An unpleasant measure theory/functional analysis problem












9












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















9












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














9












9








9


6



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 9:31









orange

675215




675215










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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















3












$begingroup$

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
$$
varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
quad (ntoinfty).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















4












$begingroup$

@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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $|phi_n(f)|$ 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











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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
    $$
    varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
    quad (ntoinfty).
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    3












    $begingroup$

    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
    $$
    varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
    quad (ntoinfty).
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    3












    3








    3





    $begingroup$

    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
    $$
    varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
    quad (ntoinfty).
    $$






    share|cite|improve this answer











    $endgroup$



    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
    $$
    varphi_n(f) geq I_0+I_1 geq C n^{2alpha-1} to infty
    quad (ntoinfty).
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 13 '18 at 17:15

























    answered Dec 13 '18 at 16:36









    supinfsupinf

    6,4261028




    6,4261028












    • $begingroup$
      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




    $begingroup$
    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











    4












    $begingroup$

    @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.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      @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.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        @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.






        share|cite|improve this answer









        $endgroup$



        @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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 18:25









        zhw.zhw.

        73.6k43175




        73.6k43175























            1












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $|phi_n(f)|$ 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
















            1












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $|phi_n(f)|$ 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














            1












            1








            1





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 13 '18 at 10:00

























            answered Dec 13 '18 at 9:36









            Bartosz MalmanBartosz Malman

            8211620




            8211620












            • $begingroup$
              $|phi_n(f)|$ 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










            • $begingroup$
              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


















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