Mixing
Exercise of sequence of continuous functions
Let $(f_n)_n$ be a sequence of continuous functions on $Dsubset mathbb{R}^{N} to mathbb{R}$ which is monotone decreasing. If $lim_{ntoinfty }f_n(c))=0$ for some $cin D$ and $ epsilon >0$ , show that there are $m in mathbb{N}$ and a neighbourhood $U$ of $c$ shuch that if $n>m$ and $x in U cap D$ , then $f_n (x)< epsilon$.
real-analysis convergence sequence-of-function
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
asked Nov 22 '18 at 8:29
ParcosParcos
506
add a comment |
Let $(f_n)_n$ be a sequence of continuous functions on $Dsubset mathbb{R}^{N} to mathbb{R}$ which is monotone decreasing. If $lim_{ntoinfty }f_n(c))=0$ for some $cin D$ and $ epsilon >0$ , show that there are $m in mathbb{N}$ and a neighbourhood $U$ of $c$ shuch that if $n>m$ and $x in U cap D$ , then $f_n (x)< epsilon$.
real-analysis convergence sequence-of-function
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
asked Nov 22 '18 at 8:29
ParcosParcos
506
Are you going to share with us your thoughts about this exercise?
– Robert Z
Nov 22 '18 at 8:39
I was thinking some like this: $|f_k(x)| leq |f_k(c)| + |f_k(c)-f_k(x)|$ so the first term is less to $epsilon$ by $f_n(c)$ converges to zero and the second is less to $epsilon$ by continuity but the problem is that the neighbourhood depend of $k$ , thank you
– Parcos
Nov 22 '18 at 12:28
Next time please include your thoughts under the question. This increases significantly the probability that some user will answer your question.
– Robert Z
Nov 22 '18 at 12:52
add a comment |
Let $(f_n)_n$ be a sequence of continuous functions on $Dsubset mathbb{R}^{N} to mathbb{R}$ which is monotone decreasing. If $lim_{ntoinfty }f_n(c))=0$ for some $cin D$ and $ epsilon >0$ , show that there are $m in mathbb{N}$ and a neighbourhood $U$ of $c$ shuch that if $n>m$ and $x in U cap D$ , then $f_n (x)< epsilon$.
real-analysis convergence sequence-of-function
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
asked Nov 22 '18 at 8:29
ParcosParcos
506
Let $(f_n)_n$ be a sequence of continuous functions on $Dsubset mathbb{R}^{N} to mathbb{R}$ which is monotone decreasing. If $lim_{ntoinfty }f_n(c))=0$ for some $cin D$ and $ epsilon >0$ , show that there are $m in mathbb{N}$ and a neighbourhood $U$ of $c$ shuch that if $n>m$ and $x in U cap D$ , then $f_n (x)< epsilon$.
real-analysis convergence sequence-of-function
real-analysis convergence sequence-of-function
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
asked Nov 22 '18 at 8:29
ParcosParcos
506
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
asked Nov 22 '18 at 8:29
ParcosParcos
506
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
edited Nov 22 '18 at 8:41
Robert Z
93.8k1061132
93.8k1061132
asked Nov 22 '18 at 8:29
ParcosParcos
506
asked Nov 22 '18 at 8:29
ParcosParcos
506
asked Nov 22 '18 at 8:29
ParcosParcos
506
506
Are you going to share with us your thoughts about this exercise?
– Robert Z
Nov 22 '18 at 8:39
I was thinking some like this: $|f_k(x)| leq |f_k(c)| + |f_k(c)-f_k(x)|$ so the first term is less to $epsilon$ by $f_n(c)$ converges to zero and the second is less to $epsilon$ by continuity but the problem is that the neighbourhood depend of $k$ , thank you
– Parcos
Nov 22 '18 at 12:28
Next time please include your thoughts under the question. This increases significantly the probability that some user will answer your question.
– Robert Z
Nov 22 '18 at 12:52
add a comment |
Are you going to share with us your thoughts about this exercise?
– Robert Z
Nov 22 '18 at 8:39
I was thinking some like this: $|f_k(x)| leq |f_k(c)| + |f_k(c)-f_k(x)|$ so the first term is less to $epsilon$ by $f_n(c)$ converges to zero and the second is less to $epsilon$ by continuity but the problem is that the neighbourhood depend of $k$ , thank you
– Parcos
Nov 22 '18 at 12:28
Next time please include your thoughts under the question. This increases significantly the probability that some user will answer your question.
– Robert Z
Nov 22 '18 at 12:52
Are you going to share with us your thoughts about this exercise?
– Robert Z
Nov 22 '18 at 8:39
Are you going to share with us your thoughts about this exercise?
– Robert Z
Nov 22 '18 at 8:39
I was thinking some like this: $|f_k(x)| leq |f_k(c)| + |f_k(c)-f_k(x)|$ so the first term is less to $epsilon$ by $f_n(c)$ converges to zero and the second is less to $epsilon$ by continuity but the problem is that the neighbourhood depend of $k$ , thank you
– Parcos
Nov 22 '18 at 12:28
I was thinking some like this: $|f_k(x)| leq |f_k(c)| + |f_k(c)-f_k(x)|$ so the first term is less to $epsilon$ by $f_n(c)$ converges to zero and the second is less to $epsilon$ by continuity but the problem is that the neighbourhood depend of $k$ , thank you
– Parcos
Nov 22 '18 at 12:28
Next time please include your thoughts under the question. This increases significantly the probability that some user will answer your question.
– Robert Z
Nov 22 '18 at 12:52
Next time please include your thoughts under the question. This increases significantly the probability that some user will answer your question.
– Robert Z
Nov 22 '18 at 12:52
add a comment |
1 Answer
1
active
oldest
votes
$f_k(c)<epsilon$ for some $k$. By continuity $f_k(x) <epsilon$ for all $x$ in some neighborhood. Since $f_n leq f_k$ for $n geq k$ we get $f_n(x) <epsilon$ for all $n geq k$ for all $x$ in the neighborhood.
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
$f_k(c)<epsilon$ for some $k$. By continuity $f_k(x) <epsilon$ for all $x$ in some neighborhood. Since $f_n leq f_k$ for $n geq k$ we get $f_n(x) <epsilon$ for all $n geq k$ for all $x$ in the neighborhood.
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
add a comment |
$f_k(c)<epsilon$ for some $k$. By continuity $f_k(x) <epsilon$ for all $x$ in some neighborhood. Since $f_n leq f_k$ for $n geq k$ we get $f_n(x) <epsilon$ for all $n geq k$ for all $x$ in the neighborhood.
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
add a comment |
$f_k(c)<epsilon$ for some $k$. By continuity $f_k(x) <epsilon$ for all $x$ in some neighborhood. Since $f_n leq f_k$ for $n geq k$ we get $f_n(x) <epsilon$ for all $n geq k$ for all $x$ in the neighborhood.
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
$f_k(c)<epsilon$ for some $k$. By continuity $f_k(x) <epsilon$ for all $x$ in some neighborhood. Since $f_n leq f_k$ for $n geq k$ we get $f_n(x) <epsilon$ for all $n geq k$ for all $x$ in the neighborhood.
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
answered Nov 22 '18 at 8:33
Kavi Rama MurthyKavi Rama Murthy
52k32055
52k32055
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
add a comment |
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
Oh thank you!, it was easier than i thought
– Parcos
Nov 22 '18 at 12:34
add a comment |
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Are you going to share with us your thoughts about this exercise?
– Robert Z
Nov 22 '18 at 8:39
I was thinking some like this: $|f_k(x)| leq |f_k(c)| + |f_k(c)-f_k(x)|$ so the first term is less to $epsilon$ by $f_n(c)$ converges to zero and the second is less to $epsilon$ by continuity but the problem is that the neighbourhood depend of $k$ , thank you
– Parcos
Nov 22 '18 at 12:28
Next time please include your thoughts under the question. This increases significantly the probability that some user will answer your question.
– Robert Z
Nov 22 '18 at 12:52