If $f$ is continuous and $int_a^b[f(x)]^2=0$ prove that $f(x)=0$ for $xin[a,b]$ [duplicate]
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
Intuitively, I know $[f(x)]^2ge0$ so the integral would be the area above the x-axis and it would result in $[f(x)]^2=0$ and then $f(x)=0$.
How do I prove this with proper terms. This is a reimann integral.
real-analysis integration
asked Nov 28 at 20:08
Albert Diaz
925
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Nov 28 at 20:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
Intuitively, I know $[f(x)]^2ge0$ so the integral would be the area above the x-axis and it would result in $[f(x)]^2=0$ and then $f(x)=0$.
How do I prove this with proper terms. This is a reimann integral.
real-analysis integration
asked Nov 28 at 20:08
Albert Diaz
925
marked as duplicate by RRL real-analysis
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Nov 28 at 20:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Is $f $ continuous ?
– hamam_Abdallah
Nov 28 at 20:10
You need to be more specific about your assumptions on $f$. This isn't true unless you add some conditions. Is $f$ continuous?
– Zestylemonzi
Nov 28 at 20:11
@hamam_Abdallah yes it is continuous, I just edited my problem
– Albert Diaz
Nov 28 at 20:12
If $f$ is not $0$ (say $f(c)neq 0$ where $cin (a,b)$), then there is a $varepsilon >0$ small s.t. $|f(x)|^2geq ell>0$ on $[c-varepsilon ,c+varepsilon ]$. Contradiction. If $f(a)neq 0$ or $f(b)neq 0$ the proof goes the same.
– Surb
Nov 28 at 20:13
add a comment |
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
Intuitively, I know $[f(x)]^2ge0$ so the integral would be the area above the x-axis and it would result in $[f(x)]^2=0$ and then $f(x)=0$.
How do I prove this with proper terms. This is a reimann integral.
real-analysis integration
asked Nov 28 at 20:08
Albert Diaz
925
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
Intuitively, I know $[f(x)]^2ge0$ so the integral would be the area above the x-axis and it would result in $[f(x)]^2=0$ and then $f(x)=0$.
How do I prove this with proper terms. This is a reimann integral.
This question already has an answer here:
Is it true that $int_{a}^b h^2=0implies h=0$?
2 answers
real-analysis integration
real-analysis integration
asked Nov 28 at 20:08
Albert Diaz
925
asked Nov 28 at 20:08
Albert Diaz
925
edited Nov 28 at 20:11
asked Nov 28 at 20:08
Albert Diaz
925
asked Nov 28 at 20:08
Albert Diaz
925
asked Nov 28 at 20:08
Albert Diaz
925
925
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Nov 28 at 20:57
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Nov 28 at 20:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Is $f $ continuous ?
– hamam_Abdallah
Nov 28 at 20:10
You need to be more specific about your assumptions on $f$. This isn't true unless you add some conditions. Is $f$ continuous?
– Zestylemonzi
Nov 28 at 20:11
@hamam_Abdallah yes it is continuous, I just edited my problem
– Albert Diaz
Nov 28 at 20:12
If $f$ is not $0$ (say $f(c)neq 0$ where $cin (a,b)$), then there is a $varepsilon >0$ small s.t. $|f(x)|^2geq ell>0$ on $[c-varepsilon ,c+varepsilon ]$. Contradiction. If $f(a)neq 0$ or $f(b)neq 0$ the proof goes the same.
– Surb
Nov 28 at 20:13
add a comment |
Is $f $ continuous ?
– hamam_Abdallah
Nov 28 at 20:10
You need to be more specific about your assumptions on $f$. This isn't true unless you add some conditions. Is $f$ continuous?
– Zestylemonzi
Nov 28 at 20:11
@hamam_Abdallah yes it is continuous, I just edited my problem
– Albert Diaz
Nov 28 at 20:12
If $f$ is not $0$ (say $f(c)neq 0$ where $cin (a,b)$), then there is a $varepsilon >0$ small s.t. $|f(x)|^2geq ell>0$ on $[c-varepsilon ,c+varepsilon ]$. Contradiction. If $f(a)neq 0$ or $f(b)neq 0$ the proof goes the same.
– Surb
Nov 28 at 20:13
Is $f $ continuous ?
– hamam_Abdallah
Nov 28 at 20:10
Is $f $ continuous ?
– hamam_Abdallah
Nov 28 at 20:10
You need to be more specific about your assumptions on $f$. This isn't true unless you add some conditions. Is $f$ continuous?
– Zestylemonzi
Nov 28 at 20:11
You need to be more specific about your assumptions on $f$. This isn't true unless you add some conditions. Is $f$ continuous?
– Zestylemonzi
Nov 28 at 20:11
@hamam_Abdallah yes it is continuous, I just edited my problem
– Albert Diaz
Nov 28 at 20:12
@hamam_Abdallah yes it is continuous, I just edited my problem
– Albert Diaz
Nov 28 at 20:12
If $f$ is not $0$ (say $f(c)neq 0$ where $cin (a,b)$), then there is a $varepsilon >0$ small s.t. $|f(x)|^2geq ell>0$ on $[c-varepsilon ,c+varepsilon ]$. Contradiction. If $f(a)neq 0$ or $f(b)neq 0$ the proof goes the same.
– Surb
Nov 28 at 20:13
If $f$ is not $0$ (say $f(c)neq 0$ where $cin (a,b)$), then there is a $varepsilon >0$ small s.t. $|f(x)|^2geq ell>0$ on $[c-varepsilon ,c+varepsilon ]$. Contradiction. If $f(a)neq 0$ or $f(b)neq 0$ the proof goes the same.
– Surb
Nov 28 at 20:13
add a comment |
2 Answers
2
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oldest
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Let $F:[a,b]to[0,+infty)$ be a continuous function with $int_a^bF(x)dx=0$.
Assume, by the way of contradiction, that $exists x_0in[a,b]$ such that $F(x_0)>0$. WLOG, we can assume that $x_one a,b$. By continuyity, $exists delta>0$ such that $F(x)>F(x_0)/2$ for every $xin (x_0-delta,x_0+delta)$.
Hence, $$int_a^bF(x)dxgeint_{x_0-delta}^{x_0+delta}F(x)dx>deltacdot F(x_0)>0,$$ which is a contradiction.
Now, for your case, if $f$ is continuous, take $F=f^2$ which is continuous an positive.
If $x_0=a$, then you'll tske the interval were the function is bigger than $F(x_0)/2$ of the form $[a,a+delta)$; If $x_0=b$, then take the interval of the form $(b-delta,b]$.
answered Nov 28 at 20:16
Tito Eliatron
1,520622
add a comment |
hint
Assume there exist $cin[a,b]$ such that
$$f^2(c)>0$$
with $epsilon=frac{f^2(c)}{2}$, and
by continuity at $c$, there exist $eta>0$ such that
$$forall xin[a,b]cap ]c-eta,c+eta[ $$
$$;; frac{f^2(c)}{2}<f^2(x)<frac{3f^2(c)}{2}$$
now observe that
$$0=int_a^bf^2(t)dtge int_{c-eta}^{c+eta}f^2(t)dt$$
$$>2etafrac{f^2(c)}{2}>0$$
this is a contradiction.
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $F:[a,b]to[0,+infty)$ be a continuous function with $int_a^bF(x)dx=0$.
Assume, by the way of contradiction, that $exists x_0in[a,b]$ such that $F(x_0)>0$. WLOG, we can assume that $x_one a,b$. By continuyity, $exists delta>0$ such that $F(x)>F(x_0)/2$ for every $xin (x_0-delta,x_0+delta)$.
Hence, $$int_a^bF(x)dxgeint_{x_0-delta}^{x_0+delta}F(x)dx>deltacdot F(x_0)>0,$$ which is a contradiction.
Now, for your case, if $f$ is continuous, take $F=f^2$ which is continuous an positive.
If $x_0=a$, then you'll tske the interval were the function is bigger than $F(x_0)/2$ of the form $[a,a+delta)$; If $x_0=b$, then take the interval of the form $(b-delta,b]$.
answered Nov 28 at 20:16
Tito Eliatron
1,520622
add a comment |
Let $F:[a,b]to[0,+infty)$ be a continuous function with $int_a^bF(x)dx=0$.
Assume, by the way of contradiction, that $exists x_0in[a,b]$ such that $F(x_0)>0$. WLOG, we can assume that $x_one a,b$. By continuyity, $exists delta>0$ such that $F(x)>F(x_0)/2$ for every $xin (x_0-delta,x_0+delta)$.
Hence, $$int_a^bF(x)dxgeint_{x_0-delta}^{x_0+delta}F(x)dx>deltacdot F(x_0)>0,$$ which is a contradiction.
Now, for your case, if $f$ is continuous, take $F=f^2$ which is continuous an positive.
If $x_0=a$, then you'll tske the interval were the function is bigger than $F(x_0)/2$ of the form $[a,a+delta)$; If $x_0=b$, then take the interval of the form $(b-delta,b]$.
answered Nov 28 at 20:16
Tito Eliatron
1,520622
add a comment |
Let $F:[a,b]to[0,+infty)$ be a continuous function with $int_a^bF(x)dx=0$.
Assume, by the way of contradiction, that $exists x_0in[a,b]$ such that $F(x_0)>0$. WLOG, we can assume that $x_one a,b$. By continuyity, $exists delta>0$ such that $F(x)>F(x_0)/2$ for every $xin (x_0-delta,x_0+delta)$.
Hence, $$int_a^bF(x)dxgeint_{x_0-delta}^{x_0+delta}F(x)dx>deltacdot F(x_0)>0,$$ which is a contradiction.
Now, for your case, if $f$ is continuous, take $F=f^2$ which is continuous an positive.
If $x_0=a$, then you'll tske the interval were the function is bigger than $F(x_0)/2$ of the form $[a,a+delta)$; If $x_0=b$, then take the interval of the form $(b-delta,b]$.
answered Nov 28 at 20:16
Tito Eliatron
1,520622
Let $F:[a,b]to[0,+infty)$ be a continuous function with $int_a^bF(x)dx=0$.
Assume, by the way of contradiction, that $exists x_0in[a,b]$ such that $F(x_0)>0$. WLOG, we can assume that $x_one a,b$. By continuyity, $exists delta>0$ such that $F(x)>F(x_0)/2$ for every $xin (x_0-delta,x_0+delta)$.
Hence, $$int_a^bF(x)dxgeint_{x_0-delta}^{x_0+delta}F(x)dx>deltacdot F(x_0)>0,$$ which is a contradiction.
Now, for your case, if $f$ is continuous, take $F=f^2$ which is continuous an positive.
If $x_0=a$, then you'll tske the interval were the function is bigger than $F(x_0)/2$ of the form $[a,a+delta)$; If $x_0=b$, then take the interval of the form $(b-delta,b]$.
answered Nov 28 at 20:16
Tito Eliatron
1,520622
edited Nov 28 at 20:47
answered Nov 28 at 20:16
Tito Eliatron
1,520622
answered Nov 28 at 20:16
Tito Eliatron
1,520622
answered Nov 28 at 20:16
Tito Eliatron
1,520622
1,520622
add a comment |
add a comment |
hint
Assume there exist $cin[a,b]$ such that
$$f^2(c)>0$$
with $epsilon=frac{f^2(c)}{2}$, and
by continuity at $c$, there exist $eta>0$ such that
$$forall xin[a,b]cap ]c-eta,c+eta[ $$
$$;; frac{f^2(c)}{2}<f^2(x)<frac{3f^2(c)}{2}$$
now observe that
$$0=int_a^bf^2(t)dtge int_{c-eta}^{c+eta}f^2(t)dt$$
$$>2etafrac{f^2(c)}{2}>0$$
this is a contradiction.
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
add a comment |
hint
Assume there exist $cin[a,b]$ such that
$$f^2(c)>0$$
with $epsilon=frac{f^2(c)}{2}$, and
by continuity at $c$, there exist $eta>0$ such that
$$forall xin[a,b]cap ]c-eta,c+eta[ $$
$$;; frac{f^2(c)}{2}<f^2(x)<frac{3f^2(c)}{2}$$
now observe that
$$0=int_a^bf^2(t)dtge int_{c-eta}^{c+eta}f^2(t)dt$$
$$>2etafrac{f^2(c)}{2}>0$$
this is a contradiction.
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
add a comment |
hint
Assume there exist $cin[a,b]$ such that
$$f^2(c)>0$$
with $epsilon=frac{f^2(c)}{2}$, and
by continuity at $c$, there exist $eta>0$ such that
$$forall xin[a,b]cap ]c-eta,c+eta[ $$
$$;; frac{f^2(c)}{2}<f^2(x)<frac{3f^2(c)}{2}$$
now observe that
$$0=int_a^bf^2(t)dtge int_{c-eta}^{c+eta}f^2(t)dt$$
$$>2etafrac{f^2(c)}{2}>0$$
this is a contradiction.
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
hint
Assume there exist $cin[a,b]$ such that
$$f^2(c)>0$$
with $epsilon=frac{f^2(c)}{2}$, and
by continuity at $c$, there exist $eta>0$ such that
$$forall xin[a,b]cap ]c-eta,c+eta[ $$
$$;; frac{f^2(c)}{2}<f^2(x)<frac{3f^2(c)}{2}$$
now observe that
$$0=int_a^bf^2(t)dtge int_{c-eta}^{c+eta}f^2(t)dt$$
$$>2etafrac{f^2(c)}{2}>0$$
this is a contradiction.
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
edited Nov 28 at 20:32
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
answered Nov 28 at 20:19
hamam_Abdallah
37.9k21634
37.9k21634
add a comment |
add a comment |
Is $f $ continuous ?
– hamam_Abdallah
Nov 28 at 20:10
You need to be more specific about your assumptions on $f$. This isn't true unless you add some conditions. Is $f$ continuous?
– Zestylemonzi
Nov 28 at 20:11
@hamam_Abdallah yes it is continuous, I just edited my problem
– Albert Diaz
Nov 28 at 20:12
If $f$ is not $0$ (say $f(c)neq 0$ where $cin (a,b)$), then there is a $varepsilon >0$ small s.t. $|f(x)|^2geq ell>0$ on $[c-varepsilon ,c+varepsilon ]$. Contradiction. If $f(a)neq 0$ or $f(b)neq 0$ the proof goes the same.
– Surb
Nov 28 at 20:13