Proof of an inequality for a function $f$ in the Hilbert space












3












$begingroup$


Prove that for any $ f in H^1(0,pi)$:



begin{equation}
int_0^pi f^2 dx leq int_0^pi left(f'right)^2 dx + left(int_0^pi f dxright)^2
end{equation}



$H$ is the Hilbert space. 1 means that the 1st (weak) derivative exists.



I am thinking of applying some inequalities involving $L^2$ norms, such as Hoelder, but right now I can't think of a way to make it work.
A further hint would be greatly appreciated!






real-analysis functional-analysis inequality pde hilbert-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Wirtinger's inequality may be helpful to you.
    $endgroup$
    – B. Mehta
    Dec 31 '18 at 5:20
















3












$begingroup$


Prove that for any $ f in H^1(0,pi)$:



begin{equation}
int_0^pi f^2 dx leq int_0^pi left(f'right)^2 dx + left(int_0^pi f dxright)^2
end{equation}



$H$ is the Hilbert space. 1 means that the 1st (weak) derivative exists.



I am thinking of applying some inequalities involving $L^2$ norms, such as Hoelder, but right now I can't think of a way to make it work.
A further hint would be greatly appreciated!






real-analysis functional-analysis inequality pde hilbert-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Wirtinger's inequality may be helpful to you.
    $endgroup$
    – B. Mehta
    Dec 31 '18 at 5:20














3












3








3


1



$begingroup$


Prove that for any $ f in H^1(0,pi)$:



begin{equation}
int_0^pi f^2 dx leq int_0^pi left(f'right)^2 dx + left(int_0^pi f dxright)^2
end{equation}



$H$ is the Hilbert space. 1 means that the 1st (weak) derivative exists.



I am thinking of applying some inequalities involving $L^2$ norms, such as Hoelder, but right now I can't think of a way to make it work.
A further hint would be greatly appreciated!






real-analysis functional-analysis inequality pde hilbert-spaces







share|cite|improve this question











$endgroup$




Prove that for any $ f in H^1(0,pi)$:



begin{equation}
int_0^pi f^2 dx leq int_0^pi left(f'right)^2 dx + left(int_0^pi f dxright)^2
end{equation}



$H$ is the Hilbert space. 1 means that the 1st (weak) derivative exists.



I am thinking of applying some inequalities involving $L^2$ norms, such as Hoelder, but right now I can't think of a way to make it work.
A further hint would be greatly appreciated!






real-analysis functional-analysis inequality pde hilbert-spaces




real-analysis functional-analysis inequality pde hilbert-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 8:31









dmtri

1,4802521




1,4802521










asked Dec 31 '18 at 4:55









math_novicemath_novice

367




367








  • 2




    $begingroup$
    Wirtinger's inequality may be helpful to you.
    $endgroup$
    – B. Mehta
    Dec 31 '18 at 5:20














  • 2




    $begingroup$
    Wirtinger's inequality may be helpful to you.
    $endgroup$
    – B. Mehta
    Dec 31 '18 at 5:20








2




2




$begingroup$
Wirtinger's inequality may be helpful to you.
$endgroup$
– B. Mehta
Dec 31 '18 at 5:20




$begingroup$
Wirtinger's inequality may be helpful to you.
$endgroup$
– B. Mehta
Dec 31 '18 at 5:20










1 Answer
1






active

oldest

votes


















4












$begingroup$

The given estimate is not the best. Indeed, we can show that
$$
int_0^pi |f|^2 dx leq frac{1}{4}int_0^pi |f'|^2 dx + frac{1}{pi}left(int_0^pi f dxright)^2.
$$This inequality can be rephrased as
$$
int_0^pi (f-overline{f})^2 dx leqfrac{1}{4} int_0^pi |f'|^2dx
$$ where $overline{f} =frac{1}{pi}int_0^pi f dx$. We can see this from
$$
int_0^pi (f-overline{f})^2 dx = int_0^pi (f^2-2overline{f}f +overline{f}^2)dx = int_0^pi f^2 dx -2pi overline{f}^2 + pi overline{f}^2 = int_0^pi f^2 dx -pi overline{f}^2.
$$Without loss of generality, we may assume $overline{f} = 0$, i.e. $int_0^pi fdx = 0$. Now, the result follows from the Fourier series of $f(x)=sum_{kneq 0} widehat{f}(k) e^{i2kx}$ and Parseval's identity:
$$
int_0^pi |f|^2 dx = pisum_{kneq 0}|widehat{f}(k)|^2leq frac{pi}{4}sum_{kinmathbb{Z}}|4k^2||widehat{f}(k)|^2=frac{1}{4}int_0^pi |f'|^2 dx.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:27






  • 1




    $begingroup$
    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:41








  • 1




    $begingroup$
    @math_novice I've edited my answer. I hope this will make it clear.
    $endgroup$
    – Song
    Dec 31 '18 at 5:46










  • $begingroup$
    That is very clear! Thank you for the patience!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The given estimate is not the best. Indeed, we can show that
$$
int_0^pi |f|^2 dx leq frac{1}{4}int_0^pi |f'|^2 dx + frac{1}{pi}left(int_0^pi f dxright)^2.
$$This inequality can be rephrased as
$$
int_0^pi (f-overline{f})^2 dx leqfrac{1}{4} int_0^pi |f'|^2dx
$$ where $overline{f} =frac{1}{pi}int_0^pi f dx$. We can see this from
$$
int_0^pi (f-overline{f})^2 dx = int_0^pi (f^2-2overline{f}f +overline{f}^2)dx = int_0^pi f^2 dx -2pi overline{f}^2 + pi overline{f}^2 = int_0^pi f^2 dx -pi overline{f}^2.
$$Without loss of generality, we may assume $overline{f} = 0$, i.e. $int_0^pi fdx = 0$. Now, the result follows from the Fourier series of $f(x)=sum_{kneq 0} widehat{f}(k) e^{i2kx}$ and Parseval's identity:
$$
int_0^pi |f|^2 dx = pisum_{kneq 0}|widehat{f}(k)|^2leq frac{pi}{4}sum_{kinmathbb{Z}}|4k^2||widehat{f}(k)|^2=frac{1}{4}int_0^pi |f'|^2 dx.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:27






  • 1




    $begingroup$
    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:41








  • 1




    $begingroup$
    @math_novice I've edited my answer. I hope this will make it clear.
    $endgroup$
    – Song
    Dec 31 '18 at 5:46










  • $begingroup$
    That is very clear! Thank you for the patience!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:49
















4












$begingroup$

The given estimate is not the best. Indeed, we can show that
$$
int_0^pi |f|^2 dx leq frac{1}{4}int_0^pi |f'|^2 dx + frac{1}{pi}left(int_0^pi f dxright)^2.
$$This inequality can be rephrased as
$$
int_0^pi (f-overline{f})^2 dx leqfrac{1}{4} int_0^pi |f'|^2dx
$$ where $overline{f} =frac{1}{pi}int_0^pi f dx$. We can see this from
$$
int_0^pi (f-overline{f})^2 dx = int_0^pi (f^2-2overline{f}f +overline{f}^2)dx = int_0^pi f^2 dx -2pi overline{f}^2 + pi overline{f}^2 = int_0^pi f^2 dx -pi overline{f}^2.
$$Without loss of generality, we may assume $overline{f} = 0$, i.e. $int_0^pi fdx = 0$. Now, the result follows from the Fourier series of $f(x)=sum_{kneq 0} widehat{f}(k) e^{i2kx}$ and Parseval's identity:
$$
int_0^pi |f|^2 dx = pisum_{kneq 0}|widehat{f}(k)|^2leq frac{pi}{4}sum_{kinmathbb{Z}}|4k^2||widehat{f}(k)|^2=frac{1}{4}int_0^pi |f'|^2 dx.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:27






  • 1




    $begingroup$
    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:41








  • 1




    $begingroup$
    @math_novice I've edited my answer. I hope this will make it clear.
    $endgroup$
    – Song
    Dec 31 '18 at 5:46










  • $begingroup$
    That is very clear! Thank you for the patience!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:49














4












4








4





$begingroup$

The given estimate is not the best. Indeed, we can show that
$$
int_0^pi |f|^2 dx leq frac{1}{4}int_0^pi |f'|^2 dx + frac{1}{pi}left(int_0^pi f dxright)^2.
$$This inequality can be rephrased as
$$
int_0^pi (f-overline{f})^2 dx leqfrac{1}{4} int_0^pi |f'|^2dx
$$ where $overline{f} =frac{1}{pi}int_0^pi f dx$. We can see this from
$$
int_0^pi (f-overline{f})^2 dx = int_0^pi (f^2-2overline{f}f +overline{f}^2)dx = int_0^pi f^2 dx -2pi overline{f}^2 + pi overline{f}^2 = int_0^pi f^2 dx -pi overline{f}^2.
$$Without loss of generality, we may assume $overline{f} = 0$, i.e. $int_0^pi fdx = 0$. Now, the result follows from the Fourier series of $f(x)=sum_{kneq 0} widehat{f}(k) e^{i2kx}$ and Parseval's identity:
$$
int_0^pi |f|^2 dx = pisum_{kneq 0}|widehat{f}(k)|^2leq frac{pi}{4}sum_{kinmathbb{Z}}|4k^2||widehat{f}(k)|^2=frac{1}{4}int_0^pi |f'|^2 dx.
$$






share|cite|improve this answer











$endgroup$



The given estimate is not the best. Indeed, we can show that
$$
int_0^pi |f|^2 dx leq frac{1}{4}int_0^pi |f'|^2 dx + frac{1}{pi}left(int_0^pi f dxright)^2.
$$This inequality can be rephrased as
$$
int_0^pi (f-overline{f})^2 dx leqfrac{1}{4} int_0^pi |f'|^2dx
$$ where $overline{f} =frac{1}{pi}int_0^pi f dx$. We can see this from
$$
int_0^pi (f-overline{f})^2 dx = int_0^pi (f^2-2overline{f}f +overline{f}^2)dx = int_0^pi f^2 dx -2pi overline{f}^2 + pi overline{f}^2 = int_0^pi f^2 dx -pi overline{f}^2.
$$Without loss of generality, we may assume $overline{f} = 0$, i.e. $int_0^pi fdx = 0$. Now, the result follows from the Fourier series of $f(x)=sum_{kneq 0} widehat{f}(k) e^{i2kx}$ and Parseval's identity:
$$
int_0^pi |f|^2 dx = pisum_{kneq 0}|widehat{f}(k)|^2leq frac{pi}{4}sum_{kinmathbb{Z}}|4k^2||widehat{f}(k)|^2=frac{1}{4}int_0^pi |f'|^2 dx.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 5:45

























answered Dec 31 '18 at 5:14









SongSong

11.7k628




11.7k628












  • $begingroup$
    Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:27






  • 1




    $begingroup$
    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:41








  • 1




    $begingroup$
    @math_novice I've edited my answer. I hope this will make it clear.
    $endgroup$
    – Song
    Dec 31 '18 at 5:46










  • $begingroup$
    That is very clear! Thank you for the patience!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:49


















  • $begingroup$
    Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:27






  • 1




    $begingroup$
    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:41








  • 1




    $begingroup$
    @math_novice I've edited my answer. I hope this will make it clear.
    $endgroup$
    – Song
    Dec 31 '18 at 5:46










  • $begingroup$
    That is very clear! Thank you for the patience!
    $endgroup$
    – math_novice
    Dec 31 '18 at 5:49
















$begingroup$
Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
$endgroup$
– math_novice
Dec 31 '18 at 5:27




$begingroup$
Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
$endgroup$
– math_novice
Dec 31 '18 at 5:27




1




1




$begingroup$
One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
$endgroup$
– math_novice
Dec 31 '18 at 5:41






$begingroup$
One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
$endgroup$
– math_novice
Dec 31 '18 at 5:41






1




1




$begingroup$
@math_novice I've edited my answer. I hope this will make it clear.
$endgroup$
– Song
Dec 31 '18 at 5:46




$begingroup$
@math_novice I've edited my answer. I hope this will make it clear.
$endgroup$
– Song
Dec 31 '18 at 5:46












$begingroup$
That is very clear! Thank you for the patience!
$endgroup$
– math_novice
Dec 31 '18 at 5:49




$begingroup$
That is very clear! Thank you for the patience!
$endgroup$
– math_novice
Dec 31 '18 at 5:49


















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