
$endgroup$
– Mindlack
Jan 21 at 17:16
1
$begingroup$
An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here.
$endgroup$
– Daniel
Jan 21 at 17:34
|
show 2 more comments
$begingroup$
Let $f: [0,1]to mathbb{R}$ be a lower semi-continuous function, then
$$ liminf_{xto a} f(x) geq f(a), forall a in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$exists x_0 in [0,1]$ such that $f(x_0) le f(x)$, $forall x in [0,1]$.
My proof:
Assume that $f$ is lower semi-continuous, but $f$ doesn't attain it's minimum on $[0,1]$,
Since $f$ is lower semicontinuous at $x_0$,
$$forall epsilon > 0, exists delta > 0 mbox{ such that } |x - x_0| < delta, Rightarrow f(x) > f(x_0) - epsilon, forall x in (x_0 - delta,x_0 + delta)$$
Now since $f$ doesn't attain it's minimum on $[0,1]$, then
$$forall x_0 in [0,1], exists x in [0,1] mbox{ s.t } f(x_0) > f(x)$$
Let $epsilon = f(x_0) - f(x) > 0, exists delta > 0$, such that $|x-x_0| < delta$ and $f(x) > f(x_0) - epsilon Rightarrow f(x_0) - f(x) > epsilon = f(x_0) - f(x)$ which is a contradiction.
I am right? Thanks.
real-analysis proof-verification upper-lower-bounds semicontinuous-functions
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
$endgroup$
Let $f: [0,1]to mathbb{R}$ be a lower semi-continuous function, then
$$ liminf_{xto a} f(x) geq f(a), forall a in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$exists x_0 in [0,1]$ such that $f(x_0) le f(x)$, $forall x in [0,1]$.
My proof:
Assume that $f$ is lower semi-continuous, but $f$ doesn't attain it's minimum on $[0,1]$,
Since $f$ is lower semicontinuous at $x_0$,
$$forall epsilon > 0, exists delta > 0 mbox{ such that } |x - x_0| < delta, Rightarrow f(x) > f(x_0) - epsilon, forall x in (x_0 - delta,x_0 + delta)$$
Now since $f$ doesn't attain it's minimum on $[0,1]$, then
$$forall x_0 in [0,1], exists x in [0,1] mbox{ s.t } f(x_0) > f(x)$$
Let $epsilon = f(x_0) - f(x) > 0, exists delta > 0$, such that $|x-x_0| < delta$ and $f(x) > f(x_0) - epsilon Rightarrow f(x_0) - f(x) > epsilon = f(x_0) - f(x)$ which is a contradiction.
I am right? Thanks.
real-analysis proof-verification upper-lower-bounds semicontinuous-functions
real-analysis proof-verification upper-lower-bounds semicontinuous-functions
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
edited Jan 21 at 17:11
Richard Clare
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
asked Jan 21 at 16:58
Richard ClareRichard Clare
1,076314
1,076314
1
$begingroup$
It's wrong. You should have said "$forallepsilon>0forall x_0in [0,1] $... such that $|x-x_0|<delta implies f(x)>f(x_0)color{red}- epsilon$".
$endgroup$
– Song
Jan 21 at 17:04
1
$begingroup$
It is $-epsilon$ in the definition, and the $forall x$ has to be after the “such that”.
$endgroup$
– Mindlack
Jan 21 at 17:04
$begingroup$
Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now.
$endgroup$
– Richard Clare
Jan 21 at 17:06
1
$begingroup$
No. The $delta$ is allowed to depend on $epsilon$ and $x_0$, not $x$.
$endgroup$
– Mindlack
Jan 21 at 17:16
1
$begingroup$
An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here.
$endgroup$
– Daniel
Jan 21 at 17:34
|
show 2 more comments
1
$begingroup$
It's wrong. You should have said "$forallepsilon>0forall x_0in [0,1] $... such that $|x-x_0|<delta implies f(x)>f(x_0)color{red}- epsilon$".
$endgroup$
– Song
Jan 21 at 17:04
1
$begingroup$
It is $-epsilon$ in the definition, and the $forall x$ has to be after the “such that”.
$endgroup$
– Mindlack
Jan 21 at 17:04
$begingroup$
Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now.
$endgroup$
– Richard Clare
Jan 21 at 17:06
1
$begingroup$
No. The $delta$ is allowed to depend on $epsilon$ and $x_0$, not $x$.
$endgroup$
– Mindlack
Jan 21 at 17:16
1
$begingroup$
An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here.
$endgroup$
– Daniel
Jan 21 at 17:34
1
1
$begingroup$
It's wrong. You should have said "$forallepsilon>0forall x_0in [0,1] $... such that $|x-x_0|<delta implies f(x)>f(x_0)color{red}- epsilon$".
$endgroup$
– Song
Jan 21 at 17:04
$begingroup$
It's wrong. You should have said "$forallepsilon>0forall x_0in [0,1] $... such that $|x-x_0|<delta implies f(x)>f(x_0)color{red}- epsilon$".
$endgroup$
– Song
Jan 21 at 17:04
1
1
$begingroup$
It is $-epsilon$ in the definition, and the $forall x$ has to be after the “such that”.
$endgroup$
– Mindlack
Jan 21 at 17:04
$begingroup$
It is $-epsilon$ in the definition, and the $forall x$ has to be after the “such that”.
$endgroup$
– Mindlack
Jan 21 at 17:04
$begingroup$
Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now.
$endgroup$
– Richard Clare
Jan 21 at 17:06
$begingroup$
Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now.
$endgroup$
– Richard Clare
Jan 21 at 17:06
1
1
$begingroup$
No. The $delta$ is allowed to depend on $epsilon$ and $x_0$, not $x$.
$endgroup$
– Mindlack
Jan 21 at 17:16
$begingroup$
No. The $delta$ is allowed to depend on $epsilon$ and $x_0$, not $x$.
$endgroup$
– Mindlack
Jan 21 at 17:16
1
1
$begingroup$
An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here.
$endgroup$
– Daniel
Jan 21 at 17:34
$begingroup$
An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here.
$endgroup$
– Daniel
Jan 21 at 17:34
|
show 2 more comments
2 Answers
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$begingroup$
Since $f$ is lower-semicontinuous, for each $xin [0,1],$ there is an open interval $I_xsubseteq [0,1]$ such that $inf{f(y):yin I_x}ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.
So, letting $y=inf{f(x):xin [0,1]}$, we can find a sequence $(x_n)subseteq [0,1]$ such that $f(x_n)to y.$ And of course, there is a subsequence $(x_{n_k})subseteq (x_n)$ such that $x_{n_k}to x_0in [0,1]$.
Then, $f(x_0)le liminf_{xto x_0} f(x)le liminf_{kto infty}f(x_{n_k})=lim_{kto infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
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Another characterization of lower semi-continuous function that might be helpful:
A function $f:[0,1] to Bbb R$ is lower semi-continuous if and only if its upper level set ${xin [0,1]:f(x)>c}$ is open in $[0,1]$ for all $cinBbb R$.
(Indeed above characterization can be applied to every topological space $X$ and $f:XtoBbb R$. And the proof below can be generalized to any compact topological space.)
Proof. Assume $x_0in X$ is such that $f(x_0)>c$. If we choose $0<epsilon< f(x_0)-c$, we can find $delta>0$ such that $$xin [0,1], xin (x_0-delta,x_0+delta)implies f(x)>f(x_0)-epsilon.$$ Since $f(x_0)-epsilon>c$, it follows
$$
[0,1]cap (x_0-delta,x_0+delta)subset {x : f(x)>c}.
$$ This proves ${x:f(x)>c}$ is open in $[0,1]$. $blacksquare$
By the above proposition, $F_c={x:f(x)le c}$ is closed in $[0,1]$ for every $cinBbb R$, hence is compact. If there is no minimum, then $F_c$ is not empty for every $c$. Consider the decreasing sequence of closed sets$$F_{-1}supset F_{-2}supset cdotssupset F_{-n}supsetcdots.$$ Since
$$
bigcap_{ninBbb N}F_{-n} = varnothing,
$$ by the finite intersection property of the compact set $[0,1]$, there exists $n_0inBbb N$ such that $F_{-n_0}=varnothing$. This implies $f(x)ge -n_0$ for all $xin [0,1]$, hence $f$ is lower bounded.
Let us define $S={cinBbb R: F_cne varnothing}$. Since $-n_0notin S$ and $-n_0le S$, we know $S$ is lower bounded. Let $m=inf S$. Then by finite intersection property,
$$
F_m=bigcap_{kinBbb N}F_{m+1/k}ne varnothing.
$$ Thus there exists $x_0$ such that $f(x_0)=m$. Finally, for all $c<m$, it holds $cnotin S$, i.e. $F_c=varnothing.$ This establishes $m$ is the minimum of $f$.
answered Jan 21 at 18:01
SongSong
16.7k11044
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Since $f$ is lower-semicontinuous, for each $xin [0,1],$ there is an open interval $I_xsubseteq [0,1]$ such that $inf{f(y):yin I_x}ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.
So, letting $y=inf{f(x):xin [0,1]}$, we can find a sequence $(x_n)subseteq [0,1]$ such that $f(x_n)to y.$ And of course, there is a subsequence $(x_{n_k})subseteq (x_n)$ such that $x_{n_k}to x_0in [0,1]$.
Then, $f(x_0)le liminf_{xto x_0} f(x)le liminf_{kto infty}f(x_{n_k})=lim_{kto infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
Since $f$ is lower-semicontinuous, for each $xin [0,1],$ there is an open interval $I_xsubseteq [0,1]$ such that $inf{f(y):yin I_x}ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.
So, letting $y=inf{f(x):xin [0,1]}$, we can find a sequence $(x_n)subseteq [0,1]$ such that $f(x_n)to y.$ And of course, there is a subsequence $(x_{n_k})subseteq (x_n)$ such that $x_{n_k}to x_0in [0,1]$.
Then, $f(x_0)le liminf_{xto x_0} f(x)le liminf_{kto infty}f(x_{n_k})=lim_{kto infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
Since $f$ is lower-semicontinuous, for each $xin [0,1],$ there is an open interval $I_xsubseteq [0,1]$ such that $inf{f(y):yin I_x}ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.
So, letting $y=inf{f(x):xin [0,1]}$, we can find a sequence $(x_n)subseteq [0,1]$ such that $f(x_n)to y.$ And of course, there is a subsequence $(x_{n_k})subseteq (x_n)$ such that $x_{n_k}to x_0in [0,1]$.
Then, $f(x_0)le liminf_{xto x_0} f(x)le liminf_{kto infty}f(x_{n_k})=lim_{kto infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
$endgroup$
Since $f$ is lower-semicontinuous, for each $xin [0,1],$ there is an open interval $I_xsubseteq [0,1]$ such that $inf{f(y):yin I_x}ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.
So, letting $y=inf{f(x):xin [0,1]}$, we can find a sequence $(x_n)subseteq [0,1]$ such that $f(x_n)to y.$ And of course, there is a subsequence $(x_{n_k})subseteq (x_n)$ such that $x_{n_k}to x_0in [0,1]$.
Then, $f(x_0)le liminf_{xto x_0} f(x)le liminf_{kto infty}f(x_{n_k})=lim_{kto infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
answered Jan 21 at 22:47
MatematletaMatematleta
11.5k2920
11.5k2920
add a comment |
add a comment |
$begingroup$
Another characterization of lower semi-continuous function that might be helpful:
A function $f:[0,1] to Bbb R$ is lower semi-continuous if and only if its upper level set ${xin [0,1]:f(x)>c}$ is open in $[0,1]$ for all $cinBbb R$.
(Indeed above characterization can be applied to every topological space $X$ and $f:XtoBbb R$. And the proof below can be generalized to any compact topological space.)
Proof. Assume $x_0in X$ is such that $f(x_0)>c$. If we choose $0<epsilon< f(x_0)-c$, we can find $delta>0$ such that $$xin [0,1], xin (x_0-delta,x_0+delta)implies f(x)>f(x_0)-epsilon.$$ Since $f(x_0)-epsilon>c$, it follows
$$
[0,1]cap (x_0-delta,x_0+delta)subset {x : f(x)>c}.
$$ This proves ${x:f(x)>c}$ is open in $[0,1]$. $blacksquare$
By the above proposition, $F_c={x:f(x)le c}$ is closed in $[0,1]$ for every $cinBbb R$, hence is compact. If there is no minimum, then $F_c$ is not empty for every $c$. Consider the decreasing sequence of closed sets$$F_{-1}supset F_{-2}supset cdotssupset F_{-n}supsetcdots.$$ Since
$$
bigcap_{ninBbb N}F_{-n} = varnothing,
$$ by the finite intersection property of the compact set $[0,1]$, there exists $n_0inBbb N$ such that $F_{-n_0}=varnothing$. This implies $f(x)ge -n_0$ for all $xin [0,1]$, hence $f$ is lower bounded.
Let us define $S={cinBbb R: F_cne varnothing}$. Since $-n_0notin S$ and $-n_0le S$, we know $S$ is lower bounded. Let $m=inf S$. Then by finite intersection property,
$$
F_m=bigcap_{kinBbb N}F_{m+1/k}ne varnothing.
$$ Thus there exists $x_0$ such that $f(x_0)=m$. Finally, for all $c<m$, it holds $cnotin S$, i.e. $F_c=varnothing.$ This establishes $m$ is the minimum of $f$.
answered Jan 21 at 18:01
SongSong
16.7k11044
$endgroup$
add a comment |
$begingroup$
Another characterization of lower semi-continuous function that might be helpful:
A function $f:[0,1] to Bbb R$ is lower semi-continuous if and only if its upper level set ${xin [0,1]:f(x)>c}$ is open in $[0,1]$ for all $cinBbb R$.
(Indeed above characterization can be applied to every topological space $X$ and $f:XtoBbb R$. And the proof below can be generalized to any compact topological space.)
Proof. Assume $x_0in X$ is such that $f(x_0)>c$. If we choose $0<epsilon< f(x_0)-c$, we can find $delta>0$ such that $$xin [0,1], xin (x_0-delta,x_0+delta)implies f(x)>f(x_0)-epsilon.$$ Since $f(x_0)-epsilon>c$, it follows
$$
[0,1]cap (x_0-delta,x_0+delta)subset {x : f(x)>c}.
$$ This proves ${x:f(x)>c}$ is open in $[0,1]$. $blacksquare$
By the above proposition, $F_c={x:f(x)le c}$ is closed in $[0,1]$ for every $cinBbb R$, hence is compact. If there is no minimum, then $F_c$ is not empty for every $c$. Consider the decreasing sequence of closed sets$$F_{-1}supset F_{-2}supset cdotssupset F_{-n}supsetcdots.$$ Since
$$
bigcap_{ninBbb N}F_{-n} = varnothing,
$$ by the finite intersection property of the compact set $[0,1]$, there exists $n_0inBbb N$ such that $F_{-n_0}=varnothing$. This implies $f(x)ge -n_0$ for all $xin [0,1]$, hence $f$ is lower bounded.
Let us define $S={cinBbb R: F_cne varnothing}$. Since $-n_0notin S$ and $-n_0le S$, we know $S$ is lower bounded. Let $m=inf S$. Then by finite intersection property,
$$
F_m=bigcap_{kinBbb N}F_{m+1/k}ne varnothing.
$$ Thus there exists $x_0$ such that $f(x_0)=m$. Finally, for all $c<m$, it holds $cnotin S$, i.e. $F_c=varnothing.$ This establishes $m$ is the minimum of $f$.
answered Jan 21 at 18:01
SongSong
16.7k11044
$endgroup$
add a comment |
$begingroup$
Another characterization of lower semi-continuous function that might be helpful:
A function $f:[0,1] to Bbb R$ is lower semi-continuous if and only if its upper level set ${xin [0,1]:f(x)>c}$ is open in $[0,1]$ for all $cinBbb R$.
(Indeed above characterization can be applied to every topological space $X$ and $f:XtoBbb R$. And the proof below can be generalized to any compact topological space.)
Proof. Assume $x_0in X$ is such that $f(x_0)>c$. If we choose $0<epsilon< f(x_0)-c$, we can find $delta>0$ such that $$xin [0,1], xin (x_0-delta,x_0+delta)implies f(x)>f(x_0)-epsilon.$$ Since $f(x_0)-epsilon>c$, it follows
$$
[0,1]cap (x_0-delta,x_0+delta)subset {x : f(x)>c}.
$$ This proves ${x:f(x)>c}$ is open in $[0,1]$. $blacksquare$
By the above proposition, $F_c={x:f(x)le c}$ is closed in $[0,1]$ for every $cinBbb R$, hence is compact. If there is no minimum, then $F_c$ is not empty for every $c$. Consider the decreasing sequence of closed sets$$F_{-1}supset F_{-2}supset cdotssupset F_{-n}supsetcdots.$$ Since
$$
bigcap_{ninBbb N}F_{-n} = varnothing,
$$ by the finite intersection property of the compact set $[0,1]$, there exists $n_0inBbb N$ such that $F_{-n_0}=varnothing$. This implies $f(x)ge -n_0$ for all $xin [0,1]$, hence $f$ is lower bounded.
Let us define $S={cinBbb R: F_cne varnothing}$. Since $-n_0notin S$ and $-n_0le S$, we know $S$ is lower bounded. Let $m=inf S$. Then by finite intersection property,
$$
F_m=bigcap_{kinBbb N}F_{m+1/k}ne varnothing.
$$ Thus there exists $x_0$ such that $f(x_0)=m$. Finally, for all $c<m$, it holds $cnotin S$, i.e. $F_c=varnothing.$ This establishes $m$ is the minimum of $f$.
answered Jan 21 at 18:01
SongSong
16.7k11044
$endgroup$
Another characterization of lower semi-continuous function that might be helpful:
A function $f:[0,1] to Bbb R$ is lower semi-continuous if and only if its upper level set ${xin [0,1]:f(x)>c}$ is open in $[0,1]$ for all $cinBbb R$.
(Indeed above characterization can be applied to every topological space $X$ and $f:XtoBbb R$. And the proof below can be generalized to any compact topological space.)
Proof. Assume $x_0in X$ is such that $f(x_0)>c$. If we choose $0<epsilon< f(x_0)-c$, we can find $delta>0$ such that $$xin [0,1], xin (x_0-delta,x_0+delta)implies f(x)>f(x_0)-epsilon.$$ Since $f(x_0)-epsilon>c$, it follows
$$
[0,1]cap (x_0-delta,x_0+delta)subset {x : f(x)>c}.
$$ This proves ${x:f(x)>c}$ is open in $[0,1]$. $blacksquare$
By the above proposition, $F_c={x:f(x)le c}$ is closed in $[0,1]$ for every $cinBbb R$, hence is compact. If there is no minimum, then $F_c$ is not empty for every $c$. Consider the decreasing sequence of closed sets$$F_{-1}supset F_{-2}supset cdotssupset F_{-n}supsetcdots.$$ Since
$$
bigcap_{ninBbb N}F_{-n} = varnothing,
$$ by the finite intersection property of the compact set $[0,1]$, there exists $n_0inBbb N$ such that $F_{-n_0}=varnothing$. This implies $f(x)ge -n_0$ for all $xin [0,1]$, hence $f$ is lower bounded.
Let us define $S={cinBbb R: F_cne varnothing}$. Since $-n_0notin S$ and $-n_0le S$, we know $S$ is lower bounded. Let $m=inf S$. Then by finite intersection property,
$$
F_m=bigcap_{kinBbb N}F_{m+1/k}ne varnothing.
$$ Thus there exists $x_0$ such that $f(x_0)=m$. Finally, for all $c<m$, it holds $cnotin S$, i.e. $F_c=varnothing.$ This establishes $m$ is the minimum of $f$.
answered Jan 21 at 18:01
SongSong
16.7k11044
answered Jan 21 at 18:01
SongSong
16.7k11044
answered Jan 21 at 18:01
SongSong
16.7k11044
answered Jan 21 at 18:01
SongSong
16.7k11044
16.7k11044
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1
$begingroup$
It's wrong. You should have said "$forallepsilon>0forall x_0in [0,1] $... such that $|x-x_0|<delta implies f(x)>f(x_0)color{red}- epsilon$".
$endgroup$
– Song
Jan 21 at 17:04
1
$begingroup$
It is $-epsilon$ in the definition, and the $forall x$ has to be after the “such that”.
$endgroup$
– Mindlack
Jan 21 at 17:04
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Yes, I fixed the problem about the epsilon, it was a typo using the upper semi-continuity. I will see the quantifier problem now.
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– Richard Clare
Jan 21 at 17:06
1
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No. The $delta$ is allowed to depend on $epsilon$ and $x_0$, not $x$.
$endgroup$
– Mindlack
Jan 21 at 17:16
1
$begingroup$
An idea to fix this proof: if $f$ does not attain a minimum, there is a sequence $x_n$ in $[0, 1]$ such that $f(x_n) > f(x_{n+1})$ for every $n$. Now, this sequence must have a convergent subsequence since $[0, 1]$ is compact. Try to reach your contradiction from here.
$endgroup$
– Daniel
Jan 21 at 17:34