Dominated convergence exercise
$begingroup$
Given a function $f in L^1(mathbb{R})$, suppose I want to evaluate the following integral
$$lim_{ntoinfty} int_{mathbb{R}}e^{-nx^2}f(x)dx$$
It looks to me like one could use dominated convergence here. If we let $f_n(x) = e^{-nx^2}f(x)$, then clearly $|f_n(x)| leq |f(x)|$ for $n geq 0$, and we know that $|f(x)|$ is an integrable function. The issue I am having is showing that $f_n(x)$ itself is integrable - it seems clear to me that it would be, but not sure how to show it (maybe it just follows from the above inequality since $|f(x)|$ is integrable?)
Also, suppose we can show dominated convergence theorem applies, the limit is a little bit interesting as it depends on $x$. Clearly, if $xneq 0$, then we have
$$lim_{ntoinfty} e^{-nx^2}f(x) = 0,$$ but if $x = 0$, then what happens?
Thanks!
integration limits convergence
asked Dec 4 '18 at 19:03
SoreySorey
577212
$endgroup$
add a comment |
$begingroup$
Given a function $f in L^1(mathbb{R})$, suppose I want to evaluate the following integral
$$lim_{ntoinfty} int_{mathbb{R}}e^{-nx^2}f(x)dx$$
It looks to me like one could use dominated convergence here. If we let $f_n(x) = e^{-nx^2}f(x)$, then clearly $|f_n(x)| leq |f(x)|$ for $n geq 0$, and we know that $|f(x)|$ is an integrable function. The issue I am having is showing that $f_n(x)$ itself is integrable - it seems clear to me that it would be, but not sure how to show it (maybe it just follows from the above inequality since $|f(x)|$ is integrable?)
Also, suppose we can show dominated convergence theorem applies, the limit is a little bit interesting as it depends on $x$. Clearly, if $xneq 0$, then we have
$$lim_{ntoinfty} e^{-nx^2}f(x) = 0,$$ but if $x = 0$, then what happens?
Thanks!
integration limits convergence
asked Dec 4 '18 at 19:03
SoreySorey
577212
$endgroup$
$begingroup$
${0}$ is of measure $0$, so it doesn't matter what happens there.
$endgroup$
– saulspatz
Dec 4 '18 at 19:13
add a comment |
$begingroup$
Given a function $f in L^1(mathbb{R})$, suppose I want to evaluate the following integral
$$lim_{ntoinfty} int_{mathbb{R}}e^{-nx^2}f(x)dx$$
It looks to me like one could use dominated convergence here. If we let $f_n(x) = e^{-nx^2}f(x)$, then clearly $|f_n(x)| leq |f(x)|$ for $n geq 0$, and we know that $|f(x)|$ is an integrable function. The issue I am having is showing that $f_n(x)$ itself is integrable - it seems clear to me that it would be, but not sure how to show it (maybe it just follows from the above inequality since $|f(x)|$ is integrable?)
Also, suppose we can show dominated convergence theorem applies, the limit is a little bit interesting as it depends on $x$. Clearly, if $xneq 0$, then we have
$$lim_{ntoinfty} e^{-nx^2}f(x) = 0,$$ but if $x = 0$, then what happens?
Thanks!
integration limits convergence
asked Dec 4 '18 at 19:03
SoreySorey
577212
$endgroup$
Given a function $f in L^1(mathbb{R})$, suppose I want to evaluate the following integral
$$lim_{ntoinfty} int_{mathbb{R}}e^{-nx^2}f(x)dx$$
It looks to me like one could use dominated convergence here. If we let $f_n(x) = e^{-nx^2}f(x)$, then clearly $|f_n(x)| leq |f(x)|$ for $n geq 0$, and we know that $|f(x)|$ is an integrable function. The issue I am having is showing that $f_n(x)$ itself is integrable - it seems clear to me that it would be, but not sure how to show it (maybe it just follows from the above inequality since $|f(x)|$ is integrable?)
Also, suppose we can show dominated convergence theorem applies, the limit is a little bit interesting as it depends on $x$. Clearly, if $xneq 0$, then we have
$$lim_{ntoinfty} e^{-nx^2}f(x) = 0,$$ but if $x = 0$, then what happens?
Thanks!
integration limits convergence
integration limits convergence
asked Dec 4 '18 at 19:03
SoreySorey
577212
asked Dec 4 '18 at 19:03
SoreySorey
577212
asked Dec 4 '18 at 19:03
SoreySorey
577212
asked Dec 4 '18 at 19:03
SoreySorey
577212
asked Dec 4 '18 at 19:03
SoreySorey
577212
577212
$begingroup$
${0}$ is of measure $0$, so it doesn't matter what happens there.
$endgroup$
– saulspatz
Dec 4 '18 at 19:13
add a comment |
$begingroup$
${0}$ is of measure $0$, so it doesn't matter what happens there.
$endgroup$
– saulspatz
Dec 4 '18 at 19:13
$begingroup$
${0}$ is of measure $0$, so it doesn't matter what happens there.
$endgroup$
– saulspatz
Dec 4 '18 at 19:13
$begingroup$
${0}$ is of measure $0$, so it doesn't matter what happens there.
$endgroup$
– saulspatz
Dec 4 '18 at 19:13
add a comment |
1 Answer
1
active
oldest
votes
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$f_n$ is integrable because $|f_n| le |f|$, as you mentioned.
We have $$lim_{n to infty} f_n(x) = begin{cases} 0 & x ne 0 \ f(0) & x = 0. end{cases}$$
The integral of this function is simply zero. So if you show that you can use the dominated convergence theorem, then $lim_{n to infty} int f_n = 0$.
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
$endgroup$
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$f_n$ is integrable because $|f_n| le |f|$, as you mentioned.
We have $$lim_{n to infty} f_n(x) = begin{cases} 0 & x ne 0 \ f(0) & x = 0. end{cases}$$
The integral of this function is simply zero. So if you show that you can use the dominated convergence theorem, then $lim_{n to infty} int f_n = 0$.
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
$endgroup$
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
add a comment |
$begingroup$
$f_n$ is integrable because $|f_n| le |f|$, as you mentioned.
We have $$lim_{n to infty} f_n(x) = begin{cases} 0 & x ne 0 \ f(0) & x = 0. end{cases}$$
The integral of this function is simply zero. So if you show that you can use the dominated convergence theorem, then $lim_{n to infty} int f_n = 0$.
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
$endgroup$
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
add a comment |
$begingroup$
$f_n$ is integrable because $|f_n| le |f|$, as you mentioned.
We have $$lim_{n to infty} f_n(x) = begin{cases} 0 & x ne 0 \ f(0) & x = 0. end{cases}$$
The integral of this function is simply zero. So if you show that you can use the dominated convergence theorem, then $lim_{n to infty} int f_n = 0$.
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
$endgroup$
$f_n$ is integrable because $|f_n| le |f|$, as you mentioned.
We have $$lim_{n to infty} f_n(x) = begin{cases} 0 & x ne 0 \ f(0) & x = 0. end{cases}$$
The integral of this function is simply zero. So if you show that you can use the dominated convergence theorem, then $lim_{n to infty} int f_n = 0$.
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
answered Dec 4 '18 at 19:15
angryavianangryavian
41.8k23381
41.8k23381
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
add a comment |
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
What else needs to be shown to use dominated convergence? I thought it was that the sequence $f_n$ had to be bounded above by an integrable function and be integrable themselves and then conditions were satisfied.
$endgroup$
– Sorey
Dec 4 '18 at 19:17
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
$begingroup$
@Sorey Sure, that's enough.
$endgroup$
– angryavian
Dec 4 '18 at 20:25
add a comment |
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$begingroup$
${0}$ is of measure $0$, so it doesn't matter what happens there.
$endgroup$
– saulspatz
Dec 4 '18 at 19:13