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












3














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







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  • 2




    Wirtinger's inequality may be helpful to you.
    – B. Mehta
    2 days ago
















3














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




















  • 2




    Wirtinger's inequality may be helpful to you.
    – B. Mehta
    2 days ago














3












3








3


1





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















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 2 days ago









dmtri

1,4201521




1,4201521










asked Dec 31 '18 at 4:55









math_novice

347




347








  • 2




    Wirtinger's inequality may be helpful to you.
    – B. Mehta
    2 days ago














  • 2




    Wirtinger's inequality may be helpful to you.
    – B. Mehta
    2 days ago








2




2




Wirtinger's inequality may be helpful to you.
– B. Mehta
2 days ago




Wirtinger's inequality may be helpful to you.
– B. Mehta
2 days ago










1 Answer
1






active

oldest

votes


















4














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























  • Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    – math_novice
    2 days ago






  • 1




    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    – math_novice
    2 days ago








  • 1




    @math_novice I've edited my answer. I hope this will make it clear.
    – Song
    2 days ago










  • That is very clear! Thank you for the patience!
    – math_novice
    2 days ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














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























  • Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    – math_novice
    2 days ago






  • 1




    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    – math_novice
    2 days ago








  • 1




    @math_novice I've edited my answer. I hope this will make it clear.
    – Song
    2 days ago










  • That is very clear! Thank you for the patience!
    – math_novice
    2 days ago
















4














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























  • Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    – math_novice
    2 days ago






  • 1




    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    – math_novice
    2 days ago








  • 1




    @math_novice I've edited my answer. I hope this will make it clear.
    – Song
    2 days ago










  • That is very clear! Thank you for the patience!
    – math_novice
    2 days ago














4












4








4






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














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 2 days ago

























answered 2 days ago









Song

5,090317




5,090317












  • Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    – math_novice
    2 days ago






  • 1




    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    – math_novice
    2 days ago








  • 1




    @math_novice I've edited my answer. I hope this will make it clear.
    – Song
    2 days ago










  • That is very clear! Thank you for the patience!
    – math_novice
    2 days ago


















  • Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
    – math_novice
    2 days ago






  • 1




    One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
    – math_novice
    2 days ago








  • 1




    @math_novice I've edited my answer. I hope this will make it clear.
    – Song
    2 days ago










  • That is very clear! Thank you for the patience!
    – math_novice
    2 days ago
















Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
– math_novice
2 days ago




Thank you very much! I should have thought about Fourier series when I saw the interval $(0,pi)$!
– math_novice
2 days ago




1




1




One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
– math_novice
2 days ago






One more question on your answer: how did you get the left hand side of the second inequality? Shouldn't it be $(f ^2- bar f ^2)$?
– math_novice
2 days ago






1




1




@math_novice I've edited my answer. I hope this will make it clear.
– Song
2 days ago




@math_novice I've edited my answer. I hope this will make it clear.
– Song
2 days ago












That is very clear! Thank you for the patience!
– math_novice
2 days ago




That is very clear! Thank you for the patience!
– math_novice
2 days ago


















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