If $f(x)$ is continuous on $[a,b]$ and $M=max ; |f(x)|$, is $M=lim limits_{ntoinfty}...
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$
Thanks!
real-analysis
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
asked Dec 4 '11 at 17:45
tomerg
738820
add a comment |
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$
Thanks!
real-analysis
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
asked Dec 4 '11 at 17:45
tomerg
738820
3
Did you forget $limlimits_{ntoinfty}$ somewhere?
– Ilya
Dec 4 '11 at 17:46
6
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
– Srivatsan
Dec 4 '11 at 19:09
add a comment |
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$
Thanks!
real-analysis
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
asked Dec 4 '11 at 17:45
tomerg
738820
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=max{|f(x)| ; :; x in [a,b]}$. Is it true that:
$$
M= lim_{ntoinfty}left(int_a^b|f(x)|^n,mathrm dxright)^{1/n} ?
$$
Thanks!
real-analysis
real-analysis
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
asked Dec 4 '11 at 17:45
tomerg
738820
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
asked Dec 4 '11 at 17:45
tomerg
738820
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
edited Dec 4 '11 at 19:04
Srivatsan
20.9k371124
20.9k371124
asked Dec 4 '11 at 17:45
tomerg
738820
asked Dec 4 '11 at 17:45
tomerg
738820
asked Dec 4 '11 at 17:45
tomerg
738820
738820
3
Did you forget $limlimits_{ntoinfty}$ somewhere?
– Ilya
Dec 4 '11 at 17:46
6
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
– Srivatsan
Dec 4 '11 at 19:09
add a comment |
3
Did you forget $limlimits_{ntoinfty}$ somewhere?
– Ilya
Dec 4 '11 at 17:46
6
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
– Srivatsan
Dec 4 '11 at 19:09
3
3
Did you forget $limlimits_{ntoinfty}$ somewhere?
– Ilya
Dec 4 '11 at 17:46
Did you forget $limlimits_{ntoinfty}$ somewhere?
– Ilya
Dec 4 '11 at 17:46
6
6
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
– Srivatsan
Dec 4 '11 at 19:09
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
– Srivatsan
Dec 4 '11 at 19:09
add a comment |
2 Answers
2
active
oldest
votes
Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
$$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
$$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
3
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
add a comment |
It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.
$$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$
Taking $exp$ of both sides gives the result.
answered Nov 29 at 21:38
becko
2,34431942
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
$$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
$$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
3
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
add a comment |
Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
$$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
$$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
3
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
add a comment |
Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
$$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
$$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
Put $S_{delta}:={xinleft[a,bright], |f(x)|geq M-delta}$ for any $delta>0$. Then we have for all $n$
$$Mcdot (b-a)^{frac 1n}geqleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq left(int_{S_{delta}}|f(x)|^ndxright)^{frac 1n}geq (M-delta)left(lambda(S_{delta})right)^{frac 1n}.$$
Since $f$ is continuous, the measure of $S_{delta}$ is positive and taking the $liminf$ and $limsup$, we can see that
$$Mgeq limsup_n left(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-deltaquad mbox{and }quad Mgeq liminf_nleft(int_a^b|f(x)|^ndxright)^{frac 1n}geq M-delta$$
for all $delta>0$, so $lim_{ntoinfty}left(int_a^b|f(x)|^ndxright)^{frac 1n}=M$.
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
edited Dec 5 '11 at 15:27
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
answered Dec 4 '11 at 18:06
Davide Giraudo
125k16150259
125k16150259
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
3
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
add a comment |
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
3
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
Why do you absolutely need to take the lim inf and lim sup? All the limits exists.
– Patrick Da Silva
Jan 5 '12 at 10:42
3
3
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
@PatrickDaSilva We don't know a priori that the limit $lim_nleft(int_a^b|f(x)|^nright)^{frac 1n}$ exists.
– Davide Giraudo
Jan 5 '12 at 12:36
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
Hm. I thought it was a sandwich argument at first, but I noticed the difference. Thanks for pointing it out
– Patrick Da Silva
Jan 5 '12 at 23:13
add a comment |
It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.
$$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$
Taking $exp$ of both sides gives the result.
answered Nov 29 at 21:38
becko
2,34431942
add a comment |
It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.
$$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$
Taking $exp$ of both sides gives the result.
answered Nov 29 at 21:38
becko
2,34431942
add a comment |
It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.
$$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$
Taking $exp$ of both sides gives the result.
answered Nov 29 at 21:38
becko
2,34431942
It can be seen as a particular case of Laplace's principle. For simplicity assume that $f(x) ge 0$.
$$lim_{nrightarrowinfty} log frac{1}{n} int_a^b exp(n log f(x))mathrm d x = max_{xin[a,b]} log f(x)$$
Taking $exp$ of both sides gives the result.
answered Nov 29 at 21:38
becko
2,34431942
answered Nov 29 at 21:38
becko
2,34431942
answered Nov 29 at 21:38
becko
2,34431942
answered Nov 29 at 21:38
becko
2,34431942
2,34431942
add a comment |
add a comment |
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3
Did you forget $limlimits_{ntoinfty}$ somewhere?
– Ilya
Dec 4 '11 at 17:46
6
In other words, you are asking if $| f |_n to | f |_infty$ for a continuous $f$.
– Srivatsan
Dec 4 '11 at 19:09