Prove $int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$












0















Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$




Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.






calculus







share|cite|improve this question


















  • 3




    This is Poincaré inequality
    – Zachary Selk
    Nov 24 at 4:39










  • @ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
    – Seewoo Lee
    Nov 24 at 4:45










  • @SeewooLee See for example here math.stackexchange.com/questions/360670/…
    – Zachary Selk
    Nov 24 at 5:03
















0















Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$




Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.






calculus







share|cite|improve this question


















  • 3




    This is Poincaré inequality
    – Zachary Selk
    Nov 24 at 4:39










  • @ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
    – Seewoo Lee
    Nov 24 at 4:45










  • @SeewooLee See for example here math.stackexchange.com/questions/360670/…
    – Zachary Selk
    Nov 24 at 5:03














0












0








0








Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$




Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.






calculus







share|cite|improve this question














Suppose $f$ is dericative and continuous on $[0,1]$, $f(0)=f(1)=0$, Prove
$$int_0^1|f(x)|^2dxlefrac{1}{4}int_0^1|f'(x)|^2dx$$




Tried to expand it as Taylor series, after trying some different points , still don't know how to connect $f$ and $f'$.






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 at 4:37









Jaqen Chou

406110




406110








  • 3




    This is Poincaré inequality
    – Zachary Selk
    Nov 24 at 4:39










  • @ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
    – Seewoo Lee
    Nov 24 at 4:45










  • @SeewooLee See for example here math.stackexchange.com/questions/360670/…
    – Zachary Selk
    Nov 24 at 5:03














  • 3




    This is Poincaré inequality
    – Zachary Selk
    Nov 24 at 4:39










  • @ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
    – Seewoo Lee
    Nov 24 at 4:45










  • @SeewooLee See for example here math.stackexchange.com/questions/360670/…
    – Zachary Selk
    Nov 24 at 5:03








3




3




This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39




This is Poincaré inequality
– Zachary Selk
Nov 24 at 4:39












@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45




@ZacharySelk Is there an effective version of the inequality? It is true that $L^{2}$ norm of the original function can be bounded by the constant times $L^{2}$ norm of $f'$, but I can't find any reference that gives the explicit value of the constant.
– Seewoo Lee
Nov 24 at 4:45












@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03




@SeewooLee See for example here math.stackexchange.com/questions/360670/…
– Zachary Selk
Nov 24 at 5:03










1 Answer
1






active

oldest

votes


















4














Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.






share|cite|improve this answer























  • [+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
    – Jean Marie
    Nov 24 at 5:24










  • why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
    – Jaqen Chou
    Nov 24 at 16:10










  • @JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
    – Seewoo Lee
    Nov 24 at 20:29










  • Thanks. And why the expansion only about $sin$ not involved $cos$.
    – Jaqen Chou
    Nov 25 at 2:30











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4














Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.






share|cite|improve this answer























  • [+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
    – Jean Marie
    Nov 24 at 5:24










  • why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
    – Jaqen Chou
    Nov 24 at 16:10










  • @JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
    – Seewoo Lee
    Nov 24 at 20:29










  • Thanks. And why the expansion only about $sin$ not involved $cos$.
    – Jaqen Chou
    Nov 25 at 2:30
















4














Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.






share|cite|improve this answer























  • [+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
    – Jean Marie
    Nov 24 at 5:24










  • why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
    – Jaqen Chou
    Nov 24 at 16:10










  • @JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
    – Seewoo Lee
    Nov 24 at 20:29










  • Thanks. And why the expansion only about $sin$ not involved $cos$.
    – Jaqen Chou
    Nov 25 at 2:30














4












4








4






Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.






share|cite|improve this answer














Consider the Fourier expansion of $f(x)$:
$$
f(x) = sum_{n=1}^{infty} a_{n}sin(npi x).
$$
We have
$$
int_{0}^{1}f(x)^{2}dx = frac{1}{2}(a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots)
$$
and
$$
int_{0}^{1}f'(x)^{2}dx = frac{pi^{2}}{2}(a_{1}^{2} + 4a_{2}^{2} + 9a_{3}^{2} + cdots)
$$
so actually we have a stronger inequality
$$
int_{0}^{1} f(x)^{2}dx leq frac{1}{pi^{2}}int_{0}^{1}f'(x)^{2}dx
$$
where the equality holds for $f(x) = csin(pi x)$ for any $cin mathbb{R}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 20:30

























answered Nov 24 at 5:08









Seewoo Lee

6,165826




6,165826












  • [+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
    – Jean Marie
    Nov 24 at 5:24










  • why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
    – Jaqen Chou
    Nov 24 at 16:10










  • @JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
    – Seewoo Lee
    Nov 24 at 20:29










  • Thanks. And why the expansion only about $sin$ not involved $cos$.
    – Jaqen Chou
    Nov 25 at 2:30


















  • [+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
    – Jean Marie
    Nov 24 at 5:24










  • why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
    – Jaqen Chou
    Nov 24 at 16:10










  • @JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
    – Seewoo Lee
    Nov 24 at 20:29










  • Thanks. And why the expansion only about $sin$ not involved $cos$.
    – Jaqen Chou
    Nov 25 at 2:30
















[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24




[+1] Excellent. Maybe say that the first formula is an "odd Fourier series extension".
– Jean Marie
Nov 24 at 5:24












why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10




why $ int_{0}^{1}f(x)^{2}dx = a_{1}^{2} + a_{2}^{2} +a_{3}^{2} + cdots $. How to calculate it directly?
– Jaqen Chou
Nov 24 at 16:10












@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29




@JaqenChou It is called Parseval's theorem. Formally, you can take a square of the Fourier expansion and use the fact that $int_{0}^{1}sin^{2}(npi x)dx = 1/2$, $int_{0}^{1}sin(npi x)sin(mpi x)dx = 0$ for $nneq m$. (I realized that we need $1/2$ on the both side. I edited.
– Seewoo Lee
Nov 24 at 20:29












Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30




Thanks. And why the expansion only about $sin$ not involved $cos$.
– Jaqen Chou
Nov 25 at 2:30


















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