Let $I=[0, infty)$ and $f:I to R$ a mensurable function such that $|f(t)| leq frac{t^alpha}{1+t}$, where $0...
$begingroup$
Let $I=[0, infty)$ and $f:I to mathbb{R}$ be a mensurable function such that $|f(t)| leq frac{t^alpha}{1+t}$, where $0 <alpha <1$. Show that the function $e^{-tx}f(t)$ is integrable in $I times I$.
I was trying to show that the module is integrable using Tonelli, but I did not succeed.
real-analysis calculus lebesgue-integral
asked Dec 3 '18 at 14:00
LucasLucas
275
$endgroup$
add a comment |
$begingroup$
Let $I=[0, infty)$ and $f:I to mathbb{R}$ be a mensurable function such that $|f(t)| leq frac{t^alpha}{1+t}$, where $0 <alpha <1$. Show that the function $e^{-tx}f(t)$ is integrable in $I times I$.
I was trying to show that the module is integrable using Tonelli, but I did not succeed.
real-analysis calculus lebesgue-integral
asked Dec 3 '18 at 14:00
LucasLucas
275
$endgroup$
add a comment |
$begingroup$
Let $I=[0, infty)$ and $f:I to mathbb{R}$ be a mensurable function such that $|f(t)| leq frac{t^alpha}{1+t}$, where $0 <alpha <1$. Show that the function $e^{-tx}f(t)$ is integrable in $I times I$.
I was trying to show that the module is integrable using Tonelli, but I did not succeed.
real-analysis calculus lebesgue-integral
asked Dec 3 '18 at 14:00
LucasLucas
275
$endgroup$
Let $I=[0, infty)$ and $f:I to mathbb{R}$ be a mensurable function such that $|f(t)| leq frac{t^alpha}{1+t}$, where $0 <alpha <1$. Show that the function $e^{-tx}f(t)$ is integrable in $I times I$.
I was trying to show that the module is integrable using Tonelli, but I did not succeed.
real-analysis calculus lebesgue-integral
real-analysis calculus lebesgue-integral
asked Dec 3 '18 at 14:00
LucasLucas
275
asked Dec 3 '18 at 14:00
LucasLucas
275
asked Dec 3 '18 at 14:00
LucasLucas
275
asked Dec 3 '18 at 14:00
LucasLucas
275
asked Dec 3 '18 at 14:00
LucasLucas
275
275
add a comment |
add a comment |
1 Answer
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$begingroup$
Just calculate:
$int_I{int_I{e^{-tx}frac{t^alpha}{1+t}dx}dt} = int_I{frac{t^alpha}{1+t}int_I{e^{-tx}dx}dt} = int_I{frac{t^alpha}{1+t}frac{1}{t}dt} < int_{0}^{1}{t^{alpha-1}dt} + int_{1}^{infty}{t^{alpha-2}dt} = frac{1}{alpha} + frac{1}{1-alpha} < infty. $
Note that I used inequality $1+t>1, 1+t>t$ respectively for two divided interval $(0,1)$ and $(1,infty)$ for $t.$
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Just calculate:
$int_I{int_I{e^{-tx}frac{t^alpha}{1+t}dx}dt} = int_I{frac{t^alpha}{1+t}int_I{e^{-tx}dx}dt} = int_I{frac{t^alpha}{1+t}frac{1}{t}dt} < int_{0}^{1}{t^{alpha-1}dt} + int_{1}^{infty}{t^{alpha-2}dt} = frac{1}{alpha} + frac{1}{1-alpha} < infty. $
Note that I used inequality $1+t>1, 1+t>t$ respectively for two divided interval $(0,1)$ and $(1,infty)$ for $t.$
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
$endgroup$
add a comment |
$begingroup$
Just calculate:
$int_I{int_I{e^{-tx}frac{t^alpha}{1+t}dx}dt} = int_I{frac{t^alpha}{1+t}int_I{e^{-tx}dx}dt} = int_I{frac{t^alpha}{1+t}frac{1}{t}dt} < int_{0}^{1}{t^{alpha-1}dt} + int_{1}^{infty}{t^{alpha-2}dt} = frac{1}{alpha} + frac{1}{1-alpha} < infty. $
Note that I used inequality $1+t>1, 1+t>t$ respectively for two divided interval $(0,1)$ and $(1,infty)$ for $t.$
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
$endgroup$
add a comment |
$begingroup$
Just calculate:
$int_I{int_I{e^{-tx}frac{t^alpha}{1+t}dx}dt} = int_I{frac{t^alpha}{1+t}int_I{e^{-tx}dx}dt} = int_I{frac{t^alpha}{1+t}frac{1}{t}dt} < int_{0}^{1}{t^{alpha-1}dt} + int_{1}^{infty}{t^{alpha-2}dt} = frac{1}{alpha} + frac{1}{1-alpha} < infty. $
Note that I used inequality $1+t>1, 1+t>t$ respectively for two divided interval $(0,1)$ and $(1,infty)$ for $t.$
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
$endgroup$
Just calculate:
$int_I{int_I{e^{-tx}frac{t^alpha}{1+t}dx}dt} = int_I{frac{t^alpha}{1+t}int_I{e^{-tx}dx}dt} = int_I{frac{t^alpha}{1+t}frac{1}{t}dt} < int_{0}^{1}{t^{alpha-1}dt} + int_{1}^{infty}{t^{alpha-2}dt} = frac{1}{alpha} + frac{1}{1-alpha} < infty. $
Note that I used inequality $1+t>1, 1+t>t$ respectively for two divided interval $(0,1)$ and $(1,infty)$ for $t.$
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
answered Dec 3 '18 at 14:30
EuduardoEuduardo
1288
1288
add a comment |
add a comment |
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