Prove that $ exists [c,d] subset [a,b], m > 0$ such that $f(x) geq m$ for $x in [c,d]$.
if $f$ is integrable on $[a,b]$ and $int _{a}^{b} f(x)dx >0$. Prove that $ exists [c,d] subset [a,b], m > 0$ such that $f(x) geq m$ for $x in [c,d].$
Could anyone give me a hint for proving this?
calculus integration
asked Nov 29 at 7:59
hopefully
129112
add a comment |
if $f$ is integrable on $[a,b]$ and $int _{a}^{b} f(x)dx >0$. Prove that $ exists [c,d] subset [a,b], m > 0$ such that $f(x) geq m$ for $x in [c,d].$
Could anyone give me a hint for proving this?
calculus integration
asked Nov 29 at 7:59
hopefully
129112
1
Darboux lower sum.
– xbh
Nov 29 at 8:04
This is not a measure course @BrevanEllefsen just advanced calculus one
– hopefully
Nov 29 at 8:13
add a comment |
if $f$ is integrable on $[a,b]$ and $int _{a}^{b} f(x)dx >0$. Prove that $ exists [c,d] subset [a,b], m > 0$ such that $f(x) geq m$ for $x in [c,d].$
Could anyone give me a hint for proving this?
calculus integration
asked Nov 29 at 7:59
hopefully
129112
if $f$ is integrable on $[a,b]$ and $int _{a}^{b} f(x)dx >0$. Prove that $ exists [c,d] subset [a,b], m > 0$ such that $f(x) geq m$ for $x in [c,d].$
Could anyone give me a hint for proving this?
calculus integration
calculus integration
asked Nov 29 at 7:59
hopefully
129112
asked Nov 29 at 7:59
hopefully
129112
asked Nov 29 at 7:59
hopefully
129112
asked Nov 29 at 7:59
hopefully
129112
asked Nov 29 at 7:59
hopefully
129112
129112
1
Darboux lower sum.
– xbh
Nov 29 at 8:04
This is not a measure course @BrevanEllefsen just advanced calculus one
– hopefully
Nov 29 at 8:13
add a comment |
1
Darboux lower sum.
– xbh
Nov 29 at 8:04
This is not a measure course @BrevanEllefsen just advanced calculus one
– hopefully
Nov 29 at 8:13
1
1
Darboux lower sum.
– xbh
Nov 29 at 8:04
Darboux lower sum.
– xbh
Nov 29 at 8:04
This is not a measure course @BrevanEllefsen just advanced calculus one
– hopefully
Nov 29 at 8:13
This is not a measure course @BrevanEllefsen just advanced calculus one
– hopefully
Nov 29 at 8:13
add a comment |
1 Answer
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Prove by contradiction. If this is not true then the minimum in each interval of any partition is $leq 0$ so each lower Riemann sum is $leq 0$ which makes the integral $leq 0$.
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
1
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
Thanks for explanation!
– xbh
Nov 29 at 8:43
add a comment |
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1 Answer
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Prove by contradiction. If this is not true then the minimum in each interval of any partition is $leq 0$ so each lower Riemann sum is $leq 0$ which makes the integral $leq 0$.
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
1
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
Thanks for explanation!
– xbh
Nov 29 at 8:43
add a comment |
Prove by contradiction. If this is not true then the minimum in each interval of any partition is $leq 0$ so each lower Riemann sum is $leq 0$ which makes the integral $leq 0$.
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
1
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
Thanks for explanation!
– xbh
Nov 29 at 8:43
add a comment |
Prove by contradiction. If this is not true then the minimum in each interval of any partition is $leq 0$ so each lower Riemann sum is $leq 0$ which makes the integral $leq 0$.
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
Prove by contradiction. If this is not true then the minimum in each interval of any partition is $leq 0$ so each lower Riemann sum is $leq 0$ which makes the integral $leq 0$.
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
edited Nov 29 at 8:44
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 8:17
Kavi Rama Murthy
48.9k31854
48.9k31854
1
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
Thanks for explanation!
– xbh
Nov 29 at 8:43
add a comment |
1
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
Thanks for explanation!
– xbh
Nov 29 at 8:43
1
1
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
If the claim is not true, does it mean $forall [c,d] subseteq [a,b], forall m > 0, exists x_0 in [c, d]colon f(x_0) < m$? Then how could we assert that the infimum of each subinterval $leqslant 0$ [I don't think this is obvious]?
– xbh
Nov 29 at 8:30
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
@xbh Just take $m=frac 1 n$ and get a points $x_n$ where $f(x_n) <frac 1 n$. Doesn't that make the infimum $leq 0$.
– Kavi Rama Murthy
Nov 29 at 8:39
Thanks for explanation!
– xbh
Nov 29 at 8:43
Thanks for explanation!
– xbh
Nov 29 at 8:43
add a comment |
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1
Darboux lower sum.
– xbh
Nov 29 at 8:04
This is not a measure course @BrevanEllefsen just advanced calculus one
– hopefully
Nov 29 at 8:13