Uniform convergence after transforming?
$begingroup$
I face a problem now, it is known that $f_n(x)$ converge to $+infty$ pointwise as $nrightarrow+infty$, but I want to prove $f_n(x)$ converge to $+infty$ uniformly.
I can prove $arctan f_n(x)$ converges to $frac{pi}{2}$ uniformly. Does this imply that $f_n(x)$ converge to $+infty$ uniformly?
real-analysis sequences-and-series uniform-convergence
edited Jan 8 at 13:01
Hayk
2,7671215
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
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add a comment |
$begingroup$
I face a problem now, it is known that $f_n(x)$ converge to $+infty$ pointwise as $nrightarrow+infty$, but I want to prove $f_n(x)$ converge to $+infty$ uniformly.
I can prove $arctan f_n(x)$ converges to $frac{pi}{2}$ uniformly. Does this imply that $f_n(x)$ converge to $+infty$ uniformly?
real-analysis sequences-and-series uniform-convergence
edited Jan 8 at 13:01
Hayk
2,7671215
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
$endgroup$
1
$begingroup$
You need two variables to talk about uniform convergence. As it stands it is difficult to understand what you are saying. And how is probability involved here?
$endgroup$
– Kavi Rama Murthy
Jan 8 at 10:24
$begingroup$
Sorry, I edit my problem again.
$endgroup$
– Xiaol.Song
Jan 8 at 11:54
add a comment |
$begingroup$
I face a problem now, it is known that $f_n(x)$ converge to $+infty$ pointwise as $nrightarrow+infty$, but I want to prove $f_n(x)$ converge to $+infty$ uniformly.
I can prove $arctan f_n(x)$ converges to $frac{pi}{2}$ uniformly. Does this imply that $f_n(x)$ converge to $+infty$ uniformly?
real-analysis sequences-and-series uniform-convergence
edited Jan 8 at 13:01
Hayk
2,7671215
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
$endgroup$
I face a problem now, it is known that $f_n(x)$ converge to $+infty$ pointwise as $nrightarrow+infty$, but I want to prove $f_n(x)$ converge to $+infty$ uniformly.
I can prove $arctan f_n(x)$ converges to $frac{pi}{2}$ uniformly. Does this imply that $f_n(x)$ converge to $+infty$ uniformly?
real-analysis sequences-and-series uniform-convergence
real-analysis sequences-and-series uniform-convergence
edited Jan 8 at 13:01
Hayk
2,7671215
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
edited Jan 8 at 13:01
Hayk
2,7671215
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
edited Jan 8 at 13:01
Hayk
2,7671215
edited Jan 8 at 13:01
Hayk
2,7671215
edited Jan 8 at 13:01
Hayk
2,7671215
2,7671215
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
asked Jan 8 at 10:22
Xiaol.SongXiaol.Song
33
33
1
$begingroup$
You need two variables to talk about uniform convergence. As it stands it is difficult to understand what you are saying. And how is probability involved here?
$endgroup$
– Kavi Rama Murthy
Jan 8 at 10:24
$begingroup$
Sorry, I edit my problem again.
$endgroup$
– Xiaol.Song
Jan 8 at 11:54
add a comment |
1
$begingroup$
You need two variables to talk about uniform convergence. As it stands it is difficult to understand what you are saying. And how is probability involved here?
$endgroup$
– Kavi Rama Murthy
Jan 8 at 10:24
$begingroup$
Sorry, I edit my problem again.
$endgroup$
– Xiaol.Song
Jan 8 at 11:54
1
1
$begingroup$
You need two variables to talk about uniform convergence. As it stands it is difficult to understand what you are saying. And how is probability involved here?
$endgroup$
– Kavi Rama Murthy
Jan 8 at 10:24
$begingroup$
You need two variables to talk about uniform convergence. As it stands it is difficult to understand what you are saying. And how is probability involved here?
$endgroup$
– Kavi Rama Murthy
Jan 8 at 10:24
$begingroup$
Sorry, I edit my problem again.
$endgroup$
– Xiaol.Song
Jan 8 at 11:54
$begingroup$
Sorry, I edit my problem again.
$endgroup$
– Xiaol.Song
Jan 8 at 11:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes it does, because $tan$ is a bijective and increasing mapping from $[0, frac pi 2)$ to $[0, infty) $.
More detailed: Given $M > 0$ we have that $0 < arctan M < frac pi 2$. The uniform convergence of $arctan f_n(x) to frac pi 2$ implies the existence of an $n_0 in Bbb N$ such that
$$
arctan f_n(x) > arctan M text{ for all $n ge n_0$ and all $x$.}
$$
It follows that
$$
f_n(x) > M text{ for all $n ge n_0$ and all $x$.}
$$
answered Jan 8 at 12:55
Martin RMartin R
31k33561
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add a comment |
$begingroup$
Assume we have a sequence of functions $f_n(x)$ defined on some set $Dsubset mathbb{R}$ such that $arctan f_n(x) to pi/2$ uniformly on $D$. We prove that $f_n $ converge to $+infty$ uniformly, i.e.
$$
forall A > 0 exists N = N(A) in mathbb{N} text{such that if } n>N text{ then } f_n(x) > A forall x in D.
$$
Assume for contradiction, that the above does not hold, that is
$$
tag{1} exists A > 0 text{s.t.} forall N in mathbb{N} exists x_N in D text{s.t.} f_N(x_N) leq A .
$$
In view of uniform convergence of $arctan f_n$ we have that for any $varepsilon > 0$ there is $N = N(varepsilon) in mathbb{N}$ such that whenever $n> N$ then $|arctan f_n(x) - pi/2| < varepsilon$ for all $xin D$ . Combining this with $(1)$ and using the fact that $arctan$ is increasing, from $(1)$ we get
$$
pi/2 - varepsilon < arctan f_n(x_n) leq arctan A < pi/2 tag{2}
$$
if $n>N(varepsilon)$.
Since $A>0$ is fixed, we can choose $varepsilon> 0$ small enough to make the inequality $(2)$ false. This contradiction proves uniform convergence of $f_n$ to $pi/2$ on $D$.
answered Jan 8 at 12:58
HaykHayk
2,7671215
$endgroup$
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
$begingroup$
Yes it does, because $tan$ is a bijective and increasing mapping from $[0, frac pi 2)$ to $[0, infty) $.
More detailed: Given $M > 0$ we have that $0 < arctan M < frac pi 2$. The uniform convergence of $arctan f_n(x) to frac pi 2$ implies the existence of an $n_0 in Bbb N$ such that
$$
arctan f_n(x) > arctan M text{ for all $n ge n_0$ and all $x$.}
$$
It follows that
$$
f_n(x) > M text{ for all $n ge n_0$ and all $x$.}
$$
answered Jan 8 at 12:55
Martin RMartin R
31k33561
$endgroup$
add a comment |
$begingroup$
Yes it does, because $tan$ is a bijective and increasing mapping from $[0, frac pi 2)$ to $[0, infty) $.
More detailed: Given $M > 0$ we have that $0 < arctan M < frac pi 2$. The uniform convergence of $arctan f_n(x) to frac pi 2$ implies the existence of an $n_0 in Bbb N$ such that
$$
arctan f_n(x) > arctan M text{ for all $n ge n_0$ and all $x$.}
$$
It follows that
$$
f_n(x) > M text{ for all $n ge n_0$ and all $x$.}
$$
answered Jan 8 at 12:55
Martin RMartin R
31k33561
$endgroup$
add a comment |
$begingroup$
Yes it does, because $tan$ is a bijective and increasing mapping from $[0, frac pi 2)$ to $[0, infty) $.
More detailed: Given $M > 0$ we have that $0 < arctan M < frac pi 2$. The uniform convergence of $arctan f_n(x) to frac pi 2$ implies the existence of an $n_0 in Bbb N$ such that
$$
arctan f_n(x) > arctan M text{ for all $n ge n_0$ and all $x$.}
$$
It follows that
$$
f_n(x) > M text{ for all $n ge n_0$ and all $x$.}
$$
answered Jan 8 at 12:55
Martin RMartin R
31k33561
$endgroup$
Yes it does, because $tan$ is a bijective and increasing mapping from $[0, frac pi 2)$ to $[0, infty) $.
More detailed: Given $M > 0$ we have that $0 < arctan M < frac pi 2$. The uniform convergence of $arctan f_n(x) to frac pi 2$ implies the existence of an $n_0 in Bbb N$ such that
$$
arctan f_n(x) > arctan M text{ for all $n ge n_0$ and all $x$.}
$$
It follows that
$$
f_n(x) > M text{ for all $n ge n_0$ and all $x$.}
$$
answered Jan 8 at 12:55
Martin RMartin R
31k33561
answered Jan 8 at 12:55
Martin RMartin R
31k33561
answered Jan 8 at 12:55
Martin RMartin R
31k33561
answered Jan 8 at 12:55
Martin RMartin R
31k33561
31k33561
add a comment |
add a comment |
$begingroup$
Assume we have a sequence of functions $f_n(x)$ defined on some set $Dsubset mathbb{R}$ such that $arctan f_n(x) to pi/2$ uniformly on $D$. We prove that $f_n $ converge to $+infty$ uniformly, i.e.
$$
forall A > 0 exists N = N(A) in mathbb{N} text{such that if } n>N text{ then } f_n(x) > A forall x in D.
$$
Assume for contradiction, that the above does not hold, that is
$$
tag{1} exists A > 0 text{s.t.} forall N in mathbb{N} exists x_N in D text{s.t.} f_N(x_N) leq A .
$$
In view of uniform convergence of $arctan f_n$ we have that for any $varepsilon > 0$ there is $N = N(varepsilon) in mathbb{N}$ such that whenever $n> N$ then $|arctan f_n(x) - pi/2| < varepsilon$ for all $xin D$ . Combining this with $(1)$ and using the fact that $arctan$ is increasing, from $(1)$ we get
$$
pi/2 - varepsilon < arctan f_n(x_n) leq arctan A < pi/2 tag{2}
$$
if $n>N(varepsilon)$.
Since $A>0$ is fixed, we can choose $varepsilon> 0$ small enough to make the inequality $(2)$ false. This contradiction proves uniform convergence of $f_n$ to $pi/2$ on $D$.
answered Jan 8 at 12:58
HaykHayk
2,7671215
$endgroup$
add a comment |
$begingroup$
Assume we have a sequence of functions $f_n(x)$ defined on some set $Dsubset mathbb{R}$ such that $arctan f_n(x) to pi/2$ uniformly on $D$. We prove that $f_n $ converge to $+infty$ uniformly, i.e.
$$
forall A > 0 exists N = N(A) in mathbb{N} text{such that if } n>N text{ then } f_n(x) > A forall x in D.
$$
Assume for contradiction, that the above does not hold, that is
$$
tag{1} exists A > 0 text{s.t.} forall N in mathbb{N} exists x_N in D text{s.t.} f_N(x_N) leq A .
$$
In view of uniform convergence of $arctan f_n$ we have that for any $varepsilon > 0$ there is $N = N(varepsilon) in mathbb{N}$ such that whenever $n> N$ then $|arctan f_n(x) - pi/2| < varepsilon$ for all $xin D$ . Combining this with $(1)$ and using the fact that $arctan$ is increasing, from $(1)$ we get
$$
pi/2 - varepsilon < arctan f_n(x_n) leq arctan A < pi/2 tag{2}
$$
if $n>N(varepsilon)$.
Since $A>0$ is fixed, we can choose $varepsilon> 0$ small enough to make the inequality $(2)$ false. This contradiction proves uniform convergence of $f_n$ to $pi/2$ on $D$.
answered Jan 8 at 12:58
HaykHayk
2,7671215
$endgroup$
add a comment |
$begingroup$
Assume we have a sequence of functions $f_n(x)$ defined on some set $Dsubset mathbb{R}$ such that $arctan f_n(x) to pi/2$ uniformly on $D$. We prove that $f_n $ converge to $+infty$ uniformly, i.e.
$$
forall A > 0 exists N = N(A) in mathbb{N} text{such that if } n>N text{ then } f_n(x) > A forall x in D.
$$
Assume for contradiction, that the above does not hold, that is
$$
tag{1} exists A > 0 text{s.t.} forall N in mathbb{N} exists x_N in D text{s.t.} f_N(x_N) leq A .
$$
In view of uniform convergence of $arctan f_n$ we have that for any $varepsilon > 0$ there is $N = N(varepsilon) in mathbb{N}$ such that whenever $n> N$ then $|arctan f_n(x) - pi/2| < varepsilon$ for all $xin D$ . Combining this with $(1)$ and using the fact that $arctan$ is increasing, from $(1)$ we get
$$
pi/2 - varepsilon < arctan f_n(x_n) leq arctan A < pi/2 tag{2}
$$
if $n>N(varepsilon)$.
Since $A>0$ is fixed, we can choose $varepsilon> 0$ small enough to make the inequality $(2)$ false. This contradiction proves uniform convergence of $f_n$ to $pi/2$ on $D$.
answered Jan 8 at 12:58
HaykHayk
2,7671215
$endgroup$
Assume we have a sequence of functions $f_n(x)$ defined on some set $Dsubset mathbb{R}$ such that $arctan f_n(x) to pi/2$ uniformly on $D$. We prove that $f_n $ converge to $+infty$ uniformly, i.e.
$$
forall A > 0 exists N = N(A) in mathbb{N} text{such that if } n>N text{ then } f_n(x) > A forall x in D.
$$
Assume for contradiction, that the above does not hold, that is
$$
tag{1} exists A > 0 text{s.t.} forall N in mathbb{N} exists x_N in D text{s.t.} f_N(x_N) leq A .
$$
In view of uniform convergence of $arctan f_n$ we have that for any $varepsilon > 0$ there is $N = N(varepsilon) in mathbb{N}$ such that whenever $n> N$ then $|arctan f_n(x) - pi/2| < varepsilon$ for all $xin D$ . Combining this with $(1)$ and using the fact that $arctan$ is increasing, from $(1)$ we get
$$
pi/2 - varepsilon < arctan f_n(x_n) leq arctan A < pi/2 tag{2}
$$
if $n>N(varepsilon)$.
Since $A>0$ is fixed, we can choose $varepsilon> 0$ small enough to make the inequality $(2)$ false. This contradiction proves uniform convergence of $f_n$ to $pi/2$ on $D$.
answered Jan 8 at 12:58
HaykHayk
2,7671215
answered Jan 8 at 12:58
HaykHayk
2,7671215
answered Jan 8 at 12:58
HaykHayk
2,7671215
answered Jan 8 at 12:58
HaykHayk
2,7671215
2,7671215
add a comment |
add a comment |
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1
$begingroup$
You need two variables to talk about uniform convergence. As it stands it is difficult to understand what you are saying. And how is probability involved here?
$endgroup$
– Kavi Rama Murthy
Jan 8 at 10:24
$begingroup$
Sorry, I edit my problem again.
$endgroup$
– Xiaol.Song
Jan 8 at 11:54