Mixing
Showing that a set of functions is dense in $L^{p}$
$begingroup$
Let $f in L^{p}(mathbb{R})$, for $h>0$ define:
$f_{h}(x)=frac{1}{h}int_{x}^{x+h}f(t)dt$
Show that $f_{h}$ is continuous and that continuous functions are dense in $L^{p}(mathbb{R})$, by showing $||f_{h}-f||_{p} to 0$ as $h to 0$.
I know this seems like a really easy question, but I don't know where to start. What would be the best way to approach this problem?
real-analysis integration continuity
asked Jan 18 at 5:25
fenetrefenetre
1
$endgroup$
add a comment |
$begingroup$
Let $f in L^{p}(mathbb{R})$, for $h>0$ define:
$f_{h}(x)=frac{1}{h}int_{x}^{x+h}f(t)dt$
Show that $f_{h}$ is continuous and that continuous functions are dense in $L^{p}(mathbb{R})$, by showing $||f_{h}-f||_{p} to 0$ as $h to 0$.
I know this seems like a really easy question, but I don't know where to start. What would be the best way to approach this problem?
real-analysis integration continuity
asked Jan 18 at 5:25
fenetrefenetre
1
$endgroup$
add a comment |
$begingroup$
Let $f in L^{p}(mathbb{R})$, for $h>0$ define:
$f_{h}(x)=frac{1}{h}int_{x}^{x+h}f(t)dt$
Show that $f_{h}$ is continuous and that continuous functions are dense in $L^{p}(mathbb{R})$, by showing $||f_{h}-f||_{p} to 0$ as $h to 0$.
I know this seems like a really easy question, but I don't know where to start. What would be the best way to approach this problem?
real-analysis integration continuity
asked Jan 18 at 5:25
fenetrefenetre
1
$endgroup$
Let $f in L^{p}(mathbb{R})$, for $h>0$ define:
$f_{h}(x)=frac{1}{h}int_{x}^{x+h}f(t)dt$
Show that $f_{h}$ is continuous and that continuous functions are dense in $L^{p}(mathbb{R})$, by showing $||f_{h}-f||_{p} to 0$ as $h to 0$.
I know this seems like a really easy question, but I don't know where to start. What would be the best way to approach this problem?
real-analysis integration continuity
real-analysis integration continuity
asked Jan 18 at 5:25
fenetrefenetre
1
asked Jan 18 at 5:25
fenetrefenetre
1
asked Jan 18 at 5:25
fenetrefenetre
1
asked Jan 18 at 5:25
fenetrefenetre
1
asked Jan 18 at 5:25
fenetrefenetre
1
1
add a comment |
add a comment |
1 Answer
1
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$begingroup$
By the integral form of Minkowski's inequality $|f_h-g_h||_p <epsilon$ if $|f-g|_p <epsilon$. Any $f$ in $L^{p}$ can be approximated in the $L^{p}$ norm by a bounded measurable function with compact support. [Just look at $fI_A$ where $A={x:|f(x)| leq N,|x|leq N}$ with $N$ large enough]. Hence it suffices to prove the result when $f$ is a bounded measurable function with compact support. In that case we can use Lebesgue's Theorem (which tells you that $f_h to f$ almost everywhere ) along with DCT to prove that $|f_h-f||_p to 0$ as $h to 0$. Let me know if a more detailed answer is needed.
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
add a comment |
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1 Answer
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1 Answer
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$begingroup$
By the integral form of Minkowski's inequality $|f_h-g_h||_p <epsilon$ if $|f-g|_p <epsilon$. Any $f$ in $L^{p}$ can be approximated in the $L^{p}$ norm by a bounded measurable function with compact support. [Just look at $fI_A$ where $A={x:|f(x)| leq N,|x|leq N}$ with $N$ large enough]. Hence it suffices to prove the result when $f$ is a bounded measurable function with compact support. In that case we can use Lebesgue's Theorem (which tells you that $f_h to f$ almost everywhere ) along with DCT to prove that $|f_h-f||_p to 0$ as $h to 0$. Let me know if a more detailed answer is needed.
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
add a comment |
$begingroup$
By the integral form of Minkowski's inequality $|f_h-g_h||_p <epsilon$ if $|f-g|_p <epsilon$. Any $f$ in $L^{p}$ can be approximated in the $L^{p}$ norm by a bounded measurable function with compact support. [Just look at $fI_A$ where $A={x:|f(x)| leq N,|x|leq N}$ with $N$ large enough]. Hence it suffices to prove the result when $f$ is a bounded measurable function with compact support. In that case we can use Lebesgue's Theorem (which tells you that $f_h to f$ almost everywhere ) along with DCT to prove that $|f_h-f||_p to 0$ as $h to 0$. Let me know if a more detailed answer is needed.
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
add a comment |
$begingroup$
By the integral form of Minkowski's inequality $|f_h-g_h||_p <epsilon$ if $|f-g|_p <epsilon$. Any $f$ in $L^{p}$ can be approximated in the $L^{p}$ norm by a bounded measurable function with compact support. [Just look at $fI_A$ where $A={x:|f(x)| leq N,|x|leq N}$ with $N$ large enough]. Hence it suffices to prove the result when $f$ is a bounded measurable function with compact support. In that case we can use Lebesgue's Theorem (which tells you that $f_h to f$ almost everywhere ) along with DCT to prove that $|f_h-f||_p to 0$ as $h to 0$. Let me know if a more detailed answer is needed.
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
By the integral form of Minkowski's inequality $|f_h-g_h||_p <epsilon$ if $|f-g|_p <epsilon$. Any $f$ in $L^{p}$ can be approximated in the $L^{p}$ norm by a bounded measurable function with compact support. [Just look at $fI_A$ where $A={x:|f(x)| leq N,|x|leq N}$ with $N$ large enough]. Hence it suffices to prove the result when $f$ is a bounded measurable function with compact support. In that case we can use Lebesgue's Theorem (which tells you that $f_h to f$ almost everywhere ) along with DCT to prove that $|f_h-f||_p to 0$ as $h to 0$. Let me know if a more detailed answer is needed.
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
edited Jan 18 at 8:14
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 18 at 6:30
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
add a comment |
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
Thanks for the answer! I think I get it. For the first equation, wouldn't it be $||f_{h}-g_{h}||_{p} < epsilon$ if $||f-g||_{p} < epsilon$? Also, why would $f_{h}$ be continuous? I still can't see that....
$endgroup$
– fenetre
Jan 18 at 8:12
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
$begingroup$
@fenetre Yes, there was a typo. $int_x^{x+h} f(t), dt=int_0^{x+h} f(t), dt-int_0^{x} f(t), dt$ and indefinite integrals of integrable functions are continuous. Hence $f_h$ is continuous.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 8:17
add a comment |
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