Uniform convergence exercise
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This was a question asked to me in an exam which I couldn't answer:
Let $g:[0,1]to Bbb R$ be a continuous function, such that $0<g(x)<1$ for all $xin[0,1]$. Let $f_n :[0,1] to Bbb R$ be a sequence of functions. Prove that if $f_n$ converges uniformly to $mathit g$ then there exists $n_0 in Bbb N$ such that $0< f_n (x)< 1$ for all $ngeqslant n_0 ,text{ for all $x in [0,1]$}$.
What I tried to do was using the definition of uniform convergence, given $varepsilon >0 $there exists $n_0$ such that for all $n geqslant n_0$, $|f_n - g|<varepsilon$. This would mean that $-varepsilon < f_n - g < varepsilon$. Therefore, $$-varepsilon + g < f_n < varepsilon + g$$ and since $0<g<1$, $$-varepsilon + 0< -varepsilon + g< f_n < varepsilon + g <varepsilon + 1.$$ So, $-varepsilon<f_n<varepsilon +1$. I don't know how to go from here. Any help would be appreciated.
continuity uniform-convergence
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
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add a comment |
$begingroup$
This was a question asked to me in an exam which I couldn't answer:
Let $g:[0,1]to Bbb R$ be a continuous function, such that $0<g(x)<1$ for all $xin[0,1]$. Let $f_n :[0,1] to Bbb R$ be a sequence of functions. Prove that if $f_n$ converges uniformly to $mathit g$ then there exists $n_0 in Bbb N$ such that $0< f_n (x)< 1$ for all $ngeqslant n_0 ,text{ for all $x in [0,1]$}$.
What I tried to do was using the definition of uniform convergence, given $varepsilon >0 $there exists $n_0$ such that for all $n geqslant n_0$, $|f_n - g|<varepsilon$. This would mean that $-varepsilon < f_n - g < varepsilon$. Therefore, $$-varepsilon + g < f_n < varepsilon + g$$ and since $0<g<1$, $$-varepsilon + 0< -varepsilon + g< f_n < varepsilon + g <varepsilon + 1.$$ So, $-varepsilon<f_n<varepsilon +1$. I don't know how to go from here. Any help would be appreciated.
continuity uniform-convergence
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
$endgroup$
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Since $g$ is continuous and $[0, 1]$ is compact, $|g| = sup g < 1$ and then, $|f_n - g| < 1 -|g|$ for all large $n,$ this implies $sup f_n < 1$ for all large $n.$ The case $0 < inf f_n$ is similar.
$endgroup$
– Will M.
Dec 17 '18 at 21:55
add a comment |
$begingroup$
This was a question asked to me in an exam which I couldn't answer:
Let $g:[0,1]to Bbb R$ be a continuous function, such that $0<g(x)<1$ for all $xin[0,1]$. Let $f_n :[0,1] to Bbb R$ be a sequence of functions. Prove that if $f_n$ converges uniformly to $mathit g$ then there exists $n_0 in Bbb N$ such that $0< f_n (x)< 1$ for all $ngeqslant n_0 ,text{ for all $x in [0,1]$}$.
What I tried to do was using the definition of uniform convergence, given $varepsilon >0 $there exists $n_0$ such that for all $n geqslant n_0$, $|f_n - g|<varepsilon$. This would mean that $-varepsilon < f_n - g < varepsilon$. Therefore, $$-varepsilon + g < f_n < varepsilon + g$$ and since $0<g<1$, $$-varepsilon + 0< -varepsilon + g< f_n < varepsilon + g <varepsilon + 1.$$ So, $-varepsilon<f_n<varepsilon +1$. I don't know how to go from here. Any help would be appreciated.
continuity uniform-convergence
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
$endgroup$
This was a question asked to me in an exam which I couldn't answer:
Let $g:[0,1]to Bbb R$ be a continuous function, such that $0<g(x)<1$ for all $xin[0,1]$. Let $f_n :[0,1] to Bbb R$ be a sequence of functions. Prove that if $f_n$ converges uniformly to $mathit g$ then there exists $n_0 in Bbb N$ such that $0< f_n (x)< 1$ for all $ngeqslant n_0 ,text{ for all $x in [0,1]$}$.
What I tried to do was using the definition of uniform convergence, given $varepsilon >0 $there exists $n_0$ such that for all $n geqslant n_0$, $|f_n - g|<varepsilon$. This would mean that $-varepsilon < f_n - g < varepsilon$. Therefore, $$-varepsilon + g < f_n < varepsilon + g$$ and since $0<g<1$, $$-varepsilon + 0< -varepsilon + g< f_n < varepsilon + g <varepsilon + 1.$$ So, $-varepsilon<f_n<varepsilon +1$. I don't know how to go from here. Any help would be appreciated.
continuity uniform-convergence
continuity uniform-convergence
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
asked Dec 17 '18 at 21:22
juan deutschjuan deutsch
156
156
$begingroup$
Since $g$ is continuous and $[0, 1]$ is compact, $|g| = sup g < 1$ and then, $|f_n - g| < 1 -|g|$ for all large $n,$ this implies $sup f_n < 1$ for all large $n.$ The case $0 < inf f_n$ is similar.
$endgroup$
– Will M.
Dec 17 '18 at 21:55
add a comment |
$begingroup$
Since $g$ is continuous and $[0, 1]$ is compact, $|g| = sup g < 1$ and then, $|f_n - g| < 1 -|g|$ for all large $n,$ this implies $sup f_n < 1$ for all large $n.$ The case $0 < inf f_n$ is similar.
$endgroup$
– Will M.
Dec 17 '18 at 21:55
$begingroup$
Since $g$ is continuous and $[0, 1]$ is compact, $|g| = sup g < 1$ and then, $|f_n - g| < 1 -|g|$ for all large $n,$ this implies $sup f_n < 1$ for all large $n.$ The case $0 < inf f_n$ is similar.
$endgroup$
– Will M.
Dec 17 '18 at 21:55
$begingroup$
Since $g$ is continuous and $[0, 1]$ is compact, $|g| = sup g < 1$ and then, $|f_n - g| < 1 -|g|$ for all large $n,$ this implies $sup f_n < 1$ for all large $n.$ The case $0 < inf f_n$ is similar.
$endgroup$
– Will M.
Dec 17 '18 at 21:55
add a comment |
2 Answers
2
active
oldest
votes
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Let $M=max g$ and let $m=min G$. Then $0<mleqslant M<1$. Take $varepsilon>0$ such that, $varepsilon<m$ and that $varepsilon<1-M$. THere is a natural $N$ such that$$(forall ninmathbb{N})(forall xin[0,1]):ngeqslant Nimpliesbigllvert g(x)-f_n(x)bigrrvert<varepsilon.$$But then, since$$(forall xin[0,1]):mleqslant g(x)leqslant M$$and since $varepsilon<m$ and $varepsilon<1-M$, we have$$(forall xin[0,1]):0<f_N(x)<1.$$
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
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I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
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Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
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By the same argument, $f_N(x)>0$.
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– José Carlos Santos
Dec 17 '18 at 21:49
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Thank you for the clarification.
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– juan deutsch
Dec 17 '18 at 21:55
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I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
add a comment |
$begingroup$
Since $g$ is continuous on a compact set we have that $0<m<g(x)<M<1$ for all $xin [0,1]$. Let $n_0$ be large enough so for $ngeq n_0$, $|f_n(x)-g(x)|<min(m,1-M)$ for all $xin [0,1]$. Then for all $xin[0,1]$
$$f_n(x)= (f_n(x)-g(x))+g(x)<(1-M)+M<1$$ and
$$-f_n(x)=(g(x)-f_n(x))-g(x)< m-m=0implies 0<f_n(x)$$
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
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add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $M=max g$ and let $m=min G$. Then $0<mleqslant M<1$. Take $varepsilon>0$ such that, $varepsilon<m$ and that $varepsilon<1-M$. THere is a natural $N$ such that$$(forall ninmathbb{N})(forall xin[0,1]):ngeqslant Nimpliesbigllvert g(x)-f_n(x)bigrrvert<varepsilon.$$But then, since$$(forall xin[0,1]):mleqslant g(x)leqslant M$$and since $varepsilon<m$ and $varepsilon<1-M$, we have$$(forall xin[0,1]):0<f_N(x)<1.$$
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
$begingroup$
I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
$begingroup$
Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
$begingroup$
By the same argument, $f_N(x)>0$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:49
$begingroup$
Thank you for the clarification.
$endgroup$
– juan deutsch
Dec 17 '18 at 21:55
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
add a comment |
$begingroup$
Let $M=max g$ and let $m=min G$. Then $0<mleqslant M<1$. Take $varepsilon>0$ such that, $varepsilon<m$ and that $varepsilon<1-M$. THere is a natural $N$ such that$$(forall ninmathbb{N})(forall xin[0,1]):ngeqslant Nimpliesbigllvert g(x)-f_n(x)bigrrvert<varepsilon.$$But then, since$$(forall xin[0,1]):mleqslant g(x)leqslant M$$and since $varepsilon<m$ and $varepsilon<1-M$, we have$$(forall xin[0,1]):0<f_N(x)<1.$$
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
$begingroup$
I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
$begingroup$
Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
$begingroup$
By the same argument, $f_N(x)>0$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:49
$begingroup$
Thank you for the clarification.
$endgroup$
– juan deutsch
Dec 17 '18 at 21:55
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
add a comment |
$begingroup$
Let $M=max g$ and let $m=min G$. Then $0<mleqslant M<1$. Take $varepsilon>0$ such that, $varepsilon<m$ and that $varepsilon<1-M$. THere is a natural $N$ such that$$(forall ninmathbb{N})(forall xin[0,1]):ngeqslant Nimpliesbigllvert g(x)-f_n(x)bigrrvert<varepsilon.$$But then, since$$(forall xin[0,1]):mleqslant g(x)leqslant M$$and since $varepsilon<m$ and $varepsilon<1-M$, we have$$(forall xin[0,1]):0<f_N(x)<1.$$
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
$endgroup$
Let $M=max g$ and let $m=min G$. Then $0<mleqslant M<1$. Take $varepsilon>0$ such that, $varepsilon<m$ and that $varepsilon<1-M$. THere is a natural $N$ such that$$(forall ninmathbb{N})(forall xin[0,1]):ngeqslant Nimpliesbigllvert g(x)-f_n(x)bigrrvert<varepsilon.$$But then, since$$(forall xin[0,1]):mleqslant g(x)leqslant M$$and since $varepsilon<m$ and $varepsilon<1-M$, we have$$(forall xin[0,1]):0<f_N(x)<1.$$
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
edited Dec 17 '18 at 21:48
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
answered Dec 17 '18 at 21:31
José Carlos SantosJosé Carlos Santos
162k22130233
162k22130233
$begingroup$
I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
$begingroup$
Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
$begingroup$
By the same argument, $f_N(x)>0$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:49
$begingroup$
Thank you for the clarification.
$endgroup$
– juan deutsch
Dec 17 '18 at 21:55
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
add a comment |
$begingroup$
I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
$begingroup$
Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
$begingroup$
By the same argument, $f_N(x)>0$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:49
$begingroup$
Thank you for the clarification.
$endgroup$
– juan deutsch
Dec 17 '18 at 21:55
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
$begingroup$
I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
$begingroup$
I did not understand the last step. Could you explain what you did? Thanks!
$endgroup$
– juan deutsch
Dec 17 '18 at 21:37
$begingroup$
Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
$begingroup$
Sure:begin{align}f_N(x)&=f_N(x)-g(x)+g(x)\&leqslantbigllvert f_N(x)-g(x)bigrrvert+g(x)\&leqslantvarepsilon+g(x)\&<1-M+M\&=1.end{align}
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:46
$begingroup$
By the same argument, $f_N(x)>0$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:49
$begingroup$
By the same argument, $f_N(x)>0$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:49
$begingroup$
Thank you for the clarification.
$endgroup$
– juan deutsch
Dec 17 '18 at 21:55
$begingroup$
Thank you for the clarification.
$endgroup$
– juan deutsch
Dec 17 '18 at 21:55
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
$begingroup$
I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 21:56
add a comment |
$begingroup$
Since $g$ is continuous on a compact set we have that $0<m<g(x)<M<1$ for all $xin [0,1]$. Let $n_0$ be large enough so for $ngeq n_0$, $|f_n(x)-g(x)|<min(m,1-M)$ for all $xin [0,1]$. Then for all $xin[0,1]$
$$f_n(x)= (f_n(x)-g(x))+g(x)<(1-M)+M<1$$ and
$$-f_n(x)=(g(x)-f_n(x))-g(x)< m-m=0implies 0<f_n(x)$$
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
$endgroup$
add a comment |
$begingroup$
Since $g$ is continuous on a compact set we have that $0<m<g(x)<M<1$ for all $xin [0,1]$. Let $n_0$ be large enough so for $ngeq n_0$, $|f_n(x)-g(x)|<min(m,1-M)$ for all $xin [0,1]$. Then for all $xin[0,1]$
$$f_n(x)= (f_n(x)-g(x))+g(x)<(1-M)+M<1$$ and
$$-f_n(x)=(g(x)-f_n(x))-g(x)< m-m=0implies 0<f_n(x)$$
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
$endgroup$
add a comment |
$begingroup$
Since $g$ is continuous on a compact set we have that $0<m<g(x)<M<1$ for all $xin [0,1]$. Let $n_0$ be large enough so for $ngeq n_0$, $|f_n(x)-g(x)|<min(m,1-M)$ for all $xin [0,1]$. Then for all $xin[0,1]$
$$f_n(x)= (f_n(x)-g(x))+g(x)<(1-M)+M<1$$ and
$$-f_n(x)=(g(x)-f_n(x))-g(x)< m-m=0implies 0<f_n(x)$$
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
$endgroup$
Since $g$ is continuous on a compact set we have that $0<m<g(x)<M<1$ for all $xin [0,1]$. Let $n_0$ be large enough so for $ngeq n_0$, $|f_n(x)-g(x)|<min(m,1-M)$ for all $xin [0,1]$. Then for all $xin[0,1]$
$$f_n(x)= (f_n(x)-g(x))+g(x)<(1-M)+M<1$$ and
$$-f_n(x)=(g(x)-f_n(x))-g(x)< m-m=0implies 0<f_n(x)$$
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
answered Dec 17 '18 at 21:31
user293794user293794
1,711613
1,711613
add a comment |
add a comment |
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$begingroup$
Since $g$ is continuous and $[0, 1]$ is compact, $|g| = sup g < 1$ and then, $|f_n - g| < 1 -|g|$ for all large $n,$ this implies $sup f_n < 1$ for all large $n.$ The case $0 < inf f_n$ is similar.
$endgroup$
– Will M.
Dec 17 '18 at 21:55