Proving that $L^p(mathbbR^d)notsubset L^infty(mathbbR^d)$ for $1leq p<infty.$Prove that the indicator function for $mathbbQcap[0,1]$ is not Riemann integrableDefine $f$ on $mathbbR$ by $f(x)=sqrtn$ if $nle x le n+(1/n^2))$ for $n=1,2,ldots$ and $f(x)=0$ otherwise.Show that $gin L^1(mathbbR)$ but $g^2notin L^1(mathbbR)$.Show that $f$ is NOT Lebesgue integrable.Converse of Lebesgue Monotone Convergence TheoremShow that the sequence $x_n_n=1^infty$ does not converge in $ ell^2 (mathbbN). $show that $int_(0,1) f dx = infty$From Donaldson's Riemann Surfaces- Existence of non-vanishing $C^infty$ function $ f(t)= 1$ for small $t$, $f(t)=frac1t$for large $t$Prove that $sum_k=1^infty a_k g(x-r_k) notin L^2(mathbbR)$Prove that if $fin L^p(E), 1leq p<infty, m(E)<infty$, then $fin L^q(E)$ for all $1leq qleq p$
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Proving that $L^p(mathbbR^d)notsubset L^infty(mathbbR^d)$ for $1leq p
Prove that the indicator function for $mathbbQcap[0,1]$ is not Riemann integrableDefine $f$ on $mathbbR$ by $f(x)=sqrtn$ if $nle x le n+(1/n^2))$ for $n=1,2,ldots$ and $f(x)=0$ otherwise.Show that $gin L^1(mathbbR)$ but $g^2notin L^1(mathbbR)$.Show that $f$ is NOT Lebesgue integrable.Converse of Lebesgue Monotone Convergence TheoremShow that the sequence $x_n_n=1^infty$ does not converge in $ ell^2 (mathbbN). $show that $int_(0,1) f dx = infty$From Donaldson's Riemann Surfaces- Existence of non-vanishing $C^infty$ function $ f(t)= 1$ for small $t$, $f(t)=frac1t$for large $t$Prove that $sum_k=1^infty a_k g(x-r_k) notin L^2(mathbbR)$Prove that if $fin L^p(E), 1leq p<infty, m(E)<infty$, then $fin L^q(E)$ for all $1leq qleq p$
$begingroup$
Show that $L^p(mathbbR^d)notsubset L^infty(mathbbR^d)$ for all $pin[1,infty).$
My attempt at this problem:
I define the following function for $xinmathbbR^d,$
$$f_a(x)=begincasesvert xvert^a&textif 0<vert xvert<1\0&textotherwise. endcases$$
I am able to prove that this function is in $L^1(mathbbR^d)$ if and only if $a>-d.$ So I define another function
$$f(x)=begincasesvert xvert^-d/(p+1)&textif 0<vert xvert<1\0&textotherwise. endcases$$
Then since $-dp/(p+1)>-d,$ we have
$$large|f|_p^p=int_mathbbR^dvert xvert^-dp/(p+1),dx<infty,$$
so that $fin L^p(mathbbR^d).$
My difficulty is with showing that $fnotin L^infty(mathbbR^d),$ since for any $M>0$ we have that $f>M$ on a set not of measure zero. However we can choose a larger $M$, and then the measure of the set where $f$ is not bounded decreases.
I my reasoning above correct, and $f$ actually works, or am I missundertanding the notion of the $||_infty$ norm?
Appreciate any feedback.
Thank you for your time.
real-analysis measure-theory lebesgue-integral lp-spaces
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
$endgroup$
add a comment |
$begingroup$
Show that $L^p(mathbbR^d)notsubset L^infty(mathbbR^d)$ for all $pin[1,infty).$
My attempt at this problem:
I define the following function for $xinmathbbR^d,$
$$f_a(x)=begincasesvert xvert^a&textif 0<vert xvert<1\0&textotherwise. endcases$$
I am able to prove that this function is in $L^1(mathbbR^d)$ if and only if $a>-d.$ So I define another function
$$f(x)=begincasesvert xvert^-d/(p+1)&textif 0<vert xvert<1\0&textotherwise. endcases$$
Then since $-dp/(p+1)>-d,$ we have
$$large|f|_p^p=int_mathbbR^dvert xvert^-dp/(p+1),dx<infty,$$
so that $fin L^p(mathbbR^d).$
My difficulty is with showing that $fnotin L^infty(mathbbR^d),$ since for any $M>0$ we have that $f>M$ on a set not of measure zero. However we can choose a larger $M$, and then the measure of the set where $f$ is not bounded decreases.
I my reasoning above correct, and $f$ actually works, or am I missundertanding the notion of the $||_infty$ norm?
Appreciate any feedback.
Thank you for your time.
real-analysis measure-theory lebesgue-integral lp-spaces
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
$endgroup$
add a comment |
$begingroup$
Show that $L^p(mathbbR^d)notsubset L^infty(mathbbR^d)$ for all $pin[1,infty).$
My attempt at this problem:
I define the following function for $xinmathbbR^d,$
$$f_a(x)=begincasesvert xvert^a&textif 0<vert xvert<1\0&textotherwise. endcases$$
I am able to prove that this function is in $L^1(mathbbR^d)$ if and only if $a>-d.$ So I define another function
$$f(x)=begincasesvert xvert^-d/(p+1)&textif 0<vert xvert<1\0&textotherwise. endcases$$
Then since $-dp/(p+1)>-d,$ we have
$$large|f|_p^p=int_mathbbR^dvert xvert^-dp/(p+1),dx<infty,$$
so that $fin L^p(mathbbR^d).$
My difficulty is with showing that $fnotin L^infty(mathbbR^d),$ since for any $M>0$ we have that $f>M$ on a set not of measure zero. However we can choose a larger $M$, and then the measure of the set where $f$ is not bounded decreases.
I my reasoning above correct, and $f$ actually works, or am I missundertanding the notion of the $||_infty$ norm?
Appreciate any feedback.
Thank you for your time.
real-analysis measure-theory lebesgue-integral lp-spaces
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
$endgroup$
Show that $L^p(mathbbR^d)notsubset L^infty(mathbbR^d)$ for all $pin[1,infty).$
My attempt at this problem:
I define the following function for $xinmathbbR^d,$
$$f_a(x)=begincasesvert xvert^a&textif 0<vert xvert<1\0&textotherwise. endcases$$
I am able to prove that this function is in $L^1(mathbbR^d)$ if and only if $a>-d.$ So I define another function
$$f(x)=begincasesvert xvert^-d/(p+1)&textif 0<vert xvert<1\0&textotherwise. endcases$$
Then since $-dp/(p+1)>-d,$ we have
$$large|f|_p^p=int_mathbbR^dvert xvert^-dp/(p+1),dx<infty,$$
so that $fin L^p(mathbbR^d).$
My difficulty is with showing that $fnotin L^infty(mathbbR^d),$ since for any $M>0$ we have that $f>M$ on a set not of measure zero. However we can choose a larger $M$, and then the measure of the set where $f$ is not bounded decreases.
I my reasoning above correct, and $f$ actually works, or am I missundertanding the notion of the $||_infty$ norm?
Appreciate any feedback.
Thank you for your time.
real-analysis measure-theory lebesgue-integral lp-spaces
real-analysis measure-theory lebesgue-integral lp-spaces
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
edited 13 hours ago
Gaby Alfonso
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
asked 14 hours ago
Gaby AlfonsoGaby Alfonso
1,127317
1,127317
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $f in L^infty$ the there exists a constant $M in (0,infty)$ such that $|f(x)| leq M$ almost everywhere. This gives $|x|^-d/(p+1) leq M$ for almost all points $x$ in the region $0<|x|<1$. Actually, since LHS is continuous the inequality must hold for all $x$ in the region $0<|x|<1$. Can you get a contradiction from here by taking $x$ close to the origin?
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
add a comment |
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1 Answer
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$begingroup$
If $f in L^infty$ the there exists a constant $M in (0,infty)$ such that $|f(x)| leq M$ almost everywhere. This gives $|x|^-d/(p+1) leq M$ for almost all points $x$ in the region $0<|x|<1$. Actually, since LHS is continuous the inequality must hold for all $x$ in the region $0<|x|<1$. Can you get a contradiction from here by taking $x$ close to the origin?
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
add a comment |
$begingroup$
If $f in L^infty$ the there exists a constant $M in (0,infty)$ such that $|f(x)| leq M$ almost everywhere. This gives $|x|^-d/(p+1) leq M$ for almost all points $x$ in the region $0<|x|<1$. Actually, since LHS is continuous the inequality must hold for all $x$ in the region $0<|x|<1$. Can you get a contradiction from here by taking $x$ close to the origin?
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
add a comment |
$begingroup$
If $f in L^infty$ the there exists a constant $M in (0,infty)$ such that $|f(x)| leq M$ almost everywhere. This gives $|x|^-d/(p+1) leq M$ for almost all points $x$ in the region $0<|x|<1$. Actually, since LHS is continuous the inequality must hold for all $x$ in the region $0<|x|<1$. Can you get a contradiction from here by taking $x$ close to the origin?
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
$endgroup$
If $f in L^infty$ the there exists a constant $M in (0,infty)$ such that $|f(x)| leq M$ almost everywhere. This gives $|x|^-d/(p+1) leq M$ for almost all points $x$ in the region $0<|x|<1$. Actually, since LHS is continuous the inequality must hold for all $x$ in the region $0<|x|<1$. Can you get a contradiction from here by taking $x$ close to the origin?
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
answered 14 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.6k42766
65.6k42766
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
add a comment |
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Thank you for your help! Choosing $vert xvert<M^-(p+1)/d$ yields the contradiction, since then $M<vert xvert^-d/(p+1)$, right?
$endgroup$
– Gaby Alfonso
13 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
$begingroup$
Yes, that's right.
$endgroup$
– Kavi Rama Murthy
12 hours ago
add a comment |
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