Mixing
Interchanging limit and integral.
$begingroup$
Suppose $(X,mu)$ is a probability space, $Win L^1(X)$, $Vin L^infty(X)$, and $V_nto V$ in $L^2(X)$ (in my situation $V_n$ is the partial Fourier sum and so the $L^2(X)$ convergence is automatic).
Can we say that $int_X WVdmu= lim_{ntoinfty}int_X WV_ndmu$?
I think so, but I am having trouble showing this.
If $W$ is square integrable then the result follows from Cauchy-Schwarz. But of course this argument does not work for less regular $W$. I have considered dominated convergence theorem and it's variation, but finding a dominating function is eluding me. Could anyone help out? Both suggestions and solutions are welcome.
Thanks in advance!
measure-theory fourier-analysis lp-spaces cauchy-schwarz-inequality holder-inequality
edited Jan 6 at 4:34
BigbearZzz
8,51721652
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
$endgroup$
add a comment |
$begingroup$
Suppose $(X,mu)$ is a probability space, $Win L^1(X)$, $Vin L^infty(X)$, and $V_nto V$ in $L^2(X)$ (in my situation $V_n$ is the partial Fourier sum and so the $L^2(X)$ convergence is automatic).
Can we say that $int_X WVdmu= lim_{ntoinfty}int_X WV_ndmu$?
I think so, but I am having trouble showing this.
If $W$ is square integrable then the result follows from Cauchy-Schwarz. But of course this argument does not work for less regular $W$. I have considered dominated convergence theorem and it's variation, but finding a dominating function is eluding me. Could anyone help out? Both suggestions and solutions are welcome.
Thanks in advance!
measure-theory fourier-analysis lp-spaces cauchy-schwarz-inequality holder-inequality
edited Jan 6 at 4:34
BigbearZzz
8,51721652
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
$endgroup$
$begingroup$
Under the hypothesis $int_X WV_n , dmu$ need not exist for any $n$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 5:13
add a comment |
$begingroup$
Suppose $(X,mu)$ is a probability space, $Win L^1(X)$, $Vin L^infty(X)$, and $V_nto V$ in $L^2(X)$ (in my situation $V_n$ is the partial Fourier sum and so the $L^2(X)$ convergence is automatic).
Can we say that $int_X WVdmu= lim_{ntoinfty}int_X WV_ndmu$?
I think so, but I am having trouble showing this.
If $W$ is square integrable then the result follows from Cauchy-Schwarz. But of course this argument does not work for less regular $W$. I have considered dominated convergence theorem and it's variation, but finding a dominating function is eluding me. Could anyone help out? Both suggestions and solutions are welcome.
Thanks in advance!
measure-theory fourier-analysis lp-spaces cauchy-schwarz-inequality holder-inequality
edited Jan 6 at 4:34
BigbearZzz
8,51721652
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
$endgroup$
Suppose $(X,mu)$ is a probability space, $Win L^1(X)$, $Vin L^infty(X)$, and $V_nto V$ in $L^2(X)$ (in my situation $V_n$ is the partial Fourier sum and so the $L^2(X)$ convergence is automatic).
Can we say that $int_X WVdmu= lim_{ntoinfty}int_X WV_ndmu$?
I think so, but I am having trouble showing this.
If $W$ is square integrable then the result follows from Cauchy-Schwarz. But of course this argument does not work for less regular $W$. I have considered dominated convergence theorem and it's variation, but finding a dominating function is eluding me. Could anyone help out? Both suggestions and solutions are welcome.
Thanks in advance!
measure-theory fourier-analysis lp-spaces cauchy-schwarz-inequality holder-inequality
measure-theory fourier-analysis lp-spaces cauchy-schwarz-inequality holder-inequality
edited Jan 6 at 4:34
BigbearZzz
8,51721652
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
edited Jan 6 at 4:34
BigbearZzz
8,51721652
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
edited Jan 6 at 4:34
BigbearZzz
8,51721652
edited Jan 6 at 4:34
BigbearZzz
8,51721652
edited Jan 6 at 4:34
BigbearZzz
8,51721652
8,51721652
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
asked Jan 6 at 1:50
Mr MartingaleMr Martingale
25217
25217
$begingroup$
Under the hypothesis $int_X WV_n , dmu$ need not exist for any $n$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 5:13
add a comment |
$begingroup$
Under the hypothesis $int_X WV_n , dmu$ need not exist for any $n$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 5:13
$begingroup$
Under the hypothesis $int_X WV_n , dmu$ need not exist for any $n$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 5:13
$begingroup$
Under the hypothesis $int_X WV_n , dmu$ need not exist for any $n$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 5:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the probability space $X=[0,1]$ with uniform probability measure $mu$. We let $V=0$ and $W$ be defined by
$$
W(x) :=sum_{n=1}^infty frac {2^n}{n^2} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is not hard to verify that $Win L^1(X,mu)$.
Now, let the sequence $V_n$ be defined as
$$
V_n(x):= 2^{n/4} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is also easy to compute that $V_nto 0$ in $L^2(X,mu)$.
However, we have
$$
int_X WV,dmu = 0 ne infty = lim_{ntoinfty} frac 1{n^2}cdot frac{2^{5n/4}}{2^n}= lim_{ntoinfty}int_X WV_n,dmu.
$$
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
$endgroup$
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
$begingroup$
Consider the probability space $X=[0,1]$ with uniform probability measure $mu$. We let $V=0$ and $W$ be defined by
$$
W(x) :=sum_{n=1}^infty frac {2^n}{n^2} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is not hard to verify that $Win L^1(X,mu)$.
Now, let the sequence $V_n$ be defined as
$$
V_n(x):= 2^{n/4} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is also easy to compute that $V_nto 0$ in $L^2(X,mu)$.
However, we have
$$
int_X WV,dmu = 0 ne infty = lim_{ntoinfty} frac 1{n^2}cdot frac{2^{5n/4}}{2^n}= lim_{ntoinfty}int_X WV_n,dmu.
$$
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
$endgroup$
add a comment |
$begingroup$
Consider the probability space $X=[0,1]$ with uniform probability measure $mu$. We let $V=0$ and $W$ be defined by
$$
W(x) :=sum_{n=1}^infty frac {2^n}{n^2} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is not hard to verify that $Win L^1(X,mu)$.
Now, let the sequence $V_n$ be defined as
$$
V_n(x):= 2^{n/4} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is also easy to compute that $V_nto 0$ in $L^2(X,mu)$.
However, we have
$$
int_X WV,dmu = 0 ne infty = lim_{ntoinfty} frac 1{n^2}cdot frac{2^{5n/4}}{2^n}= lim_{ntoinfty}int_X WV_n,dmu.
$$
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
$endgroup$
add a comment |
$begingroup$
Consider the probability space $X=[0,1]$ with uniform probability measure $mu$. We let $V=0$ and $W$ be defined by
$$
W(x) :=sum_{n=1}^infty frac {2^n}{n^2} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is not hard to verify that $Win L^1(X,mu)$.
Now, let the sequence $V_n$ be defined as
$$
V_n(x):= 2^{n/4} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is also easy to compute that $V_nto 0$ in $L^2(X,mu)$.
However, we have
$$
int_X WV,dmu = 0 ne infty = lim_{ntoinfty} frac 1{n^2}cdot frac{2^{5n/4}}{2^n}= lim_{ntoinfty}int_X WV_n,dmu.
$$
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
$endgroup$
Consider the probability space $X=[0,1]$ with uniform probability measure $mu$. We let $V=0$ and $W$ be defined by
$$
W(x) :=sum_{n=1}^infty frac {2^n}{n^2} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is not hard to verify that $Win L^1(X,mu)$.
Now, let the sequence $V_n$ be defined as
$$
V_n(x):= 2^{n/4} chi_{[2^{-n}, 2^{-n+1}]},
$$
it is also easy to compute that $V_nto 0$ in $L^2(X,mu)$.
However, we have
$$
int_X WV,dmu = 0 ne infty = lim_{ntoinfty} frac 1{n^2}cdot frac{2^{5n/4}}{2^n}= lim_{ntoinfty}int_X WV_n,dmu.
$$
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
answered Jan 6 at 4:26
BigbearZzzBigbearZzz
8,51721652
8,51721652
add a comment |
add a comment |
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$begingroup$
Under the hypothesis $int_X WV_n , dmu$ need not exist for any $n$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 5:13