Mixing
Poincaré constant for subspace of Hilbert space
$begingroup$
Let $V={h in H^1(0,L)mid h(0)=0}$. I wish to show a Poincaré-type inequality for $vin V$ with a specific constant:
$$| v|_{L^2(0,L)} le frac{2L}{pi} | v' |_{L^2(0,L)}$$
I already have a sketch for a proof involving $L/pi$ that uses Fourier series. I can't seem to get the better constant with the same approach.
inequality hilbert-spaces normed-spaces
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
$begingroup$
Let $V={h in H^1(0,L)mid h(0)=0}$. I wish to show a Poincaré-type inequality for $vin V$ with a specific constant:
$$| v|_{L^2(0,L)} le frac{2L}{pi} | v' |_{L^2(0,L)}$$
I already have a sketch for a proof involving $L/pi$ that uses Fourier series. I can't seem to get the better constant with the same approach.
inequality hilbert-spaces normed-spaces
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
$endgroup$
$begingroup$
see: mis.mpg.de/preprints/2003/preprint2003_24.pdf
$endgroup$
– Aleksas Domarkas
Dec 29 '18 at 22:50
$begingroup$
@Aleksas Domarkas it looks useful, but I can't quite extract the information I need.
$endgroup$
– Niki Di Giano
Dec 30 '18 at 0:20
add a comment |
$begingroup$
Let $V={h in H^1(0,L)mid h(0)=0}$. I wish to show a Poincaré-type inequality for $vin V$ with a specific constant:
$$| v|_{L^2(0,L)} le frac{2L}{pi} | v' |_{L^2(0,L)}$$
I already have a sketch for a proof involving $L/pi$ that uses Fourier series. I can't seem to get the better constant with the same approach.
inequality hilbert-spaces normed-spaces
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
$endgroup$
Let $V={h in H^1(0,L)mid h(0)=0}$. I wish to show a Poincaré-type inequality for $vin V$ with a specific constant:
$$| v|_{L^2(0,L)} le frac{2L}{pi} | v' |_{L^2(0,L)}$$
I already have a sketch for a proof involving $L/pi$ that uses Fourier series. I can't seem to get the better constant with the same approach.
inequality hilbert-spaces normed-spaces
inequality hilbert-spaces normed-spaces
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
edited Dec 30 '18 at 10:36
Niki Di Giano
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
asked Dec 29 '18 at 19:42
Niki Di GianoNiki Di Giano
941211
941211
$begingroup$
see: mis.mpg.de/preprints/2003/preprint2003_24.pdf
$endgroup$
– Aleksas Domarkas
Dec 29 '18 at 22:50
$begingroup$
@Aleksas Domarkas it looks useful, but I can't quite extract the information I need.
$endgroup$
– Niki Di Giano
Dec 30 '18 at 0:20
add a comment |
$begingroup$
see: mis.mpg.de/preprints/2003/preprint2003_24.pdf
$endgroup$
– Aleksas Domarkas
Dec 29 '18 at 22:50
$begingroup$
@Aleksas Domarkas it looks useful, but I can't quite extract the information I need.
$endgroup$
– Niki Di Giano
Dec 30 '18 at 0:20
$begingroup$
see: mis.mpg.de/preprints/2003/preprint2003_24.pdf
$endgroup$
– Aleksas Domarkas
Dec 29 '18 at 22:50
$begingroup$
see: mis.mpg.de/preprints/2003/preprint2003_24.pdf
$endgroup$
– Aleksas Domarkas
Dec 29 '18 at 22:50
$begingroup$
@Aleksas Domarkas it looks useful, but I can't quite extract the information I need.
$endgroup$
– Niki Di Giano
Dec 30 '18 at 0:20
$begingroup$
@Aleksas Domarkas it looks useful, but I can't quite extract the information I need.
$endgroup$
– Niki Di Giano
Dec 30 '18 at 0:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I've found a way. First, consider the space $H_0^1(0,L)$. This is a subspace of $V$ and $H^1$. Let $hin H_0^1(0,L)$, thus we can define:
$$tilde{h}=h quad text{for} ; xge 0, quad -h quad text{for} ; x<0$$
$tilde{h}$ is an odd, $2L$-periodic function. Its Fourier series will only contain odd terms:
$$tilde{h} = sum_n b_n sin(pi nx/L)$$
And it's easy to see that this Fourier series converges uniformly to the function $tilde{h}$, since the function is continuous on $Bbb{R}$ and its derivative is piecewise continuous. Then $tilde{h}'$ can be obtained by differentiating each term of the series:
$$tilde{h} = sum_n frac{pi n b_n}{L} cos(pi nx/L)$$
Now applying Bessel's identity we get:
$$||tilde{h}||_{L^2(-L,L)}^2 = sum_n b_n^2 \\
||tilde{h}'||_{L^2(-L,L)}^2 = sum_n Big(frac{pi n b_n}{L}Big)^2 \\
implies ||tilde{h}'||_{L^2(-L,L)}^2 ge sum_n Big(frac{pi b_n}{L}Big)^2 = frac{pi^2}{L^2} sum_n b_n^2 \\
implies||tilde{h}'||_{L^2(-L,L)} ge frac{pi}{L} ||tilde{h}||_{L^2(-L,L)}$$
We will now notice that:
$$||tilde{h}||_{L^2(-L,L)}^2 = 2||h||_{L^2(0,L)}^2 \
||tilde{h}' ||_{L^2(-L,L)}^2 = 2||h'||_{L^2(0,L)}^2$$
And this allows us to say, finally, that:
$$||h||_{L^2(0,L)} le frac{L}{pi}||h'||_{L^2(0,L)}$$
To get the final inequality, note that for any function $vin V$ we may define a function $bar{v} in H_0^1(0, 2L)$ as follows:
$$bar{v}(x) = v(x)quad text{for} ; 0le x < L, quad v(2L - x) quad text{for} ; Lle x le 2L$$
Which is symmetric with respect to the vertical axis centered in $L$, thus $bar{v} in H_0^1(0, 2L)$. As before:
$$||bar{v}||_{L^2(0,2L)}^2 = 2||v||_{L^2(0,L)}^2 \
||bar{v}' ||_{L^2(0,2L)}^2 = 2||v'||_{L^2(0,L)}^2$$
By the argument above, it immediately follows that:
$$||v||_{L^2(0,L)} le frac{2L}{pi}||v'||_{L^2(0,L)}$$
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I've found a way. First, consider the space $H_0^1(0,L)$. This is a subspace of $V$ and $H^1$. Let $hin H_0^1(0,L)$, thus we can define:
$$tilde{h}=h quad text{for} ; xge 0, quad -h quad text{for} ; x<0$$
$tilde{h}$ is an odd, $2L$-periodic function. Its Fourier series will only contain odd terms:
$$tilde{h} = sum_n b_n sin(pi nx/L)$$
And it's easy to see that this Fourier series converges uniformly to the function $tilde{h}$, since the function is continuous on $Bbb{R}$ and its derivative is piecewise continuous. Then $tilde{h}'$ can be obtained by differentiating each term of the series:
$$tilde{h} = sum_n frac{pi n b_n}{L} cos(pi nx/L)$$
Now applying Bessel's identity we get:
$$||tilde{h}||_{L^2(-L,L)}^2 = sum_n b_n^2 \\
||tilde{h}'||_{L^2(-L,L)}^2 = sum_n Big(frac{pi n b_n}{L}Big)^2 \\
implies ||tilde{h}'||_{L^2(-L,L)}^2 ge sum_n Big(frac{pi b_n}{L}Big)^2 = frac{pi^2}{L^2} sum_n b_n^2 \\
implies||tilde{h}'||_{L^2(-L,L)} ge frac{pi}{L} ||tilde{h}||_{L^2(-L,L)}$$
We will now notice that:
$$||tilde{h}||_{L^2(-L,L)}^2 = 2||h||_{L^2(0,L)}^2 \
||tilde{h}' ||_{L^2(-L,L)}^2 = 2||h'||_{L^2(0,L)}^2$$
And this allows us to say, finally, that:
$$||h||_{L^2(0,L)} le frac{L}{pi}||h'||_{L^2(0,L)}$$
To get the final inequality, note that for any function $vin V$ we may define a function $bar{v} in H_0^1(0, 2L)$ as follows:
$$bar{v}(x) = v(x)quad text{for} ; 0le x < L, quad v(2L - x) quad text{for} ; Lle x le 2L$$
Which is symmetric with respect to the vertical axis centered in $L$, thus $bar{v} in H_0^1(0, 2L)$. As before:
$$||bar{v}||_{L^2(0,2L)}^2 = 2||v||_{L^2(0,L)}^2 \
||bar{v}' ||_{L^2(0,2L)}^2 = 2||v'||_{L^2(0,L)}^2$$
By the argument above, it immediately follows that:
$$||v||_{L^2(0,L)} le frac{2L}{pi}||v'||_{L^2(0,L)}$$
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
$begingroup$
I've found a way. First, consider the space $H_0^1(0,L)$. This is a subspace of $V$ and $H^1$. Let $hin H_0^1(0,L)$, thus we can define:
$$tilde{h}=h quad text{for} ; xge 0, quad -h quad text{for} ; x<0$$
$tilde{h}$ is an odd, $2L$-periodic function. Its Fourier series will only contain odd terms:
$$tilde{h} = sum_n b_n sin(pi nx/L)$$
And it's easy to see that this Fourier series converges uniformly to the function $tilde{h}$, since the function is continuous on $Bbb{R}$ and its derivative is piecewise continuous. Then $tilde{h}'$ can be obtained by differentiating each term of the series:
$$tilde{h} = sum_n frac{pi n b_n}{L} cos(pi nx/L)$$
Now applying Bessel's identity we get:
$$||tilde{h}||_{L^2(-L,L)}^2 = sum_n b_n^2 \\
||tilde{h}'||_{L^2(-L,L)}^2 = sum_n Big(frac{pi n b_n}{L}Big)^2 \\
implies ||tilde{h}'||_{L^2(-L,L)}^2 ge sum_n Big(frac{pi b_n}{L}Big)^2 = frac{pi^2}{L^2} sum_n b_n^2 \\
implies||tilde{h}'||_{L^2(-L,L)} ge frac{pi}{L} ||tilde{h}||_{L^2(-L,L)}$$
We will now notice that:
$$||tilde{h}||_{L^2(-L,L)}^2 = 2||h||_{L^2(0,L)}^2 \
||tilde{h}' ||_{L^2(-L,L)}^2 = 2||h'||_{L^2(0,L)}^2$$
And this allows us to say, finally, that:
$$||h||_{L^2(0,L)} le frac{L}{pi}||h'||_{L^2(0,L)}$$
To get the final inequality, note that for any function $vin V$ we may define a function $bar{v} in H_0^1(0, 2L)$ as follows:
$$bar{v}(x) = v(x)quad text{for} ; 0le x < L, quad v(2L - x) quad text{for} ; Lle x le 2L$$
Which is symmetric with respect to the vertical axis centered in $L$, thus $bar{v} in H_0^1(0, 2L)$. As before:
$$||bar{v}||_{L^2(0,2L)}^2 = 2||v||_{L^2(0,L)}^2 \
||bar{v}' ||_{L^2(0,2L)}^2 = 2||v'||_{L^2(0,L)}^2$$
By the argument above, it immediately follows that:
$$||v||_{L^2(0,L)} le frac{2L}{pi}||v'||_{L^2(0,L)}$$
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
$endgroup$
add a comment |
$begingroup$
I've found a way. First, consider the space $H_0^1(0,L)$. This is a subspace of $V$ and $H^1$. Let $hin H_0^1(0,L)$, thus we can define:
$$tilde{h}=h quad text{for} ; xge 0, quad -h quad text{for} ; x<0$$
$tilde{h}$ is an odd, $2L$-periodic function. Its Fourier series will only contain odd terms:
$$tilde{h} = sum_n b_n sin(pi nx/L)$$
And it's easy to see that this Fourier series converges uniformly to the function $tilde{h}$, since the function is continuous on $Bbb{R}$ and its derivative is piecewise continuous. Then $tilde{h}'$ can be obtained by differentiating each term of the series:
$$tilde{h} = sum_n frac{pi n b_n}{L} cos(pi nx/L)$$
Now applying Bessel's identity we get:
$$||tilde{h}||_{L^2(-L,L)}^2 = sum_n b_n^2 \\
||tilde{h}'||_{L^2(-L,L)}^2 = sum_n Big(frac{pi n b_n}{L}Big)^2 \\
implies ||tilde{h}'||_{L^2(-L,L)}^2 ge sum_n Big(frac{pi b_n}{L}Big)^2 = frac{pi^2}{L^2} sum_n b_n^2 \\
implies||tilde{h}'||_{L^2(-L,L)} ge frac{pi}{L} ||tilde{h}||_{L^2(-L,L)}$$
We will now notice that:
$$||tilde{h}||_{L^2(-L,L)}^2 = 2||h||_{L^2(0,L)}^2 \
||tilde{h}' ||_{L^2(-L,L)}^2 = 2||h'||_{L^2(0,L)}^2$$
And this allows us to say, finally, that:
$$||h||_{L^2(0,L)} le frac{L}{pi}||h'||_{L^2(0,L)}$$
To get the final inequality, note that for any function $vin V$ we may define a function $bar{v} in H_0^1(0, 2L)$ as follows:
$$bar{v}(x) = v(x)quad text{for} ; 0le x < L, quad v(2L - x) quad text{for} ; Lle x le 2L$$
Which is symmetric with respect to the vertical axis centered in $L$, thus $bar{v} in H_0^1(0, 2L)$. As before:
$$||bar{v}||_{L^2(0,2L)}^2 = 2||v||_{L^2(0,L)}^2 \
||bar{v}' ||_{L^2(0,2L)}^2 = 2||v'||_{L^2(0,L)}^2$$
By the argument above, it immediately follows that:
$$||v||_{L^2(0,L)} le frac{2L}{pi}||v'||_{L^2(0,L)}$$
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
$endgroup$
I've found a way. First, consider the space $H_0^1(0,L)$. This is a subspace of $V$ and $H^1$. Let $hin H_0^1(0,L)$, thus we can define:
$$tilde{h}=h quad text{for} ; xge 0, quad -h quad text{for} ; x<0$$
$tilde{h}$ is an odd, $2L$-periodic function. Its Fourier series will only contain odd terms:
$$tilde{h} = sum_n b_n sin(pi nx/L)$$
And it's easy to see that this Fourier series converges uniformly to the function $tilde{h}$, since the function is continuous on $Bbb{R}$ and its derivative is piecewise continuous. Then $tilde{h}'$ can be obtained by differentiating each term of the series:
$$tilde{h} = sum_n frac{pi n b_n}{L} cos(pi nx/L)$$
Now applying Bessel's identity we get:
$$||tilde{h}||_{L^2(-L,L)}^2 = sum_n b_n^2 \\
||tilde{h}'||_{L^2(-L,L)}^2 = sum_n Big(frac{pi n b_n}{L}Big)^2 \\
implies ||tilde{h}'||_{L^2(-L,L)}^2 ge sum_n Big(frac{pi b_n}{L}Big)^2 = frac{pi^2}{L^2} sum_n b_n^2 \\
implies||tilde{h}'||_{L^2(-L,L)} ge frac{pi}{L} ||tilde{h}||_{L^2(-L,L)}$$
We will now notice that:
$$||tilde{h}||_{L^2(-L,L)}^2 = 2||h||_{L^2(0,L)}^2 \
||tilde{h}' ||_{L^2(-L,L)}^2 = 2||h'||_{L^2(0,L)}^2$$
And this allows us to say, finally, that:
$$||h||_{L^2(0,L)} le frac{L}{pi}||h'||_{L^2(0,L)}$$
To get the final inequality, note that for any function $vin V$ we may define a function $bar{v} in H_0^1(0, 2L)$ as follows:
$$bar{v}(x) = v(x)quad text{for} ; 0le x < L, quad v(2L - x) quad text{for} ; Lle x le 2L$$
Which is symmetric with respect to the vertical axis centered in $L$, thus $bar{v} in H_0^1(0, 2L)$. As before:
$$||bar{v}||_{L^2(0,2L)}^2 = 2||v||_{L^2(0,L)}^2 \
||bar{v}' ||_{L^2(0,2L)}^2 = 2||v'||_{L^2(0,L)}^2$$
By the argument above, it immediately follows that:
$$||v||_{L^2(0,L)} le frac{2L}{pi}||v'||_{L^2(0,L)}$$
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
edited Jan 1 at 11:59
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
answered Dec 30 '18 at 9:56
Niki Di GianoNiki Di Giano
941211
941211
add a comment |
add a comment |
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$begingroup$
see: mis.mpg.de/preprints/2003/preprint2003_24.pdf
$endgroup$
– Aleksas Domarkas
Dec 29 '18 at 22:50
$begingroup$
@Aleksas Domarkas it looks useful, but I can't quite extract the information I need.
$endgroup$
– Niki Di Giano
Dec 30 '18 at 0:20