Convergence in $(L_P(a,b),|.|_p)$ doesn't imply convergence in $(C(a,b),sup)$
$begingroup$
Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?
It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.
And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?
functional-analysis convergence proof-explanation normed-spaces lp-spaces
$endgroup$
add a comment |
$begingroup$
Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?
It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.
And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?
functional-analysis convergence proof-explanation normed-spaces lp-spaces
$endgroup$
add a comment |
$begingroup$
Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?
It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.
And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?
functional-analysis convergence proof-explanation normed-spaces lp-spaces
$endgroup$
Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?
It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.
And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?
functional-analysis convergence proof-explanation normed-spaces lp-spaces
functional-analysis convergence proof-explanation normed-spaces lp-spaces
edited Dec 26 '18 at 16:55
Davide Giraudo
127k17154268
127k17154268
asked Dec 26 '18 at 9:32
Dreamer123Dreamer123
32729
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2 Answers
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$begingroup$
Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)
It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.
$endgroup$
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$begingroup$
First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.
$f_n to f$ with respect to $||, ., ||_p$ means only that
begin{align}
int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
end{align}
It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.
Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.
There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.
I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).
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2 Answers
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2 Answers
2
active
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$begingroup$
Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)
It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.
$endgroup$
add a comment |
$begingroup$
Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)
It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.
$endgroup$
add a comment |
$begingroup$
Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)
It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.
$endgroup$
Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)
It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.
answered Dec 26 '18 at 11:24
Julián AguirreJulián Aguirre
69.4k24096
69.4k24096
add a comment |
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$begingroup$
First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.
$f_n to f$ with respect to $||, ., ||_p$ means only that
begin{align}
int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
end{align}
It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.
Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.
There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.
I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).
$endgroup$
add a comment |
$begingroup$
First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.
$f_n to f$ with respect to $||, ., ||_p$ means only that
begin{align}
int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
end{align}
It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.
Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.
There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.
I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).
$endgroup$
add a comment |
$begingroup$
First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.
$f_n to f$ with respect to $||, ., ||_p$ means only that
begin{align}
int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
end{align}
It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.
Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.
There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.
I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).
$endgroup$
First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.
$f_n to f$ with respect to $||, ., ||_p$ means only that
begin{align}
int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
end{align}
It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.
Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.
There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.
I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).
edited Dec 26 '18 at 11:32
answered Dec 26 '18 at 11:26
N.BeckN.Beck
2937
2937
add a comment |
add a comment |
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