Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
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Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
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add a comment |
$begingroup$
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
$endgroup$
add a comment |
$begingroup$
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
$endgroup$
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
measure-theory lp-spaces
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
asked Nov 29 '18 at 3:00
jefe_16jefe_16
113
113
add a comment |
add a comment |
1 Answer
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Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
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1 Answer
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1 Answer
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$begingroup$
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Nov 29 '18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
57.7k42160
add a comment |
add a comment |
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