Mixing
Proof that $fg geq 1 implies int_E f dmuint_E g dmu geq mu^2(E)$
$begingroup$
$(E,A,mu)$ is a finite measure space, $f$ and $g$ are both positive and measurable functions from $E$ to $mathbb{R}$ such that $fg geq 1$
by holder we have $$(int_E f^2 dmu)(int_E g^2 dmu) geq (int_E fg dmu)^2 geq (int_E dmu)^2 geq mu^2(E)$$
how about this one though : $int_E f dmuint_E g dmu geq mu^2(E)$
if $f = h^2, , g = w^2$ then $|h||w| geq 1$ and $|h|,,|w| geq 0$ and measurable therefore by what precedes
$int_E f dmuint_E g dmu = (int_E h^2 dmu)(int_E w^2 dmu) geq mu^2(E)$
Is my approach correct?
integration analysis measure-theory proof-verification
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
$endgroup$
add a comment |
$begingroup$
$(E,A,mu)$ is a finite measure space, $f$ and $g$ are both positive and measurable functions from $E$ to $mathbb{R}$ such that $fg geq 1$
by holder we have $$(int_E f^2 dmu)(int_E g^2 dmu) geq (int_E fg dmu)^2 geq (int_E dmu)^2 geq mu^2(E)$$
how about this one though : $int_E f dmuint_E g dmu geq mu^2(E)$
if $f = h^2, , g = w^2$ then $|h||w| geq 1$ and $|h|,,|w| geq 0$ and measurable therefore by what precedes
$int_E f dmuint_E g dmu = (int_E h^2 dmu)(int_E w^2 dmu) geq mu^2(E)$
Is my approach correct?
integration analysis measure-theory proof-verification
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
$endgroup$
add a comment |
$begingroup$
$(E,A,mu)$ is a finite measure space, $f$ and $g$ are both positive and measurable functions from $E$ to $mathbb{R}$ such that $fg geq 1$
by holder we have $$(int_E f^2 dmu)(int_E g^2 dmu) geq (int_E fg dmu)^2 geq (int_E dmu)^2 geq mu^2(E)$$
how about this one though : $int_E f dmuint_E g dmu geq mu^2(E)$
if $f = h^2, , g = w^2$ then $|h||w| geq 1$ and $|h|,,|w| geq 0$ and measurable therefore by what precedes
$int_E f dmuint_E g dmu = (int_E h^2 dmu)(int_E w^2 dmu) geq mu^2(E)$
Is my approach correct?
integration analysis measure-theory proof-verification
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
$endgroup$
$(E,A,mu)$ is a finite measure space, $f$ and $g$ are both positive and measurable functions from $E$ to $mathbb{R}$ such that $fg geq 1$
by holder we have $$(int_E f^2 dmu)(int_E g^2 dmu) geq (int_E fg dmu)^2 geq (int_E dmu)^2 geq mu^2(E)$$
how about this one though : $int_E f dmuint_E g dmu geq mu^2(E)$
if $f = h^2, , g = w^2$ then $|h||w| geq 1$ and $|h|,,|w| geq 0$ and measurable therefore by what precedes
$int_E f dmuint_E g dmu = (int_E h^2 dmu)(int_E w^2 dmu) geq mu^2(E)$
Is my approach correct?
integration analysis measure-theory proof-verification
integration analysis measure-theory proof-verification
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
edited Jan 7 at 23:27
Martin Sleziak
44.7k9117272
44.7k9117272
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
asked Jan 7 at 22:50
rapidracimrapidracim
1,5671319
1,5671319
add a comment |
add a comment |
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