Approximation of a lower semi continuous function by monotone sequence of continuous functions
$begingroup$
Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
$$ c: X times Y to [0, +infty]$$
a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
$$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
Then $c_k$ are continuous functions, monotonically increasing converging to $c$.
Proof:
Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
$$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
$$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
and then
$$ c_{k}(x,y) le c_{k+1}(x,y) $$Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
$$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
To do that I find $k_{epsilon} in mathbb{N}$ s.t.
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
for any $(x',y') in X times Y$.
Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
$$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$
Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$
If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
$$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
I obtain
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
QED
I think this proof is too long and involved.
So
is this proof correct? Does there exist a shorter proof?
real-analysis proof-verification convergence
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
$endgroup$
add a comment |
$begingroup$
Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
$$ c: X times Y to [0, +infty]$$
a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
$$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
Then $c_k$ are continuous functions, monotonically increasing converging to $c$.
Proof:
Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
$$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
$$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
and then
$$ c_{k}(x,y) le c_{k+1}(x,y) $$Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
$$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
To do that I find $k_{epsilon} in mathbb{N}$ s.t.
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
for any $(x',y') in X times Y$.
Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
$$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$
Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$
If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
$$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
I obtain
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
QED
I think this proof is too long and involved.
So
is this proof correct? Does there exist a shorter proof?
real-analysis proof-verification convergence
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
$endgroup$
add a comment |
$begingroup$
Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
$$ c: X times Y to [0, +infty]$$
a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
$$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
Then $c_k$ are continuous functions, monotonically increasing converging to $c$.
Proof:
Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
$$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
$$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
and then
$$ c_{k}(x,y) le c_{k+1}(x,y) $$Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
$$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
To do that I find $k_{epsilon} in mathbb{N}$ s.t.
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
for any $(x',y') in X times Y$.
Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
$$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$
Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$
If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
$$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
I obtain
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
QED
I think this proof is too long and involved.
So
is this proof correct? Does there exist a shorter proof?
real-analysis proof-verification convergence
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
$endgroup$
Let $ (X, d_X)$ and $ (Y, d_Y) $ be metric spaces and let
$$ c: X times Y to [0, +infty]$$
a lower semicontinuous function. Then define for each $k in mathbb{N}$ the functions
$$ c_k (x,y): = inf_{x' in X, y' in Y} { c(x',y') wedge k + kd_X(x,x')+kd_Y(y,y') } $$
Then $c_k$ are continuous functions, monotonically increasing converging to $c$.
Proof:
Continuity: take $(x,y) in X times Y$ and a sequence ${ x_n,y_n }_{n ge 1} $ converging to $(x,y)$. Then
$$ | c_k(x,y)-c_k(x_n,y_n) | le sup_{x' in X, y' in Y} | kd_X(x,x')+kd_Y(y,y') -kd_X(x_n,x')-kd_Y(y_n,y') | le sup_{x' in X, y' in Y} k(d_X(x_n, x)|+d_Y(y_n,y)) to 0$$Monotonically increasing: fix $(x,y) in X times Y$ and consider any $(x',y') in X times Y$, then
$$ c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y') le c(x',y') wedge (k+1) + (k+1)d_X(x',x)+ (k+1)d_Y(y,y') $$
and then
$$ c_{k}(x,y) le c_{k+1}(x,y) $$Convergence: First I observe that if $ (x',y') = (x,y)$ and $c(x,y) = + infty$ I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k = k$$
Then $c_k(x,y) ge k$ and then $c_k to + infty$ when $k to + infty$.
From now on I only consider the case in which $c(x,y) < + infty$. Then if $ (x',y') = (x,y)$, I have
$$c(x',y') wedge k + kd_X(x',x)+ kd_Y(y,y')= c(x,y) wedge k le c(x,y) $$
which implies $c_k(x,y) le c(x,y)$. Then it is enough to prove that $c(x,y)$ is the supremum of the sequence $c_k(x,y)$. Fix $0< epsilon < c(x,y)/2$.
I want to prove that there exists a $k_{epsilon} in mathbb{N}$ s.t.
$$c_{k_{epsilon}}(x,y) > c(x,y) - 2epsilon $$
To do that I find $k_{epsilon} in mathbb{N}$ s.t.
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) -epsilon $$
for any $(x',y') in X times Y$.
Since $c$ is l.s.c. there exists a $delta= delta(epsilon)>0$ s.t.
$$ (x',y') in B_{delta}((x,y)) Rightarrow c(x',y') > c(x,y) - epsilon$$
Then if $(x',y') in B_{delta}((x,y))$ I take $k_{epsilon} > c(x,y)- epsilon$ and then
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > c(x,y) - epsilon + 0 $$
If $(x',y') notin B_{delta}((x,y))$ then $d_X(x',x) +d(y',y) > c(epsilon) >0$ and then, taking
$$k_{epsilon} > frac{c(x,y)-epsilon}{c(epsilon)}$$
I obtain
$$ c(x',y') wedge k_{epsilon} + k_{epsilon}d_X(x',x)+ k_{epsilon}d_Y(y,y') > 0 + c(x,y) - epsilon$$
QED
I think this proof is too long and involved.
So
is this proof correct? Does there exist a shorter proof?
real-analysis proof-verification convergence
real-analysis proof-verification convergence
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
edited Dec 3 '18 at 18:10
Bremen000
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
asked Dec 1 '18 at 23:44
Bremen000Bremen000
424210
424210
add a comment |
add a comment |
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