exchange max and limit
$begingroup$
Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?
I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?
Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.
Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.
real-analysis
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
$endgroup$
add a comment |
$begingroup$
Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?
I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?
Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.
Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.
real-analysis
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
$endgroup$
$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32
$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33
1
$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38
add a comment |
$begingroup$
Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?
I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?
Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.
Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.
real-analysis
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
$endgroup$
Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?
I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?
Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.
Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.
real-analysis
real-analysis
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
asked Mar 27 '15 at 12:30
simonzacksimonzack
728619
728619
$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32
$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33
1
$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38
add a comment |
$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32
$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33
1
$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38
$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32
$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32
$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33
$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33
1
1
$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38
$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For posterity: On the set of real numbers, $max$ and $min$ are given by
$$
max(a, b) = frac{a + b + |b - a|}{2},qquad
min(a, b) = frac{a + b - |b - a|}{2}.
$$
(Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")
These functions are (almost obviously) continuous as functions of two variables, so
$$
lim_{n to infty} max(a_{n}, b_{n})
= maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
= max(a, b),
$$
and similarly for $min$.
In fact, continuity guarantees stronger "double limit" assertions:
$$
max(a, b)
= lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
= lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
$$
etc.
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
$endgroup$
add a comment |
$begingroup$
A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.
To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For posterity: On the set of real numbers, $max$ and $min$ are given by
$$
max(a, b) = frac{a + b + |b - a|}{2},qquad
min(a, b) = frac{a + b - |b - a|}{2}.
$$
(Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")
These functions are (almost obviously) continuous as functions of two variables, so
$$
lim_{n to infty} max(a_{n}, b_{n})
= maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
= max(a, b),
$$
and similarly for $min$.
In fact, continuity guarantees stronger "double limit" assertions:
$$
max(a, b)
= lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
= lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
$$
etc.
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
$endgroup$
add a comment |
$begingroup$
For posterity: On the set of real numbers, $max$ and $min$ are given by
$$
max(a, b) = frac{a + b + |b - a|}{2},qquad
min(a, b) = frac{a + b - |b - a|}{2}.
$$
(Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")
These functions are (almost obviously) continuous as functions of two variables, so
$$
lim_{n to infty} max(a_{n}, b_{n})
= maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
= max(a, b),
$$
and similarly for $min$.
In fact, continuity guarantees stronger "double limit" assertions:
$$
max(a, b)
= lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
= lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
$$
etc.
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
$endgroup$
add a comment |
$begingroup$
For posterity: On the set of real numbers, $max$ and $min$ are given by
$$
max(a, b) = frac{a + b + |b - a|}{2},qquad
min(a, b) = frac{a + b - |b - a|}{2}.
$$
(Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")
These functions are (almost obviously) continuous as functions of two variables, so
$$
lim_{n to infty} max(a_{n}, b_{n})
= maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
= max(a, b),
$$
and similarly for $min$.
In fact, continuity guarantees stronger "double limit" assertions:
$$
max(a, b)
= lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
= lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
$$
etc.
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
$endgroup$
For posterity: On the set of real numbers, $max$ and $min$ are given by
$$
max(a, b) = frac{a + b + |b - a|}{2},qquad
min(a, b) = frac{a + b - |b - a|}{2}.
$$
(Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")
These functions are (almost obviously) continuous as functions of two variables, so
$$
lim_{n to infty} max(a_{n}, b_{n})
= maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
= max(a, b),
$$
and similarly for $min$.
In fact, continuity guarantees stronger "double limit" assertions:
$$
max(a, b)
= lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
= lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
$$
etc.
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
answered Mar 27 '15 at 14:16
Andrew D. HwangAndrew D. Hwang
52.8k447113
52.8k447113
add a comment |
add a comment |
$begingroup$
A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.
To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
$begingroup$
A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.
To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
$begingroup$
A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.
To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.
To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
answered Dec 10 '18 at 21:30
Paolo LeonettiPaolo Leonetti
11.5k21550
11.5k21550
add a comment |
add a comment |
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$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32
$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33
1
$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38