Mixing
Lebesgue integrable function involving distance function
$begingroup$
$f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$
$M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$
Prove $M_f(x)in L(mathbb{R})$
My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.
real-analysis lebesgue-integral
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
asked Jan 10 at 12:08
NickNick
63
$endgroup$
add a comment |
$begingroup$
$f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$
$M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$
Prove $M_f(x)in L(mathbb{R})$
My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.
real-analysis lebesgue-integral
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
asked Jan 10 at 12:08
NickNick
63
$endgroup$
add a comment |
$begingroup$
$f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$
$M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$
Prove $M_f(x)in L(mathbb{R})$
My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.
real-analysis lebesgue-integral
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
asked Jan 10 at 12:08
NickNick
63
$endgroup$
$f$ is a bounded measurable function with compact support in $mathbb{R}, int_mathbb{R}f(x)dx=0$
$M_f(x)=sup_{t>0}{|int_mathbb{R}f(y)frac{t}{t^2+(x-y)^2}dy|}$
Prove $M_f(x)in L(mathbb{R})$
My attempt to solve the problem: I saw Lebesgue integral involving distance function. I tried to use Tonelli Theorem but didn't succeed.
real-analysis lebesgue-integral
real-analysis lebesgue-integral
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
asked Jan 10 at 12:08
NickNick
63
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
asked Jan 10 at 12:08
NickNick
63
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
edited Jan 14 at 14:03
David C. Ullrich
60.5k43994
60.5k43994
asked Jan 10 at 12:08
NickNick
63
asked Jan 10 at 12:08
NickNick
63
asked Jan 10 at 12:08
NickNick
63
63
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:
Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.
Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple
Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show
(i) $|M_f|lepi c_1$
(ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$
it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.
Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
\&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
\&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
$endgroup$
add a comment |
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$begingroup$
In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:
Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.
Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple
Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show
(i) $|M_f|lepi c_1$
(ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$
it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.
Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
\&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
\&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
$endgroup$
add a comment |
$begingroup$
In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:
Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.
Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple
Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show
(i) $|M_f|lepi c_1$
(ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$
it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.
Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
\&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
\&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
$endgroup$
add a comment |
$begingroup$
In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:
Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.
Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple
Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show
(i) $|M_f|lepi c_1$
(ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$
it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.
Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
\&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
\&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
$endgroup$
In case anyone wonders where the question comes from: It's a standard thing, essentially the trivial half of the "atomic decomposition" for the "real" Hardy space $H^1(Bbb R)$:
Theorem. Suppose $fin L^1(Bbb R)$. Then $M_f$ is inegrable if and only if $f=sum c_j f_j$ where $sum|c_j|<infty$, $f_j$ is supported in $[a_j,b_j]$,$||f_j||_inftyle1/(b_j-a_j)$ and $int f_j=0$.
Functions $f_j$ as above are "atoms"; for one direction it's enough to show that if $f$ is an atom then $||M_f||_1le c$, which is essentially proved below. The other direction is not so simple
Anyway: Say $|f|le c_1$ and $f$ is supported in $[-A,A]$. If you can show
(i) $|M_f|lepi c_1$
(ii) $|M_f(x)|le c_2/x^2$ for $|x|ge 2A$
it follows that $M_f$ is integrable. (i) should be easy. For (ii), note that since $int f=0$ you have $$int f(y)frac t{t^2+(x-y)^2}dy=
int_{-A}^A f(y)left(frac t{t^2+(x-y)^2}-a_{t,x}right)dy$$for any choice of constants $a_{t,x}$. I think $a_{t,x}=t/(t^2+x^2)$ works; subtract the fractions, and note that if $|y|le A$ and $|x|ge 2A$ then $|x-y|ge|x|/2$.
Edit: Since the OP seems to have lost interest I may as well fill in the details. Suppose $t>0$, $|y|le A$ and $x>2A$. s noted above, $x-yge x-x/2=x/2$. Now $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&=tfrac{|2xy-y^2|}{(t^2+(x-y)^2)(t^2+x^2)}
\&le tfrac{2Ax+A^2}{(t^2+(x-y)^2)(t^2+x^2)}.end{align}$$
Since $A^2le Ax/2$ this gives $$begin{align}left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|&le cfrac{tx}{(t^2+(x-y)^2)(t^2+x^2)}
\&le cfrac{tx}{(t^2+x^2)^2}.end{align}$$Now $x/(t^2+x^2)le x/x^2=1/x$. If $t>x$ then $t/(t^2+x^2)le t/t^2=1/tle 1/x$, whhile if $t<x$ then $t/(t^2+x^2)le x/x^2=1/x$; so we have $$left|frac t{t^2+(x-y)^2}-frac t{t^2+x^2}right|lefrac{c}{x^2}quad(x>2A, |y|le A).$$Similarly for $x<-2A$; hence $$M_f(x)lefrac{c||f||_1}{x^2}quad(|x|>2A),$$as required.
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
edited Jan 14 at 14:23
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
answered Jan 10 at 15:01
David C. UllrichDavid C. Ullrich
60.5k43994
60.5k43994
add a comment |
add a comment |
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