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Proving a product of two functions is integrable over [0,1] when one of them isn't
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Let $f$ be a function defined on [0, 1] that is integrable over [0, 1], differentiable at x = 0, with f(0) = 0. Define the piecewise function g(x) to be $x^-3/2 f(x)$ if $0 < x leq 1$ and 0 if $x = 0$. Prove $g$ is integrable over [0, 1].
The fact that the exponent on $x$ is less than -1 is tripping me up here. I know that in general, $x^alpha$ is integrable over (0, 1] only when $alpha > -1$, so if I had an exponent in that range, I would have a product of integrable functions. But as it is the integral diverges. I can't apply Holder's inequality either, since clearly $x^-3/2$ is not essentially bounded. Am I missing something, or might the problem have a typo in the exponent?
real-analysis lebesgue-integral
asked Jul 25 at 19:52
BMac
24517
add a comment |Â
up vote
1
down vote
favorite
Let $f$ be a function defined on [0, 1] that is integrable over [0, 1], differentiable at x = 0, with f(0) = 0. Define the piecewise function g(x) to be $x^-3/2 f(x)$ if $0 < x leq 1$ and 0 if $x = 0$. Prove $g$ is integrable over [0, 1].
The fact that the exponent on $x$ is less than -1 is tripping me up here. I know that in general, $x^alpha$ is integrable over (0, 1] only when $alpha > -1$, so if I had an exponent in that range, I would have a product of integrable functions. But as it is the integral diverges. I can't apply Holder's inequality either, since clearly $x^-3/2$ is not essentially bounded. Am I missing something, or might the problem have a typo in the exponent?
real-analysis lebesgue-integral
asked Jul 25 at 19:52
BMac
24517
A hint: use differentiability of $f$ at $x=0$.
– user539887
Jul 25 at 19:59
Hmmm. So this implies f is continuous on [0, 1], specifically is approaching 0 as x approaches 0, yes? So do I have some general result that I'm forgetting when one of the functions in a product is continuous?
– BMac
Jul 25 at 20:35
Differentiability at $0$ implies continuity at $0$, not necessarily on $[0,1]$.
– user539887
Jul 26 at 7:08
I think that Daniel Fischer gave in his comment a full answer.
– user539887
Jul 26 at 9:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f$ be a function defined on [0, 1] that is integrable over [0, 1], differentiable at x = 0, with f(0) = 0. Define the piecewise function g(x) to be $x^-3/2 f(x)$ if $0 < x leq 1$ and 0 if $x = 0$. Prove $g$ is integrable over [0, 1].
The fact that the exponent on $x$ is less than -1 is tripping me up here. I know that in general, $x^alpha$ is integrable over (0, 1] only when $alpha > -1$, so if I had an exponent in that range, I would have a product of integrable functions. But as it is the integral diverges. I can't apply Holder's inequality either, since clearly $x^-3/2$ is not essentially bounded. Am I missing something, or might the problem have a typo in the exponent?
real-analysis lebesgue-integral
asked Jul 25 at 19:52
BMac
24517
Let $f$ be a function defined on [0, 1] that is integrable over [0, 1], differentiable at x = 0, with f(0) = 0. Define the piecewise function g(x) to be $x^-3/2 f(x)$ if $0 < x leq 1$ and 0 if $x = 0$. Prove $g$ is integrable over [0, 1].
The fact that the exponent on $x$ is less than -1 is tripping me up here. I know that in general, $x^alpha$ is integrable over (0, 1] only when $alpha > -1$, so if I had an exponent in that range, I would have a product of integrable functions. But as it is the integral diverges. I can't apply Holder's inequality either, since clearly $x^-3/2$ is not essentially bounded. Am I missing something, or might the problem have a typo in the exponent?
real-analysis lebesgue-integral
asked Jul 25 at 19:52
BMac
24517
asked Jul 25 at 19:52
BMac
24517
asked Jul 25 at 19:52
BMac
24517
asked Jul 25 at 19:52
BMac
24517
24517
A hint: use differentiability of $f$ at $x=0$.
– user539887
Jul 25 at 19:59
Hmmm. So this implies f is continuous on [0, 1], specifically is approaching 0 as x approaches 0, yes? So do I have some general result that I'm forgetting when one of the functions in a product is continuous?
– BMac
Jul 25 at 20:35
Differentiability at $0$ implies continuity at $0$, not necessarily on $[0,1]$.
– user539887
Jul 26 at 7:08
I think that Daniel Fischer gave in his comment a full answer.
– user539887
Jul 26 at 9:33
add a comment |Â
A hint: use differentiability of $f$ at $x=0$.
– user539887
Jul 25 at 19:59
Hmmm. So this implies f is continuous on [0, 1], specifically is approaching 0 as x approaches 0, yes? So do I have some general result that I'm forgetting when one of the functions in a product is continuous?
– BMac
Jul 25 at 20:35
Differentiability at $0$ implies continuity at $0$, not necessarily on $[0,1]$.
– user539887
Jul 26 at 7:08
I think that Daniel Fischer gave in his comment a full answer.
– user539887
Jul 26 at 9:33
A hint: use differentiability of $f$ at $x=0$.
– user539887
Jul 25 at 19:59
A hint: use differentiability of $f$ at $x=0$.
– user539887
Jul 25 at 19:59
Hmmm. So this implies f is continuous on [0, 1], specifically is approaching 0 as x approaches 0, yes? So do I have some general result that I'm forgetting when one of the functions in a product is continuous?
– BMac
Jul 25 at 20:35
Hmmm. So this implies f is continuous on [0, 1], specifically is approaching 0 as x approaches 0, yes? So do I have some general result that I'm forgetting when one of the functions in a product is continuous?
– BMac
Jul 25 at 20:35
Differentiability at $0$ implies continuity at $0$, not necessarily on $[0,1]$.
– user539887
Jul 26 at 7:08
Differentiability at $0$ implies continuity at $0$, not necessarily on $[0,1]$.
– user539887
Jul 26 at 7:08
I think that Daniel Fischer gave in his comment a full answer.
– user539887
Jul 26 at 9:33
I think that Daniel Fischer gave in his comment a full answer.
– user539887
Jul 26 at 9:33
add a comment |Â
1 Answer
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Putting this here since it's too long for a comment. This is only a rough outline of an approach, you have to fill in the details yourself.
I asssume you talk about Riemann integrability here. That means $f$ is bounded and by the Lebesgue criteria it is continous almost everywhere. Now use the fact's that $f(0)=0$ and $f$ is differentiable at $0$ to show that the function
$$ h(x)=dfracf(x)x $$
is integrable. Hint: It is bounded on $[0,1]$ since it is continous at $0$. And clearly it is continous almost everywhere since $f$ is. Now use Lebesgue's criteria again.
But now $g(x)=x^-1/2h(x)$, can you carry on from here?
answered Jul 25 at 21:33
Maik Pickl
1,577316
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
4
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Putting this here since it's too long for a comment. This is only a rough outline of an approach, you have to fill in the details yourself.
I asssume you talk about Riemann integrability here. That means $f$ is bounded and by the Lebesgue criteria it is continous almost everywhere. Now use the fact's that $f(0)=0$ and $f$ is differentiable at $0$ to show that the function
$$ h(x)=dfracf(x)x $$
is integrable. Hint: It is bounded on $[0,1]$ since it is continous at $0$. And clearly it is continous almost everywhere since $f$ is. Now use Lebesgue's criteria again.
But now $g(x)=x^-1/2h(x)$, can you carry on from here?
answered Jul 25 at 21:33
Maik Pickl
1,577316
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
4
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
 |Â
show 3 more comments
up vote
0
down vote
Putting this here since it's too long for a comment. This is only a rough outline of an approach, you have to fill in the details yourself.
I asssume you talk about Riemann integrability here. That means $f$ is bounded and by the Lebesgue criteria it is continous almost everywhere. Now use the fact's that $f(0)=0$ and $f$ is differentiable at $0$ to show that the function
$$ h(x)=dfracf(x)x $$
is integrable. Hint: It is bounded on $[0,1]$ since it is continous at $0$. And clearly it is continous almost everywhere since $f$ is. Now use Lebesgue's criteria again.
But now $g(x)=x^-1/2h(x)$, can you carry on from here?
answered Jul 25 at 21:33
Maik Pickl
1,577316
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
4
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
 |Â
show 3 more comments
up vote
0
down vote
up vote
0
down vote
Putting this here since it's too long for a comment. This is only a rough outline of an approach, you have to fill in the details yourself.
I asssume you talk about Riemann integrability here. That means $f$ is bounded and by the Lebesgue criteria it is continous almost everywhere. Now use the fact's that $f(0)=0$ and $f$ is differentiable at $0$ to show that the function
$$ h(x)=dfracf(x)x $$
is integrable. Hint: It is bounded on $[0,1]$ since it is continous at $0$. And clearly it is continous almost everywhere since $f$ is. Now use Lebesgue's criteria again.
But now $g(x)=x^-1/2h(x)$, can you carry on from here?
answered Jul 25 at 21:33
Maik Pickl
1,577316
Putting this here since it's too long for a comment. This is only a rough outline of an approach, you have to fill in the details yourself.
I asssume you talk about Riemann integrability here. That means $f$ is bounded and by the Lebesgue criteria it is continous almost everywhere. Now use the fact's that $f(0)=0$ and $f$ is differentiable at $0$ to show that the function
$$ h(x)=dfracf(x)x $$
is integrable. Hint: It is bounded on $[0,1]$ since it is continous at $0$. And clearly it is continous almost everywhere since $f$ is. Now use Lebesgue's criteria again.
But now $g(x)=x^-1/2h(x)$, can you carry on from here?
answered Jul 25 at 21:33
Maik Pickl
1,577316
answered Jul 25 at 21:33
Maik Pickl
1,577316
answered Jul 25 at 21:33
Maik Pickl
1,577316
answered Jul 25 at 21:33
Maik Pickl
1,577316
1,577316
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
4
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
 |Â
show 3 more comments
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
4
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Actually, the problem isn't clear if it means Riemann or Lebesgue integrable. It just says "integrable," which could mean Lebesgue, in which case your answer wouldn't work if $f$ is unbounded.
– BMac
Jul 25 at 21:39
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
Indeed, I leave it here anyway. Maybe someone else can give an explanation for the Lebesgue integrable case.
– Maik Pickl
Jul 25 at 21:41
4
4
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
@BMac Pick $varepsilon > 0$ such that $lvert f(x) - f(0)rvert leqslant (1 + lvert f'(0)rvert)cdot x$ for $0 leqslant varepsilon$. On $[varepsilon, 1]$, the factor $x^-3/2$ is bounded.
– Daniel Fischer♦
Jul 25 at 21:50
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
So that whole chain $|f(x) - f(0)| leq (1 + |f'(0)|) cdot x$ should end with less than $epsilon$? Or where does $epsilon$ come in there?
– BMac
Jul 25 at 22:05
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
@MaikPickl There's no way this can work in the realm of the Riemann integral. $f(x) = x$ is a counterexample.
– zhw.
Jul 26 at 0:12
 |Â
show 3 more comments
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A hint: use differentiability of $f$ at $x=0$.
– user539887
Jul 25 at 19:59
Hmmm. So this implies f is continuous on [0, 1], specifically is approaching 0 as x approaches 0, yes? So do I have some general result that I'm forgetting when one of the functions in a product is continuous?
– BMac
Jul 25 at 20:35
Differentiability at $0$ implies continuity at $0$, not necessarily on $[0,1]$.
– user539887
Jul 26 at 7:08
I think that Daniel Fischer gave in his comment a full answer.
– user539887
Jul 26 at 9:33