An unpleasant measure theory/functional analysis problem
Multi tool use
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I am currently taking a functional analysis course, and at the moment every student on the course is stumped by a specific question. We're looking at the bounded linear map
$$
varphi_n:L^1left([0,1]right)to mathbb{R},quad varphi_n(f) = int^1_0 f(x)(nsin(n^2 x)) dx,quad|varphi_n|=n.
$$
We are attempting to prove the existence of an $fin L^1$ with respect to the Lebesgue measure, so that $lim_{nrightarrow infty} |varphi_n(f)|rightarrow infty$, and from an earlier assignment we know that this function cannot be continuously differentiable. However, nobody has been able to make even the slightest progress. I, and all of us would appreciate a hint or a sketch of proof!
functional-analysis measure-theory
$endgroup$
|
show 5 more comments
$begingroup$
I am currently taking a functional analysis course, and at the moment every student on the course is stumped by a specific question. We're looking at the bounded linear map
$$
varphi_n:L^1left([0,1]right)to mathbb{R},quad varphi_n(f) = int^1_0 f(x)(nsin(n^2 x)) dx,quad|varphi_n|=n.
$$
We are attempting to prove the existence of an $fin L^1$ with respect to the Lebesgue measure, so that $lim_{nrightarrow infty} |varphi_n(f)|rightarrow infty$, and from an earlier assignment we know that this function cannot be continuously differentiable. However, nobody has been able to make even the slightest progress. I, and all of us would appreciate a hint or a sketch of proof!
functional-analysis measure-theory
$endgroup$
$begingroup$
Can you be more specific with the notation? Do you mean that $varphi_n$ is a linear map from, what, $L_1[0,1]$ to $mathbb{R}$? Do you mean that $[varphi_n(f)]=|int_0^1f(x)nsin(n^2x)dx|$?
$endgroup$
– Ben W
Dec 13 '18 at 9:16
$begingroup$
Have you tried for $L^2$ functions such that their Fourier series is a sum of sines? (Provided you take the integral over the full period of the sine and not $[0,1]$)
$endgroup$
– Mindlack
Dec 13 '18 at 9:18
1
$begingroup$
Have you tried uniform boundedness principle?
$endgroup$
– Song
Dec 13 '18 at 9:35
3
$begingroup$
Elswyyr: are you computing integrals of $sin(n^2x)$ over $[0,1]$ ? Meaning, there is no $2pi$ anywhere?
$endgroup$
– Mindlack
Dec 13 '18 at 10:08
1
$begingroup$
The assignment is, specifically, to prove the existence of such an $f$, so that would surprise me greatly.
$endgroup$
– Elswyyr
Dec 13 '18 at 13:37
|
show 5 more comments
$begingroup$
I am currently taking a functional analysis course, and at the moment every student on the course is stumped by a specific question. We're looking at the bounded linear map
$$
varphi_n:L^1left([0,1]right)to mathbb{R},quad varphi_n(f) = int^1_0 f(x)(nsin(n^2 x)) dx,quad|varphi_n|=n.
$$
We are attempting to prove the existence of an $fin L^1$ with respect to the Lebesgue measure, so that $lim_{nrightarrow infty} |varphi_n(f)|rightarrow infty$, and from an earlier assignment we know that this function cannot be continuously differentiable. However, nobody has been able to make even the slightest progress. I, and all of us would appreciate a hint or a sketch of proof!
functional-analysis measure-theory
$endgroup$
I am currently taking a functional analysis course, and at the moment every student on the course is stumped by a specific question. We're looking at the bounded linear map
$$
varphi_n:L^1left([0,1]right)to mathbb{R},quad varphi_n(f) = int^1_0 f(x)(nsin(n^2 x)) dx,quad|varphi_n|=n.
$$
We are attempting to prove the existence of an $fin L^1$ with respect to the Lebesgue measure, so that $lim_{nrightarrow infty} |varphi_n(f)|rightarrow infty$, and from an earlier assignment we know that this function cannot be continuously differentiable. However, nobody has been able to make even the slightest progress. I, and all of us would appreciate a hint or a sketch of proof!
functional-analysis measure-theory
functional-analysis measure-theory
edited Dec 13 '18 at 9:31
orange
687316
687316
asked Dec 13 '18 at 8:58
ElswyyrElswyyr
1307
1307
$begingroup$
Can you be more specific with the notation? Do you mean that $varphi_n$ is a linear map from, what, $L_1[0,1]$ to $mathbb{R}$? Do you mean that $[varphi_n(f)]=|int_0^1f(x)nsin(n^2x)dx|$?
$endgroup$
– Ben W
Dec 13 '18 at 9:16
$begingroup$
Have you tried for $L^2$ functions such that their Fourier series is a sum of sines? (Provided you take the integral over the full period of the sine and not $[0,1]$)
$endgroup$
– Mindlack
Dec 13 '18 at 9:18
1
$begingroup$
Have you tried uniform boundedness principle?
$endgroup$
– Song
Dec 13 '18 at 9:35
3
$begingroup$
Elswyyr: are you computing integrals of $sin(n^2x)$ over $[0,1]$ ? Meaning, there is no $2pi$ anywhere?
$endgroup$
– Mindlack
Dec 13 '18 at 10:08
1
$begingroup$
The assignment is, specifically, to prove the existence of such an $f$, so that would surprise me greatly.
$endgroup$
– Elswyyr
Dec 13 '18 at 13:37
|
show 5 more comments
$begingroup$
Can you be more specific with the notation? Do you mean that $varphi_n$ is a linear map from, what, $L_1[0,1]$ to $mathbb{R}$? Do you mean that $[varphi_n(f)]=|int_0^1f(x)nsin(n^2x)dx|$?
$endgroup$
– Ben W
Dec 13 '18 at 9:16
$begingroup$
Have you tried for $L^2$ functions such that their Fourier series is a sum of sines? (Provided you take the integral over the full period of the sine and not $[0,1]$)
$endgroup$
– Mindlack
Dec 13 '18 at 9:18
1
$begingroup$
Have you tried uniform boundedness principle?
$endgroup$
– Song
Dec 13 '18 at 9:35
3
$begingroup$
Elswyyr: are you computing integrals of $sin(n^2x)$ over $[0,1]$ ? Meaning, there is no $2pi$ anywhere?
$endgroup$
– Mindlack
Dec 13 '18 at 10:08
1
$begingroup$
The assignment is, specifically, to prove the existence of such an $f$, so that would surprise me greatly.
$endgroup$
– Elswyyr
Dec 13 '18 at 13:37
$begingroup$
Can you be more specific with the notation? Do you mean that $varphi_n$ is a linear map from, what, $L_1[0,1]$ to $mathbb{R}$? Do you mean that $[varphi_n(f)]=|int_0^1f(x)nsin(n^2x)dx|$?
$endgroup$
– Ben W
Dec 13 '18 at 9:16
$begingroup$
Can you be more specific with the notation? Do you mean that $varphi_n$ is a linear map from, what, $L_1[0,1]$ to $mathbb{R}$? Do you mean that $[varphi_n(f)]=|int_0^1f(x)nsin(n^2x)dx|$?
$endgroup$
– Ben W
Dec 13 '18 at 9:16
$begingroup$
Have you tried for $L^2$ functions such that their Fourier series is a sum of sines? (Provided you take the integral over the full period of the sine and not $[0,1]$)
$endgroup$
– Mindlack
Dec 13 '18 at 9:18
$begingroup$
Have you tried for $L^2$ functions such that their Fourier series is a sum of sines? (Provided you take the integral over the full period of the sine and not $[0,1]$)
$endgroup$
– Mindlack
Dec 13 '18 at 9:18
1
1
$begingroup$
Have you tried uniform boundedness principle?
$endgroup$
– Song
Dec 13 '18 at 9:35
$begingroup$
Have you tried uniform boundedness principle?
$endgroup$
– Song
Dec 13 '18 at 9:35
3
3
$begingroup$
Elswyyr: are you computing integrals of $sin(n^2x)$ over $[0,1]$ ? Meaning, there is no $2pi$ anywhere?
$endgroup$
– Mindlack
Dec 13 '18 at 10:08
$begingroup$
Elswyyr: are you computing integrals of $sin(n^2x)$ over $[0,1]$ ? Meaning, there is no $2pi$ anywhere?
$endgroup$
– Mindlack
Dec 13 '18 at 10:08
1
1
$begingroup$
The assignment is, specifically, to prove the existence of such an $f$, so that would surprise me greatly.
$endgroup$
– Elswyyr
Dec 13 '18 at 13:37
$begingroup$
The assignment is, specifically, to prove the existence of such an $f$, so that would surprise me greatly.
$endgroup$
– Elswyyr
Dec 13 '18 at 13:37
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I think that for $alphain (tfrac12,1)$ the function
$$
f(x) =
begin{cases}
x^{-alpha} &text{if};xin(0,1)
\
0 &text{if};xnotin (0,1)
end{cases}
$$
satisfies
$$
varphi_n(f)toinfty
qquad(ntoinfty).
$$
The proof requires some estimations because it is probably rather difficult to calculate the integral exactly.
Let $ngeq 3$ be fixed.
For $kgeq 0$, we define
$$
I_k := int_{kpi n^{-2}}^{(k+1)pi n^{-2}} f(x)(nsin(n^2 x)) mathrm dx.
$$
It is clear that $varphi_n(f) = sum_{kgeq 0} I_k$ and that $I_k=0$ for large $k$.
It is also easy to see that $I_kgeq0$ if $k$ is even and $I_kleq0$ if $k$ is odd.
Due to the monotonicity of $f$ it can be shown that
$
I_k+I_{k+1}geq 0
$
if $k$ is even.
Let us calculate a lower estimate for $I_0+I_1$.
Again, for $k=0,1,2,3$ we define the integrals
$$
J_k := int_{kpi n^{-2}/2}^{(k+1)pi n^{-2}/2} f(x)(n |sin(n^2 x)|) mathrm dx.
$$
It is clear that $I_0=J_0+J_1$ and $I_1=-J_2-J_3$.
For $k=0,1,2,3$ we know that $f(x)geq ((k+1)pi n^{-2}/2)^{-alpha}$.
Therefore we have the lower estimate
$$
J_k geq int_0^{pi n^{-2}/2}((k+1)pi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = ((k+1)pi/2)^{-alpha} n^{2alpha-1}.
$$
On the other hand, for $k=1,2,3$ we know that $f(x)leq (kpi n^{-2}/2)^{-alpha}$
Therefore we have the upper estimate
$$
J_k leq int_0^{pi n^{-2}/2}(kpi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = (kpi/2)^{-alpha} n^{2alpha-1}.
$$
Using these estimates it can be seen that
$$
I_0-I_1
=
J_0+J_1-J_2-J_3
geq n^{2alpha-1} ( (pi/2)^{-alpha} +(2pi/2)^{-alpha}-(2pi/2)^{-alpha}-(3pi/2)^{-alpha})
= C n^{2alpha-1}
$$
for some constant $C>0$.
To summarize, we have
$$
varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
quad (ntoinfty).
$$
$endgroup$
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
add a comment |
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@supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see
$$tag 1 nint^1_0 x^{-p}sin(n^2 x), dx = n^{2p-1}int_0^{n^2} y^{-p}sin(y), dy.$$
If we show $int_0^{infty} y^{-p}sin(y), dy >0,$ then it will follow that $(1)to infty$ on the order of $n^{2p-1}.$ But this integral equals
$$sum_{k=0}^{infty}int_0^{pi}sin yleft(frac{1}{(2pi k+y)^p} - frac{1}{(2pi k+pi +y)^p}right),dy.$$
Each of the integrands here is positive on $(0,pi),$ hence so is the sum, and we're done.
$endgroup$
add a comment |
$begingroup$
Uniform boundedness principle will show that a function $f$ must exist such that $|phi_n(f)|$ is unbounded. Else $phi_n(f)$ is bounded for each $f$ in a Banach space, and so $|phi_n|$ would be a bounded sequence of functionals, which it is not. I do not know however how an explicit example would look.
$endgroup$
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$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
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As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
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– Elswyyr
Dec 13 '18 at 9:56
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Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
I think that for $alphain (tfrac12,1)$ the function
$$
f(x) =
begin{cases}
x^{-alpha} &text{if};xin(0,1)
\
0 &text{if};xnotin (0,1)
end{cases}
$$
satisfies
$$
varphi_n(f)toinfty
qquad(ntoinfty).
$$
The proof requires some estimations because it is probably rather difficult to calculate the integral exactly.
Let $ngeq 3$ be fixed.
For $kgeq 0$, we define
$$
I_k := int_{kpi n^{-2}}^{(k+1)pi n^{-2}} f(x)(nsin(n^2 x)) mathrm dx.
$$
It is clear that $varphi_n(f) = sum_{kgeq 0} I_k$ and that $I_k=0$ for large $k$.
It is also easy to see that $I_kgeq0$ if $k$ is even and $I_kleq0$ if $k$ is odd.
Due to the monotonicity of $f$ it can be shown that
$
I_k+I_{k+1}geq 0
$
if $k$ is even.
Let us calculate a lower estimate for $I_0+I_1$.
Again, for $k=0,1,2,3$ we define the integrals
$$
J_k := int_{kpi n^{-2}/2}^{(k+1)pi n^{-2}/2} f(x)(n |sin(n^2 x)|) mathrm dx.
$$
It is clear that $I_0=J_0+J_1$ and $I_1=-J_2-J_3$.
For $k=0,1,2,3$ we know that $f(x)geq ((k+1)pi n^{-2}/2)^{-alpha}$.
Therefore we have the lower estimate
$$
J_k geq int_0^{pi n^{-2}/2}((k+1)pi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = ((k+1)pi/2)^{-alpha} n^{2alpha-1}.
$$
On the other hand, for $k=1,2,3$ we know that $f(x)leq (kpi n^{-2}/2)^{-alpha}$
Therefore we have the upper estimate
$$
J_k leq int_0^{pi n^{-2}/2}(kpi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = (kpi/2)^{-alpha} n^{2alpha-1}.
$$
Using these estimates it can be seen that
$$
I_0-I_1
=
J_0+J_1-J_2-J_3
geq n^{2alpha-1} ( (pi/2)^{-alpha} +(2pi/2)^{-alpha}-(2pi/2)^{-alpha}-(3pi/2)^{-alpha})
= C n^{2alpha-1}
$$
for some constant $C>0$.
To summarize, we have
$$
varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
quad (ntoinfty).
$$
$endgroup$
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
add a comment |
$begingroup$
I think that for $alphain (tfrac12,1)$ the function
$$
f(x) =
begin{cases}
x^{-alpha} &text{if};xin(0,1)
\
0 &text{if};xnotin (0,1)
end{cases}
$$
satisfies
$$
varphi_n(f)toinfty
qquad(ntoinfty).
$$
The proof requires some estimations because it is probably rather difficult to calculate the integral exactly.
Let $ngeq 3$ be fixed.
For $kgeq 0$, we define
$$
I_k := int_{kpi n^{-2}}^{(k+1)pi n^{-2}} f(x)(nsin(n^2 x)) mathrm dx.
$$
It is clear that $varphi_n(f) = sum_{kgeq 0} I_k$ and that $I_k=0$ for large $k$.
It is also easy to see that $I_kgeq0$ if $k$ is even and $I_kleq0$ if $k$ is odd.
Due to the monotonicity of $f$ it can be shown that
$
I_k+I_{k+1}geq 0
$
if $k$ is even.
Let us calculate a lower estimate for $I_0+I_1$.
Again, for $k=0,1,2,3$ we define the integrals
$$
J_k := int_{kpi n^{-2}/2}^{(k+1)pi n^{-2}/2} f(x)(n |sin(n^2 x)|) mathrm dx.
$$
It is clear that $I_0=J_0+J_1$ and $I_1=-J_2-J_3$.
For $k=0,1,2,3$ we know that $f(x)geq ((k+1)pi n^{-2}/2)^{-alpha}$.
Therefore we have the lower estimate
$$
J_k geq int_0^{pi n^{-2}/2}((k+1)pi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = ((k+1)pi/2)^{-alpha} n^{2alpha-1}.
$$
On the other hand, for $k=1,2,3$ we know that $f(x)leq (kpi n^{-2}/2)^{-alpha}$
Therefore we have the upper estimate
$$
J_k leq int_0^{pi n^{-2}/2}(kpi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = (kpi/2)^{-alpha} n^{2alpha-1}.
$$
Using these estimates it can be seen that
$$
I_0-I_1
=
J_0+J_1-J_2-J_3
geq n^{2alpha-1} ( (pi/2)^{-alpha} +(2pi/2)^{-alpha}-(2pi/2)^{-alpha}-(3pi/2)^{-alpha})
= C n^{2alpha-1}
$$
for some constant $C>0$.
To summarize, we have
$$
varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
quad (ntoinfty).
$$
$endgroup$
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
add a comment |
$begingroup$
I think that for $alphain (tfrac12,1)$ the function
$$
f(x) =
begin{cases}
x^{-alpha} &text{if};xin(0,1)
\
0 &text{if};xnotin (0,1)
end{cases}
$$
satisfies
$$
varphi_n(f)toinfty
qquad(ntoinfty).
$$
The proof requires some estimations because it is probably rather difficult to calculate the integral exactly.
Let $ngeq 3$ be fixed.
For $kgeq 0$, we define
$$
I_k := int_{kpi n^{-2}}^{(k+1)pi n^{-2}} f(x)(nsin(n^2 x)) mathrm dx.
$$
It is clear that $varphi_n(f) = sum_{kgeq 0} I_k$ and that $I_k=0$ for large $k$.
It is also easy to see that $I_kgeq0$ if $k$ is even and $I_kleq0$ if $k$ is odd.
Due to the monotonicity of $f$ it can be shown that
$
I_k+I_{k+1}geq 0
$
if $k$ is even.
Let us calculate a lower estimate for $I_0+I_1$.
Again, for $k=0,1,2,3$ we define the integrals
$$
J_k := int_{kpi n^{-2}/2}^{(k+1)pi n^{-2}/2} f(x)(n |sin(n^2 x)|) mathrm dx.
$$
It is clear that $I_0=J_0+J_1$ and $I_1=-J_2-J_3$.
For $k=0,1,2,3$ we know that $f(x)geq ((k+1)pi n^{-2}/2)^{-alpha}$.
Therefore we have the lower estimate
$$
J_k geq int_0^{pi n^{-2}/2}((k+1)pi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = ((k+1)pi/2)^{-alpha} n^{2alpha-1}.
$$
On the other hand, for $k=1,2,3$ we know that $f(x)leq (kpi n^{-2}/2)^{-alpha}$
Therefore we have the upper estimate
$$
J_k leq int_0^{pi n^{-2}/2}(kpi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = (kpi/2)^{-alpha} n^{2alpha-1}.
$$
Using these estimates it can be seen that
$$
I_0-I_1
=
J_0+J_1-J_2-J_3
geq n^{2alpha-1} ( (pi/2)^{-alpha} +(2pi/2)^{-alpha}-(2pi/2)^{-alpha}-(3pi/2)^{-alpha})
= C n^{2alpha-1}
$$
for some constant $C>0$.
To summarize, we have
$$
varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
quad (ntoinfty).
$$
$endgroup$
I think that for $alphain (tfrac12,1)$ the function
$$
f(x) =
begin{cases}
x^{-alpha} &text{if};xin(0,1)
\
0 &text{if};xnotin (0,1)
end{cases}
$$
satisfies
$$
varphi_n(f)toinfty
qquad(ntoinfty).
$$
The proof requires some estimations because it is probably rather difficult to calculate the integral exactly.
Let $ngeq 3$ be fixed.
For $kgeq 0$, we define
$$
I_k := int_{kpi n^{-2}}^{(k+1)pi n^{-2}} f(x)(nsin(n^2 x)) mathrm dx.
$$
It is clear that $varphi_n(f) = sum_{kgeq 0} I_k$ and that $I_k=0$ for large $k$.
It is also easy to see that $I_kgeq0$ if $k$ is even and $I_kleq0$ if $k$ is odd.
Due to the monotonicity of $f$ it can be shown that
$
I_k+I_{k+1}geq 0
$
if $k$ is even.
Let us calculate a lower estimate for $I_0+I_1$.
Again, for $k=0,1,2,3$ we define the integrals
$$
J_k := int_{kpi n^{-2}/2}^{(k+1)pi n^{-2}/2} f(x)(n |sin(n^2 x)|) mathrm dx.
$$
It is clear that $I_0=J_0+J_1$ and $I_1=-J_2-J_3$.
For $k=0,1,2,3$ we know that $f(x)geq ((k+1)pi n^{-2}/2)^{-alpha}$.
Therefore we have the lower estimate
$$
J_k geq int_0^{pi n^{-2}/2}((k+1)pi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = ((k+1)pi/2)^{-alpha} n^{2alpha-1}.
$$
On the other hand, for $k=1,2,3$ we know that $f(x)leq (kpi n^{-2}/2)^{-alpha}$
Therefore we have the upper estimate
$$
J_k leq int_0^{pi n^{-2}/2}(kpi n^{-2}/2)^{-alpha} n sin(n^2 x)mathrm dx
= dots = (kpi/2)^{-alpha} n^{2alpha-1}.
$$
Using these estimates it can be seen that
$$
I_0-I_1
=
J_0+J_1-J_2-J_3
geq n^{2alpha-1} ( (pi/2)^{-alpha} +(2pi/2)^{-alpha}-(2pi/2)^{-alpha}-(3pi/2)^{-alpha})
= C n^{2alpha-1}
$$
for some constant $C>0$.
To summarize, we have
$$
varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
quad (ntoinfty).
$$
edited Dec 13 '18 at 17:15
answered Dec 13 '18 at 16:36
supinfsupinf
6,7411029
6,7411029
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
add a comment |
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
$begingroup$
Nice examples,+1. I was surprised such elementary examples worked. I gave another proof, which seemed to me simpler.
$endgroup$
– zhw.
Dec 13 '18 at 18:28
add a comment |
$begingroup$
@supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see
$$tag 1 nint^1_0 x^{-p}sin(n^2 x), dx = n^{2p-1}int_0^{n^2} y^{-p}sin(y), dy.$$
If we show $int_0^{infty} y^{-p}sin(y), dy >0,$ then it will follow that $(1)to infty$ on the order of $n^{2p-1}.$ But this integral equals
$$sum_{k=0}^{infty}int_0^{pi}sin yleft(frac{1}{(2pi k+y)^p} - frac{1}{(2pi k+pi +y)^p}right),dy.$$
Each of the integrands here is positive on $(0,pi),$ hence so is the sum, and we're done.
$endgroup$
add a comment |
$begingroup$
@supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see
$$tag 1 nint^1_0 x^{-p}sin(n^2 x), dx = n^{2p-1}int_0^{n^2} y^{-p}sin(y), dy.$$
If we show $int_0^{infty} y^{-p}sin(y), dy >0,$ then it will follow that $(1)to infty$ on the order of $n^{2p-1}.$ But this integral equals
$$sum_{k=0}^{infty}int_0^{pi}sin yleft(frac{1}{(2pi k+y)^p} - frac{1}{(2pi k+pi +y)^p}right),dy.$$
Each of the integrands here is positive on $(0,pi),$ hence so is the sum, and we're done.
$endgroup$
add a comment |
$begingroup$
@supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see
$$tag 1 nint^1_0 x^{-p}sin(n^2 x), dx = n^{2p-1}int_0^{n^2} y^{-p}sin(y), dy.$$
If we show $int_0^{infty} y^{-p}sin(y), dy >0,$ then it will follow that $(1)to infty$ on the order of $n^{2p-1}.$ But this integral equals
$$sum_{k=0}^{infty}int_0^{pi}sin yleft(frac{1}{(2pi k+y)^p} - frac{1}{(2pi k+pi +y)^p}right),dy.$$
Each of the integrands here is positive on $(0,pi),$ hence so is the sum, and we're done.
$endgroup$
@supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see
$$tag 1 nint^1_0 x^{-p}sin(n^2 x), dx = n^{2p-1}int_0^{n^2} y^{-p}sin(y), dy.$$
If we show $int_0^{infty} y^{-p}sin(y), dy >0,$ then it will follow that $(1)to infty$ on the order of $n^{2p-1}.$ But this integral equals
$$sum_{k=0}^{infty}int_0^{pi}sin yleft(frac{1}{(2pi k+y)^p} - frac{1}{(2pi k+pi +y)^p}right),dy.$$
Each of the integrands here is positive on $(0,pi),$ hence so is the sum, and we're done.
answered Dec 13 '18 at 18:25
zhw.zhw.
74.8k43275
74.8k43275
add a comment |
add a comment |
$begingroup$
Uniform boundedness principle will show that a function $f$ must exist such that $|phi_n(f)|$ is unbounded. Else $phi_n(f)$ is bounded for each $f$ in a Banach space, and so $|phi_n|$ would be a bounded sequence of functionals, which it is not. I do not know however how an explicit example would look.
$endgroup$
$begingroup$
$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
$begingroup$
As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
$endgroup$
– Elswyyr
Dec 13 '18 at 9:56
$begingroup$
Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
add a comment |
$begingroup$
Uniform boundedness principle will show that a function $f$ must exist such that $|phi_n(f)|$ is unbounded. Else $phi_n(f)$ is bounded for each $f$ in a Banach space, and so $|phi_n|$ would be a bounded sequence of functionals, which it is not. I do not know however how an explicit example would look.
$endgroup$
$begingroup$
$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
$begingroup$
As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
$endgroup$
– Elswyyr
Dec 13 '18 at 9:56
$begingroup$
Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
add a comment |
$begingroup$
Uniform boundedness principle will show that a function $f$ must exist such that $|phi_n(f)|$ is unbounded. Else $phi_n(f)$ is bounded for each $f$ in a Banach space, and so $|phi_n|$ would be a bounded sequence of functionals, which it is not. I do not know however how an explicit example would look.
$endgroup$
Uniform boundedness principle will show that a function $f$ must exist such that $|phi_n(f)|$ is unbounded. Else $phi_n(f)$ is bounded for each $f$ in a Banach space, and so $|phi_n|$ would be a bounded sequence of functionals, which it is not. I do not know however how an explicit example would look.
edited Dec 13 '18 at 10:00
answered Dec 13 '18 at 9:36
Bartosz MalmanBartosz Malman
8311620
8311620
$begingroup$
$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
$begingroup$
As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
$endgroup$
– Elswyyr
Dec 13 '18 at 9:56
$begingroup$
Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
add a comment |
$begingroup$
$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
$begingroup$
As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
$endgroup$
– Elswyyr
Dec 13 '18 at 9:56
$begingroup$
Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
$begingroup$
$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
$begingroup$
$|phi_n(f)|$ unbounded is not the same thins as $|phi_n(f)| to infty$, right?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 9:39
$begingroup$
As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
$endgroup$
– Elswyyr
Dec 13 '18 at 9:56
$begingroup$
As near as I can tell, this proves that there's some f so that $|varphi_n(f)|$ is unbounded, but, as Kavi commented, not that $|varphi_n(f)|rightarrow infty$. I don't think it's quite sufficient without more arguments.
$endgroup$
– Elswyyr
Dec 13 '18 at 9:56
$begingroup$
Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
$begingroup$
Oh, I misinterpreted. You guys are of course correct.
$endgroup$
– Bartosz Malman
Dec 13 '18 at 10:00
add a comment |
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tHbpz0X2xeDVK
$begingroup$
Can you be more specific with the notation? Do you mean that $varphi_n$ is a linear map from, what, $L_1[0,1]$ to $mathbb{R}$? Do you mean that $[varphi_n(f)]=|int_0^1f(x)nsin(n^2x)dx|$?
$endgroup$
– Ben W
Dec 13 '18 at 9:16
$begingroup$
Have you tried for $L^2$ functions such that their Fourier series is a sum of sines? (Provided you take the integral over the full period of the sine and not $[0,1]$)
$endgroup$
– Mindlack
Dec 13 '18 at 9:18
1
$begingroup$
Have you tried uniform boundedness principle?
$endgroup$
– Song
Dec 13 '18 at 9:35
3
$begingroup$
Elswyyr: are you computing integrals of $sin(n^2x)$ over $[0,1]$ ? Meaning, there is no $2pi$ anywhere?
$endgroup$
– Mindlack
Dec 13 '18 at 10:08
1
$begingroup$
The assignment is, specifically, to prove the existence of such an $f$, so that would surprise me greatly.
$endgroup$
– Elswyyr
Dec 13 '18 at 13:37