Proof that Good Kernels are Approximations of Identity in $L^p(mathbb R^d)$ The 2019 Stack Overflow Developer Survey Results Are InHow show that $lim_varepsilon rightarrow 0int_A h_varepsilon(x)dx =0$, whenever $bigg|int_I hbigg|leq |I|^1/2$?Is $f(x)exp(-x^2)$ summable if $f$ is square summable?Upper bound for the norm of convolutions: $Vert f_1astcdotsast f_NVert_rleqVert f_1Vert_p_1cdotsVert f_NVert_p_N$Prove characterization of the finiteness of $E left[ |X| right]$ by the behaviour of expectations $E left[ |X| mathbb1_Aright]$Approximation by convolution and cut-off functionpointwise convergence of mollification combined with a cut-off functionCheck proof that $int_0^inftyleft(int_x^infty|f(t)|dtright)^px^b-1dxleleft(frac pbright)^pint_0^infty|f(t)|^pt^p+b-1dt$Prove that if $fin L^p(E), 1leq p<infty, m(E)<infty$, then $fin L^q(E)$ for all $1leq qleq p$Convolution of $L^1$ functions is well-definedInequality for bounded locally integrable functions
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Proof that Good Kernels are Approximations of Identity in $L^p(mathbb R^d)$
The 2019 Stack Overflow Developer Survey Results Are InHow show that $lim_varepsilon rightarrow 0int_A h_varepsilon(x)dx =0$, whenever $bigg|int_I hbigg|leq |I|^1/2$?Is $f(x)exp(-x^2)$ summable if $f$ is square summable?Upper bound for the norm of convolutions: $Vert f_1astcdotsast f_NVert_rleqVert f_1Vert_p_1cdotsVert f_NVert_p_N$Prove characterization of the finiteness of $E left[ |X| right]$ by the behaviour of expectations $E left[ |X| mathbb1_Aright]$Approximation by convolution and cut-off functionpointwise convergence of mollification combined with a cut-off functionCheck proof that $int_0^inftyleft(int_x^infty|f(t)|dtright)^px^b-1dxleleft(frac pbright)^pint_0^infty|f(t)|^pt^p+b-1dt$Prove that if $fin L^p(E), 1leq p<infty, m(E)<infty$, then $fin L^q(E)$ for all $1leq qleq p$Convolution of $L^1$ functions is well-definedInequality for bounded locally integrable functions
$begingroup$
$textbfThe Problem:$ Suppose that $(K_delta)_delta>0$ is a family of integrable functions such that there exists a constant $Cin(0,infty)$ such that $int K_delta=1,intvert K_deltavertleq C$ for every $delta>0$ and for every $eta>0$ we have $$colorbluelargelimlimits_deltato0^+int_vert xvertgeqetavert K_delta(x)vert dx=0.$$
Let $pin[1,infty)$. Prove that for every $fin L^p(mathbb R^d)$ we have $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
$textbfMy Thoughts:$ Here we go. For $pin[1,infty)$ we use Minkowski's inequality for integrals, for reference, this is $6.19$ on page $194$ of Folland's Real Analysis, $2$nd Edition. With this in mind we have, let $varepsilon>0$ be given, then we have that there is $eta>0$ such that $|tau_yf-f|_p<varepsilon$ for all $vert yvert<eta.$ Putting these together we have
beginalign*large|fast K_delta-f|_p&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-f(x)Biggvert^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-int_mathbb R^df(x)K_delta(y)dyBiggvert ^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^d[f(x-y)-f(x)]K_delta(y)dyBiggvert ^p dxright)^1/p\
&leqlargeint_mathbb R^dleft(int_mathbb R^dvert f(x-y)-f(x)vert^pdxright)^1/pvert K_delta(y)vert dy\
&leqlargeint_mathbb R^d|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlargeint_vert yvertgeqeta2|f|_pvert K_delta(y)vert dy+int_vert yvert<eta|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlarge2|f|_pint_vert yvertgeqetavert K_delta(y)vert dy+varepsilon\
&largeoversetdeltato0^+longrightarrowvarepsilon.
endalign*
It follows that $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
Do you agree with the proof presented above?
Any feedback is much appreciated.
Thank you for your time.
As a reference, here is the statement of Minkowski's Inequality for Integrals; Suppose that $(X,frakM,mu)$ and $(Y,frakN,nu)$ are $sigma$-finite measure spaces, and let $f$ be an $(frakMotimesfrakN)$-measurable function on $Xtimes Y$. Then if $fgeq0$ and $1leq p<infty,$ we have
$$left[intleft(int f(x,y)dnu(y)right)^p dmu(x)right]^1/pleqintleft[int f(x,y)^p dmu(x)right]^1/pdnu(y).$$
real-analysis proof-verification lebesgue-integral lp-spaces convolution
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
$endgroup$
add a comment |
$begingroup$
$textbfThe Problem:$ Suppose that $(K_delta)_delta>0$ is a family of integrable functions such that there exists a constant $Cin(0,infty)$ such that $int K_delta=1,intvert K_deltavertleq C$ for every $delta>0$ and for every $eta>0$ we have $$colorbluelargelimlimits_deltato0^+int_vert xvertgeqetavert K_delta(x)vert dx=0.$$
Let $pin[1,infty)$. Prove that for every $fin L^p(mathbb R^d)$ we have $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
$textbfMy Thoughts:$ Here we go. For $pin[1,infty)$ we use Minkowski's inequality for integrals, for reference, this is $6.19$ on page $194$ of Folland's Real Analysis, $2$nd Edition. With this in mind we have, let $varepsilon>0$ be given, then we have that there is $eta>0$ such that $|tau_yf-f|_p<varepsilon$ for all $vert yvert<eta.$ Putting these together we have
beginalign*large|fast K_delta-f|_p&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-f(x)Biggvert^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-int_mathbb R^df(x)K_delta(y)dyBiggvert ^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^d[f(x-y)-f(x)]K_delta(y)dyBiggvert ^p dxright)^1/p\
&leqlargeint_mathbb R^dleft(int_mathbb R^dvert f(x-y)-f(x)vert^pdxright)^1/pvert K_delta(y)vert dy\
&leqlargeint_mathbb R^d|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlargeint_vert yvertgeqeta2|f|_pvert K_delta(y)vert dy+int_vert yvert<eta|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlarge2|f|_pint_vert yvertgeqetavert K_delta(y)vert dy+varepsilon\
&largeoversetdeltato0^+longrightarrowvarepsilon.
endalign*
It follows that $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
Do you agree with the proof presented above?
Any feedback is much appreciated.
Thank you for your time.
As a reference, here is the statement of Minkowski's Inequality for Integrals; Suppose that $(X,frakM,mu)$ and $(Y,frakN,nu)$ are $sigma$-finite measure spaces, and let $f$ be an $(frakMotimesfrakN)$-measurable function on $Xtimes Y$. Then if $fgeq0$ and $1leq p<infty,$ we have
$$left[intleft(int f(x,y)dnu(y)right)^p dmu(x)right]^1/pleqintleft[int f(x,y)^p dmu(x)right]^1/pdnu(y).$$
real-analysis proof-verification lebesgue-integral lp-spaces convolution
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
$endgroup$
1
$begingroup$
How are you going from the third line to the fourth line?
$endgroup$
– Dionel Jaime
Apr 7 at 17:24
$begingroup$
@DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well.
$endgroup$
– Gaby Alfonso
Apr 8 at 1:57
1
$begingroup$
@GabyAlfonso With your correction, the proof looks correct.
$endgroup$
– Gyu Eun Lee
Apr 8 at 8:49
add a comment |
$begingroup$
$textbfThe Problem:$ Suppose that $(K_delta)_delta>0$ is a family of integrable functions such that there exists a constant $Cin(0,infty)$ such that $int K_delta=1,intvert K_deltavertleq C$ for every $delta>0$ and for every $eta>0$ we have $$colorbluelargelimlimits_deltato0^+int_vert xvertgeqetavert K_delta(x)vert dx=0.$$
Let $pin[1,infty)$. Prove that for every $fin L^p(mathbb R^d)$ we have $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
$textbfMy Thoughts:$ Here we go. For $pin[1,infty)$ we use Minkowski's inequality for integrals, for reference, this is $6.19$ on page $194$ of Folland's Real Analysis, $2$nd Edition. With this in mind we have, let $varepsilon>0$ be given, then we have that there is $eta>0$ such that $|tau_yf-f|_p<varepsilon$ for all $vert yvert<eta.$ Putting these together we have
beginalign*large|fast K_delta-f|_p&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-f(x)Biggvert^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-int_mathbb R^df(x)K_delta(y)dyBiggvert ^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^d[f(x-y)-f(x)]K_delta(y)dyBiggvert ^p dxright)^1/p\
&leqlargeint_mathbb R^dleft(int_mathbb R^dvert f(x-y)-f(x)vert^pdxright)^1/pvert K_delta(y)vert dy\
&leqlargeint_mathbb R^d|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlargeint_vert yvertgeqeta2|f|_pvert K_delta(y)vert dy+int_vert yvert<eta|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlarge2|f|_pint_vert yvertgeqetavert K_delta(y)vert dy+varepsilon\
&largeoversetdeltato0^+longrightarrowvarepsilon.
endalign*
It follows that $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
Do you agree with the proof presented above?
Any feedback is much appreciated.
Thank you for your time.
As a reference, here is the statement of Minkowski's Inequality for Integrals; Suppose that $(X,frakM,mu)$ and $(Y,frakN,nu)$ are $sigma$-finite measure spaces, and let $f$ be an $(frakMotimesfrakN)$-measurable function on $Xtimes Y$. Then if $fgeq0$ and $1leq p<infty,$ we have
$$left[intleft(int f(x,y)dnu(y)right)^p dmu(x)right]^1/pleqintleft[int f(x,y)^p dmu(x)right]^1/pdnu(y).$$
real-analysis proof-verification lebesgue-integral lp-spaces convolution
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
$endgroup$
$textbfThe Problem:$ Suppose that $(K_delta)_delta>0$ is a family of integrable functions such that there exists a constant $Cin(0,infty)$ such that $int K_delta=1,intvert K_deltavertleq C$ for every $delta>0$ and for every $eta>0$ we have $$colorbluelargelimlimits_deltato0^+int_vert xvertgeqetavert K_delta(x)vert dx=0.$$
Let $pin[1,infty)$. Prove that for every $fin L^p(mathbb R^d)$ we have $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
$textbfMy Thoughts:$ Here we go. For $pin[1,infty)$ we use Minkowski's inequality for integrals, for reference, this is $6.19$ on page $194$ of Folland's Real Analysis, $2$nd Edition. With this in mind we have, let $varepsilon>0$ be given, then we have that there is $eta>0$ such that $|tau_yf-f|_p<varepsilon$ for all $vert yvert<eta.$ Putting these together we have
beginalign*large|fast K_delta-f|_p&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-f(x)Biggvert^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^df(x-y)K_delta(y)dy-int_mathbb R^df(x)K_delta(y)dyBiggvert ^p dxright)^1/p\
&=largeleft(int_mathbb R^dBiggvertint_mathbb R^d[f(x-y)-f(x)]K_delta(y)dyBiggvert ^p dxright)^1/p\
&leqlargeint_mathbb R^dleft(int_mathbb R^dvert f(x-y)-f(x)vert^pdxright)^1/pvert K_delta(y)vert dy\
&leqlargeint_mathbb R^d|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlargeint_vert yvertgeqeta2|f|_pvert K_delta(y)vert dy+int_vert yvert<eta|tau_yf-f|_pvert K_delta(y)vert dy\
&leqlarge2|f|_pint_vert yvertgeqetavert K_delta(y)vert dy+varepsilon\
&largeoversetdeltato0^+longrightarrowvarepsilon.
endalign*
It follows that $fast K_deltato f$ in $L^p(mathbb R^d)$ as $deltato0^+.$
Do you agree with the proof presented above?
Any feedback is much appreciated.
Thank you for your time.
As a reference, here is the statement of Minkowski's Inequality for Integrals; Suppose that $(X,frakM,mu)$ and $(Y,frakN,nu)$ are $sigma$-finite measure spaces, and let $f$ be an $(frakMotimesfrakN)$-measurable function on $Xtimes Y$. Then if $fgeq0$ and $1leq p<infty,$ we have
$$left[intleft(int f(x,y)dnu(y)right)^p dmu(x)right]^1/pleqintleft[int f(x,y)^p dmu(x)right]^1/pdnu(y).$$
real-analysis proof-verification lebesgue-integral lp-spaces convolution
real-analysis proof-verification lebesgue-integral lp-spaces convolution
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
edited Apr 8 at 3:24
Gaby Alfonso
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
asked Apr 7 at 8:35
Gaby AlfonsoGaby Alfonso
1,2071418
1,2071418
1
$begingroup$
How are you going from the third line to the fourth line?
$endgroup$
– Dionel Jaime
Apr 7 at 17:24
$begingroup$
@DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well.
$endgroup$
– Gaby Alfonso
Apr 8 at 1:57
1
$begingroup$
@GabyAlfonso With your correction, the proof looks correct.
$endgroup$
– Gyu Eun Lee
Apr 8 at 8:49
add a comment |
1
$begingroup$
How are you going from the third line to the fourth line?
$endgroup$
– Dionel Jaime
Apr 7 at 17:24
$begingroup$
@DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well.
$endgroup$
– Gaby Alfonso
Apr 8 at 1:57
1
$begingroup$
@GabyAlfonso With your correction, the proof looks correct.
$endgroup$
– Gyu Eun Lee
Apr 8 at 8:49
1
1
$begingroup$
How are you going from the third line to the fourth line?
$endgroup$
– Dionel Jaime
Apr 7 at 17:24
$begingroup$
How are you going from the third line to the fourth line?
$endgroup$
– Dionel Jaime
Apr 7 at 17:24
$begingroup$
@DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well.
$endgroup$
– Gaby Alfonso
Apr 8 at 1:57
$begingroup$
@DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well.
$endgroup$
– Gaby Alfonso
Apr 8 at 1:57
1
1
$begingroup$
@GabyAlfonso With your correction, the proof looks correct.
$endgroup$
– Gyu Eun Lee
Apr 8 at 8:49
$begingroup$
@GabyAlfonso With your correction, the proof looks correct.
$endgroup$
– Gyu Eun Lee
Apr 8 at 8:49
add a comment |
1 Answer
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$begingroup$
It seems to me, fourth inequality is doubt, though I don't remember Minkowski's inequality and I don't have folland's book.
answered Apr 7 at 12:26
quickybrownquickybrown
13
$endgroup$
add a comment |
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$begingroup$
It seems to me, fourth inequality is doubt, though I don't remember Minkowski's inequality and I don't have folland's book.
answered Apr 7 at 12:26
quickybrownquickybrown
13
$endgroup$
add a comment |
$begingroup$
It seems to me, fourth inequality is doubt, though I don't remember Minkowski's inequality and I don't have folland's book.
answered Apr 7 at 12:26
quickybrownquickybrown
13
$endgroup$
add a comment |
$begingroup$
It seems to me, fourth inequality is doubt, though I don't remember Minkowski's inequality and I don't have folland's book.
answered Apr 7 at 12:26
quickybrownquickybrown
13
$endgroup$
It seems to me, fourth inequality is doubt, though I don't remember Minkowski's inequality and I don't have folland's book.
answered Apr 7 at 12:26
quickybrownquickybrown
13
answered Apr 7 at 12:26
quickybrownquickybrown
13
answered Apr 7 at 12:26
quickybrownquickybrown
13
answered Apr 7 at 12:26
quickybrownquickybrown
13
13
add a comment |
add a comment |
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$begingroup$
How are you going from the third line to the fourth line?
$endgroup$
– Dionel Jaime
Apr 7 at 17:24
$begingroup$
@DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well.
$endgroup$
– Gaby Alfonso
Apr 8 at 1:57
1
$begingroup$
@GabyAlfonso With your correction, the proof looks correct.
$endgroup$
– Gyu Eun Lee
Apr 8 at 8:49