Help in proving this statement (Analysis 2)
$begingroup$
Let $Omega$ be a limited domain of $mathbb{R}^n$.
Suppose $u$ to be measurable over $Omega$, and suppose that for every $p geq 1$
$$|u|^p in L^1(Omega)$$
Now, defined
$$phi_p(u) = left(frac{1}{text{measure}(Omega)}int_{Omega}|u|^p text{d}xright)^{1/p}$$
prove that for every $(p, q) geq 1$ with $p leq q$ we have
$$Phi_p(u) leq Phi_q(u)$$
Any hint? Thank you so much!
real-analysis calculus functional-analysis multivariable-calculus
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a limited domain of $mathbb{R}^n$.
Suppose $u$ to be measurable over $Omega$, and suppose that for every $p geq 1$
$$|u|^p in L^1(Omega)$$
Now, defined
$$phi_p(u) = left(frac{1}{text{measure}(Omega)}int_{Omega}|u|^p text{d}xright)^{1/p}$$
prove that for every $(p, q) geq 1$ with $p leq q$ we have
$$Phi_p(u) leq Phi_q(u)$$
Any hint? Thank you so much!
real-analysis calculus functional-analysis multivariable-calculus
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a limited domain of $mathbb{R}^n$.
Suppose $u$ to be measurable over $Omega$, and suppose that for every $p geq 1$
$$|u|^p in L^1(Omega)$$
Now, defined
$$phi_p(u) = left(frac{1}{text{measure}(Omega)}int_{Omega}|u|^p text{d}xright)^{1/p}$$
prove that for every $(p, q) geq 1$ with $p leq q$ we have
$$Phi_p(u) leq Phi_q(u)$$
Any hint? Thank you so much!
real-analysis calculus functional-analysis multivariable-calculus
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
$endgroup$
Let $Omega$ be a limited domain of $mathbb{R}^n$.
Suppose $u$ to be measurable over $Omega$, and suppose that for every $p geq 1$
$$|u|^p in L^1(Omega)$$
Now, defined
$$phi_p(u) = left(frac{1}{text{measure}(Omega)}int_{Omega}|u|^p text{d}xright)^{1/p}$$
prove that for every $(p, q) geq 1$ with $p leq q$ we have
$$Phi_p(u) leq Phi_q(u)$$
Any hint? Thank you so much!
real-analysis calculus functional-analysis multivariable-calculus
real-analysis calculus functional-analysis multivariable-calculus
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
asked Dec 26 '18 at 19:59
Von NeumannVon Neumann
16.5k72545
16.5k72545
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint:
If $p = q$, there is nothing to prove, so suppose $p < q$. Apply Hölder's inequality with conjugate exponents $q/p$ and $q/(q-p)$.
edited Feb 8 at 10:24
Von Neumann
16.5k72545
answered Dec 26 '18 at 20:10
kobekobe
35k22248
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint:
If $p = q$, there is nothing to prove, so suppose $p < q$. Apply Hölder's inequality with conjugate exponents $q/p$ and $q/(q-p)$.
edited Feb 8 at 10:24
Von Neumann
16.5k72545
answered Dec 26 '18 at 20:10
kobekobe
35k22248
$endgroup$
add a comment |
$begingroup$
Hint:
If $p = q$, there is nothing to prove, so suppose $p < q$. Apply Hölder's inequality with conjugate exponents $q/p$ and $q/(q-p)$.
edited Feb 8 at 10:24
Von Neumann
16.5k72545
answered Dec 26 '18 at 20:10
kobekobe
35k22248
$endgroup$
add a comment |
$begingroup$
Hint:
If $p = q$, there is nothing to prove, so suppose $p < q$. Apply Hölder's inequality with conjugate exponents $q/p$ and $q/(q-p)$.
edited Feb 8 at 10:24
Von Neumann
16.5k72545
answered Dec 26 '18 at 20:10
kobekobe
35k22248
$endgroup$
Hint:
If $p = q$, there is nothing to prove, so suppose $p < q$. Apply Hölder's inequality with conjugate exponents $q/p$ and $q/(q-p)$.
edited Feb 8 at 10:24
Von Neumann
16.5k72545
answered Dec 26 '18 at 20:10
kobekobe
35k22248
edited Feb 8 at 10:24
Von Neumann
16.5k72545
edited Feb 8 at 10:24
Von Neumann
16.5k72545
edited Feb 8 at 10:24
Von Neumann
16.5k72545
16.5k72545
answered Dec 26 '18 at 20:10
kobekobe
35k22248
answered Dec 26 '18 at 20:10
kobekobe
35k22248
answered Dec 26 '18 at 20:10
kobekobe
35k22248
35k22248
add a comment |
add a comment |
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