If $int_mathbb R^dK=1$ then $K_delta(x)=delta^-dK(delta^-1x)$ is an approximate identity The 2019 Stack Overflow Developer Survey Results Are InHow to prove Dilation property of Lebesgue integralShow that $lim_t to 0 int_mathbbR^d|f(x)-f(x-t)|dx = 0$Find $EsubseteqmathbbR$ such that $liminf_deltato 0fracm(Ecap(-delta,delta))2delta=alpha$Lebesgue integral of Dirac deltaBoundedness of an operator with kernelUse definition of Lebesgue integral for nonnegative functions to show $int_Af=int_Echi_Af$.Dilation convergence in $L^1$Compute $limlimits_nto infty int_0^infty exp(-x)(cos(x))^n dx$Translation function of Lebesgue integral is continuousProof that Good Kernels are Approximations of Identity in $L^p(mathbb R^d)$Inequality for bounded locally integrable functions
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If $int_mathbb R^dK=1$ then $K_delta(x)=delta^-dK(delta^-1x)$ is an approximate identity
The 2019 Stack Overflow Developer Survey Results Are InHow to prove Dilation property of Lebesgue integralShow that $lim_t to 0 int_mathbbR^d|f(x)-f(x-t)|dx = 0$Find $EsubseteqmathbbR$ such that $liminf_deltato 0fracm(Ecap(-delta,delta))2delta=alpha$Lebesgue integral of Dirac deltaBoundedness of an operator with kernelUse definition of Lebesgue integral for nonnegative functions to show $int_Af=int_Echi_Af$.Dilation convergence in $L^1$Compute $limlimits_nto infty int_0^infty exp(-x)(cos(x))^n dx$Translation function of Lebesgue integral is continuousProof that Good Kernels are Approximations of Identity in $L^p(mathbb R^d)$Inequality for bounded locally integrable functions
$begingroup$
Let $Kin L^1(mathbb R^d)$ be bounded and supported on a set $E$ of finite measure and let $int_mathbb R^dK=1.$ Define $K_delta(x)=delta^-dK(delta^-1x)$ for every $delta>0$. Show that $(K_delta)_delta$ is an approximation of the identity, that is, it satisfies:
$$texti) int_mathbb R^dK_delta(x)dx=1quadquadtextii) vert K_delta(x)vertleq Adelta^-dquadquadtextiii) vert K_delta(x)vertleq Adeltavert xvert^-(d+1)$$
for all $delta>0$ and $xinmathbb R^d$.
$textbfMy Attempt & Concerns:$ To verify (i), I use the dilation property of the Lebesgue integral in $mathbb R^d,$ which can be found here How to prove Dilation property of Lebesgue integral. Mainly, we have
$$int_mathbb R^dK_delta(x)dx=int_mathbb R^ddelta^-dK(delta^-1x)dx=delta^-ddelta^dint_mathbb R^dK(x)dx=1.$$
To verify (ii), we use the hypothesis that $K$ is bounded and supported on a set of finite measure, so that if $delta^-1xnotin E$ then we have that $vert K_delta(x)vert=0,$ so it is surely bounded by the claimed quatity, with say $A$ being the bound for $K$. If $delta^-1xin E$ then we use that $vert Kvertleq A$ in $mathbb R^d,$ to obtain that
$$vert K_delta(x)vert=vertdelta^-dK(delta^-1x)vertleq Adelta^-d.$$
The third property is the one I do not see a way to prove. If $delta^-1xnotin E$ then it is fine, as $K$ is supported on $E$ so we certainly have the bound. However, if $delta^-1xin E$, then we have that
$$vert K_delta(x)vert=delta^-dvert K(delta^-1x)vert=fracdeltavert K(delta^-1x)vertdelta^d+1leqfracdelta Adelta^d+1,$$
but I do not see a way to get rid of $delta^d+1$ at the bottom and end up with $vert xvert^d+1.$
Any hints on how to proceed on the third part? Please, only hints. Any feedback on the other parts are also welcomed.
Thank you for time.
real-analysis lebesgue-integral
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
$endgroup$
add a comment |
$begingroup$
Let $Kin L^1(mathbb R^d)$ be bounded and supported on a set $E$ of finite measure and let $int_mathbb R^dK=1.$ Define $K_delta(x)=delta^-dK(delta^-1x)$ for every $delta>0$. Show that $(K_delta)_delta$ is an approximation of the identity, that is, it satisfies:
$$texti) int_mathbb R^dK_delta(x)dx=1quadquadtextii) vert K_delta(x)vertleq Adelta^-dquadquadtextiii) vert K_delta(x)vertleq Adeltavert xvert^-(d+1)$$
for all $delta>0$ and $xinmathbb R^d$.
$textbfMy Attempt & Concerns:$ To verify (i), I use the dilation property of the Lebesgue integral in $mathbb R^d,$ which can be found here How to prove Dilation property of Lebesgue integral. Mainly, we have
$$int_mathbb R^dK_delta(x)dx=int_mathbb R^ddelta^-dK(delta^-1x)dx=delta^-ddelta^dint_mathbb R^dK(x)dx=1.$$
To verify (ii), we use the hypothesis that $K$ is bounded and supported on a set of finite measure, so that if $delta^-1xnotin E$ then we have that $vert K_delta(x)vert=0,$ so it is surely bounded by the claimed quatity, with say $A$ being the bound for $K$. If $delta^-1xin E$ then we use that $vert Kvertleq A$ in $mathbb R^d,$ to obtain that
$$vert K_delta(x)vert=vertdelta^-dK(delta^-1x)vertleq Adelta^-d.$$
The third property is the one I do not see a way to prove. If $delta^-1xnotin E$ then it is fine, as $K$ is supported on $E$ so we certainly have the bound. However, if $delta^-1xin E$, then we have that
$$vert K_delta(x)vert=delta^-dvert K(delta^-1x)vert=fracdeltavert K(delta^-1x)vertdelta^d+1leqfracdelta Adelta^d+1,$$
but I do not see a way to get rid of $delta^d+1$ at the bottom and end up with $vert xvert^d+1.$
Any hints on how to proceed on the third part? Please, only hints. Any feedback on the other parts are also welcomed.
Thank you for time.
real-analysis lebesgue-integral
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
$endgroup$
add a comment |
$begingroup$
Let $Kin L^1(mathbb R^d)$ be bounded and supported on a set $E$ of finite measure and let $int_mathbb R^dK=1.$ Define $K_delta(x)=delta^-dK(delta^-1x)$ for every $delta>0$. Show that $(K_delta)_delta$ is an approximation of the identity, that is, it satisfies:
$$texti) int_mathbb R^dK_delta(x)dx=1quadquadtextii) vert K_delta(x)vertleq Adelta^-dquadquadtextiii) vert K_delta(x)vertleq Adeltavert xvert^-(d+1)$$
for all $delta>0$ and $xinmathbb R^d$.
$textbfMy Attempt & Concerns:$ To verify (i), I use the dilation property of the Lebesgue integral in $mathbb R^d,$ which can be found here How to prove Dilation property of Lebesgue integral. Mainly, we have
$$int_mathbb R^dK_delta(x)dx=int_mathbb R^ddelta^-dK(delta^-1x)dx=delta^-ddelta^dint_mathbb R^dK(x)dx=1.$$
To verify (ii), we use the hypothesis that $K$ is bounded and supported on a set of finite measure, so that if $delta^-1xnotin E$ then we have that $vert K_delta(x)vert=0,$ so it is surely bounded by the claimed quatity, with say $A$ being the bound for $K$. If $delta^-1xin E$ then we use that $vert Kvertleq A$ in $mathbb R^d,$ to obtain that
$$vert K_delta(x)vert=vertdelta^-dK(delta^-1x)vertleq Adelta^-d.$$
The third property is the one I do not see a way to prove. If $delta^-1xnotin E$ then it is fine, as $K$ is supported on $E$ so we certainly have the bound. However, if $delta^-1xin E$, then we have that
$$vert K_delta(x)vert=delta^-dvert K(delta^-1x)vert=fracdeltavert K(delta^-1x)vertdelta^d+1leqfracdelta Adelta^d+1,$$
but I do not see a way to get rid of $delta^d+1$ at the bottom and end up with $vert xvert^d+1.$
Any hints on how to proceed on the third part? Please, only hints. Any feedback on the other parts are also welcomed.
Thank you for time.
real-analysis lebesgue-integral
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
$endgroup$
Let $Kin L^1(mathbb R^d)$ be bounded and supported on a set $E$ of finite measure and let $int_mathbb R^dK=1.$ Define $K_delta(x)=delta^-dK(delta^-1x)$ for every $delta>0$. Show that $(K_delta)_delta$ is an approximation of the identity, that is, it satisfies:
$$texti) int_mathbb R^dK_delta(x)dx=1quadquadtextii) vert K_delta(x)vertleq Adelta^-dquadquadtextiii) vert K_delta(x)vertleq Adeltavert xvert^-(d+1)$$
for all $delta>0$ and $xinmathbb R^d$.
$textbfMy Attempt & Concerns:$ To verify (i), I use the dilation property of the Lebesgue integral in $mathbb R^d,$ which can be found here How to prove Dilation property of Lebesgue integral. Mainly, we have
$$int_mathbb R^dK_delta(x)dx=int_mathbb R^ddelta^-dK(delta^-1x)dx=delta^-ddelta^dint_mathbb R^dK(x)dx=1.$$
To verify (ii), we use the hypothesis that $K$ is bounded and supported on a set of finite measure, so that if $delta^-1xnotin E$ then we have that $vert K_delta(x)vert=0,$ so it is surely bounded by the claimed quatity, with say $A$ being the bound for $K$. If $delta^-1xin E$ then we use that $vert Kvertleq A$ in $mathbb R^d,$ to obtain that
$$vert K_delta(x)vert=vertdelta^-dK(delta^-1x)vertleq Adelta^-d.$$
The third property is the one I do not see a way to prove. If $delta^-1xnotin E$ then it is fine, as $K$ is supported on $E$ so we certainly have the bound. However, if $delta^-1xin E$, then we have that
$$vert K_delta(x)vert=delta^-dvert K(delta^-1x)vert=fracdeltavert K(delta^-1x)vertdelta^d+1leqfracdelta Adelta^d+1,$$
but I do not see a way to get rid of $delta^d+1$ at the bottom and end up with $vert xvert^d+1.$
Any hints on how to proceed on the third part? Please, only hints. Any feedback on the other parts are also welcomed.
Thank you for time.
real-analysis lebesgue-integral
real-analysis lebesgue-integral
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
edited Apr 7 at 20:56
Gaby Alfonso
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
asked Apr 7 at 5:44
Gaby AlfonsoGaby Alfonso
1,2071418
1,2071418
add a comment |
add a comment |
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Prove it for the case when $K$ is compactly supported, say $supp(K)subset [-n,n]$ . In general case for any $epsilon$ you can find $n$ big enough such that the integral of $K$ outside $[-n,n]$ is smaller than $epsilon$, so $K$ is essentially supported on $[-n,n]$ choosing a good $epsilon$ and big $n$ should give you the result.
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
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$begingroup$
Prove it for the case when $K$ is compactly supported, say $supp(K)subset [-n,n]$ . In general case for any $epsilon$ you can find $n$ big enough such that the integral of $K$ outside $[-n,n]$ is smaller than $epsilon$, so $K$ is essentially supported on $[-n,n]$ choosing a good $epsilon$ and big $n$ should give you the result.
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
$endgroup$
add a comment |
$begingroup$
Prove it for the case when $K$ is compactly supported, say $supp(K)subset [-n,n]$ . In general case for any $epsilon$ you can find $n$ big enough such that the integral of $K$ outside $[-n,n]$ is smaller than $epsilon$, so $K$ is essentially supported on $[-n,n]$ choosing a good $epsilon$ and big $n$ should give you the result.
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
$endgroup$
add a comment |
$begingroup$
Prove it for the case when $K$ is compactly supported, say $supp(K)subset [-n,n]$ . In general case for any $epsilon$ you can find $n$ big enough such that the integral of $K$ outside $[-n,n]$ is smaller than $epsilon$, so $K$ is essentially supported on $[-n,n]$ choosing a good $epsilon$ and big $n$ should give you the result.
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
$endgroup$
Prove it for the case when $K$ is compactly supported, say $supp(K)subset [-n,n]$ . In general case for any $epsilon$ you can find $n$ big enough such that the integral of $K$ outside $[-n,n]$ is smaller than $epsilon$, so $K$ is essentially supported on $[-n,n]$ choosing a good $epsilon$ and big $n$ should give you the result.
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
edited Apr 8 at 1:28
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
answered Apr 7 at 6:32
Julian MejiaJulian Mejia
54229
54229
add a comment |
add a comment |
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