How to show there exists a $delta>0$ such that $|f(t,x)-f(s,x)|<epsilon$ for all $xin [c,d]$ and for...
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Suppose $f:(a,b)times [c,d]to Bbb R$ is a continuous function, how to prove the following conclusion?
$forall sin (a,b),forall epsilon>0$, there exists a $delta>0$ such that $|f(t,x)-f(s,x)|<epsilon$ for all $xin [c,d]$ and for all $tin (s-delta,s+delta)$.
analysis
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
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add a comment |
$begingroup$
Suppose $f:(a,b)times [c,d]to Bbb R$ is a continuous function, how to prove the following conclusion?
$forall sin (a,b),forall epsilon>0$, there exists a $delta>0$ such that $|f(t,x)-f(s,x)|<epsilon$ for all $xin [c,d]$ and for all $tin (s-delta,s+delta)$.
analysis
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
$endgroup$
add a comment |
$begingroup$
Suppose $f:(a,b)times [c,d]to Bbb R$ is a continuous function, how to prove the following conclusion?
$forall sin (a,b),forall epsilon>0$, there exists a $delta>0$ such that $|f(t,x)-f(s,x)|<epsilon$ for all $xin [c,d]$ and for all $tin (s-delta,s+delta)$.
analysis
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
$endgroup$
Suppose $f:(a,b)times [c,d]to Bbb R$ is a continuous function, how to prove the following conclusion?
$forall sin (a,b),forall epsilon>0$, there exists a $delta>0$ such that $|f(t,x)-f(s,x)|<epsilon$ for all $xin [c,d]$ and for all $tin (s-delta,s+delta)$.
analysis
analysis
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
asked Dec 7 '18 at 9:39
Born to be proudBorn to be proud
775510
775510
add a comment |
add a comment |
1 Answer
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$f$ is uniformly continuous on the compact set $[a+epsilon, b-epsilon] times [c,d]$ (assuming that $epsilon <(b-a)/2$). Your statement follows immediately from this. [ Choose $delta$ as in definition of uniformly continuity and just make sure that $(s-delta, s+delta) subset [a+epsilon, b-epsilon]$].
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
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1 Answer
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active
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1 Answer
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active
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active
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$begingroup$
$f$ is uniformly continuous on the compact set $[a+epsilon, b-epsilon] times [c,d]$ (assuming that $epsilon <(b-a)/2$). Your statement follows immediately from this. [ Choose $delta$ as in definition of uniformly continuity and just make sure that $(s-delta, s+delta) subset [a+epsilon, b-epsilon]$].
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
$endgroup$
add a comment |
$begingroup$
$f$ is uniformly continuous on the compact set $[a+epsilon, b-epsilon] times [c,d]$ (assuming that $epsilon <(b-a)/2$). Your statement follows immediately from this. [ Choose $delta$ as in definition of uniformly continuity and just make sure that $(s-delta, s+delta) subset [a+epsilon, b-epsilon]$].
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
$endgroup$
add a comment |
$begingroup$
$f$ is uniformly continuous on the compact set $[a+epsilon, b-epsilon] times [c,d]$ (assuming that $epsilon <(b-a)/2$). Your statement follows immediately from this. [ Choose $delta$ as in definition of uniformly continuity and just make sure that $(s-delta, s+delta) subset [a+epsilon, b-epsilon]$].
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
$endgroup$
$f$ is uniformly continuous on the compact set $[a+epsilon, b-epsilon] times [c,d]$ (assuming that $epsilon <(b-a)/2$). Your statement follows immediately from this. [ Choose $delta$ as in definition of uniformly continuity and just make sure that $(s-delta, s+delta) subset [a+epsilon, b-epsilon]$].
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
answered Dec 7 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
53.6k32055
53.6k32055
add a comment |
add a comment |
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