Composition of convex and $L^1$ function is not $L^1$ function.
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Give an example of functions $fin L^1$ and $varphi$ is convex such that $varphicirc fnotin L^1$ but $$varphi left(int limits_{X}f dmuright)leq intlimits_{X} varphi circ f dmu.$$ I suspect that $f$ should be some function which blows up and $varphi$ some power of its. But I am not able to make it rigorous. Can anyone help with this example, please.
real-analysis
asked Nov 15 at 19:03
ZFR
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up vote
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down vote
favorite
Give an example of functions $fin L^1$ and $varphi$ is convex such that $varphicirc fnotin L^1$ but $$varphi left(int limits_{X}f dmuright)leq intlimits_{X} varphi circ f dmu.$$ I suspect that $f$ should be some function which blows up and $varphi$ some power of its. But I am not able to make it rigorous. Can anyone help with this example, please.
real-analysis
asked Nov 15 at 19:03
ZFR
4,89831337
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Give an example of functions $fin L^1$ and $varphi$ is convex such that $varphicirc fnotin L^1$ but $$varphi left(int limits_{X}f dmuright)leq intlimits_{X} varphi circ f dmu.$$ I suspect that $f$ should be some function which blows up and $varphi$ some power of its. But I am not able to make it rigorous. Can anyone help with this example, please.
real-analysis
asked Nov 15 at 19:03
ZFR
4,89831337
Give an example of functions $fin L^1$ and $varphi$ is convex such that $varphicirc fnotin L^1$ but $$varphi left(int limits_{X}f dmuright)leq intlimits_{X} varphi circ f dmu.$$ I suspect that $f$ should be some function which blows up and $varphi$ some power of its. But I am not able to make it rigorous. Can anyone help with this example, please.
real-analysis
real-analysis
asked Nov 15 at 19:03
ZFR
4,89831337
asked Nov 15 at 19:03
ZFR
4,89831337
asked Nov 15 at 19:03
ZFR
4,89831337
asked Nov 15 at 19:03
ZFR
4,89831337
asked Nov 15 at 19:03
ZFR
4,89831337
4,89831337
add a comment |
add a comment |
1 Answer
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How about $X = [0,1]$, $mu$ is Lebesgue measure, $f(x) = dfrac 1{sqrt x}$, and $phi(x) = x^2$.
answered Nov 15 at 19:07
Umberto P.
38.1k13063
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
How about $X = [0,1]$, $mu$ is Lebesgue measure, $f(x) = dfrac 1{sqrt x}$, and $phi(x) = x^2$.
answered Nov 15 at 19:07
Umberto P.
38.1k13063
add a comment |
up vote
0
down vote
How about $X = [0,1]$, $mu$ is Lebesgue measure, $f(x) = dfrac 1{sqrt x}$, and $phi(x) = x^2$.
answered Nov 15 at 19:07
Umberto P.
38.1k13063
add a comment |
up vote
0
down vote
up vote
0
down vote
How about $X = [0,1]$, $mu$ is Lebesgue measure, $f(x) = dfrac 1{sqrt x}$, and $phi(x) = x^2$.
answered Nov 15 at 19:07
Umberto P.
38.1k13063
How about $X = [0,1]$, $mu$ is Lebesgue measure, $f(x) = dfrac 1{sqrt x}$, and $phi(x) = x^2$.
answered Nov 15 at 19:07
Umberto P.
38.1k13063
answered Nov 15 at 19:07
Umberto P.
38.1k13063
answered Nov 15 at 19:07
Umberto P.
38.1k13063
answered Nov 15 at 19:07
Umberto P.
38.1k13063
38.1k13063
add a comment |
add a comment |
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