Mixing
Mikhlin multiplier theorem
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
The Mikhlin multiplier states the following:
Let $m : mathbbR^n backslash 0 rightarrow mathbbC$ satisfy the following:
$$|partial^alpham(xi)|leq C_0|xi|^, forall alpha in mathbbN_0^n text i.e. alpha is a multi-index with |alpha| leq n+2.$$ $$text Then, for all 1 < p < infty, exists B=B(m,n,p) > 0 text such that ||T_mf||_L^p leq B||f||_L^p, forall f in mathcalS(mathbbR^n).$$
In proving this one considers a partition of unity as follows $Psi in C_c^infty(mathbbR^n),$ with support of $Psi subseteq mathbbR^n backslash 0$ and $ sum_j=- infty^infty Psi(fracx2^j) =1 ,forall x neq 0$. This can be choosen radial and non-negative.
Then defining $Psi_j(x):= Psi(fracx2^j)$ for $Psi$ as above and observing that each $Psi_j$ is supported in a dyadic annulus of size $2^j$ and given $j in mathbbZ, m_j(xi):=Psi_j(xi)m(xi)$
How the The Fourier multiplier of the above $m_j(xi)$ is $m_j(D)$
How do we obtain $m(D)f = sum_jm_j(D)f$ with convergence in $L^2$ (I know we have to use Plancherel and the Dominated Convergence Theorem somehow)
fourier-analysis proof-explanation lp-spaces harmonic-analysis
edited Jul 16 at 20:03
Hugocito
1,6451019
asked Jul 16 at 19:49
VBACODER
748
 |Â
show 3 more comments
up vote
0
down vote
favorite
The Mikhlin multiplier states the following:
Let $m : mathbbR^n backslash 0 rightarrow mathbbC$ satisfy the following:
$$|partial^alpham(xi)|leq C_0|xi|^, forall alpha in mathbbN_0^n text i.e. alpha is a multi-index with |alpha| leq n+2.$$ $$text Then, for all 1 < p < infty, exists B=B(m,n,p) > 0 text such that ||T_mf||_L^p leq B||f||_L^p, forall f in mathcalS(mathbbR^n).$$
In proving this one considers a partition of unity as follows $Psi in C_c^infty(mathbbR^n),$ with support of $Psi subseteq mathbbR^n backslash 0$ and $ sum_j=- infty^infty Psi(fracx2^j) =1 ,forall x neq 0$. This can be choosen radial and non-negative.
Then defining $Psi_j(x):= Psi(fracx2^j)$ for $Psi$ as above and observing that each $Psi_j$ is supported in a dyadic annulus of size $2^j$ and given $j in mathbbZ, m_j(xi):=Psi_j(xi)m(xi)$
How the The Fourier multiplier of the above $m_j(xi)$ is $m_j(D)$
How do we obtain $m(D)f = sum_jm_j(D)f$ with convergence in $L^2$ (I know we have to use Plancherel and the Dominated Convergence Theorem somehow)
fourier-analysis proof-explanation lp-spaces harmonic-analysis
edited Jul 16 at 20:03
Hugocito
1,6451019
asked Jul 16 at 19:49
VBACODER
748
Who is the operator $T_m$?
– Hugocito
Jul 16 at 19:55
1
@Hugocito Given a function $m : mathbbR^n rightarrow mathbbC$ the operator $(T_mf)(x):= int_mathbbR^nm(xi)f^hat(xi)e^2 pi i x xid xi$ where $f in mathcalS(mathbbR^n)$ For ease of notation we denote $T_m = m(D) $
– VBACODER
Jul 16 at 20:02
As your hints say: Use the Plancherel theorem to express both $m(D)f$ and the sum of the $m_i(D) f$ as functions in the frequency domain. By the identity $sum_j Psi(x 2^j) = 1$ you have pointwise convergence, Use the dominated convergence theorem to pass from a pointwise limit to a integral one.
– Adrián González-Pérez
Jul 17 at 11:02
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet.
– VBACODER
Jul 17 at 18:56
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet. So $m(D)f = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ $ m_j(D)f = int_mathbbR^nm_j(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n Psi_j(xi)m(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ Therefore $m(D)f = sum_jm_j(D)f$ For the convergence I need to use the $L^p$ version of the dominated convergence theorem with $p=2.$
– VBACODER
Jul 17 at 19:02
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The Mikhlin multiplier states the following:
Let $m : mathbbR^n backslash 0 rightarrow mathbbC$ satisfy the following:
$$|partial^alpham(xi)|leq C_0|xi|^, forall alpha in mathbbN_0^n text i.e. alpha is a multi-index with |alpha| leq n+2.$$ $$text Then, for all 1 < p < infty, exists B=B(m,n,p) > 0 text such that ||T_mf||_L^p leq B||f||_L^p, forall f in mathcalS(mathbbR^n).$$
In proving this one considers a partition of unity as follows $Psi in C_c^infty(mathbbR^n),$ with support of $Psi subseteq mathbbR^n backslash 0$ and $ sum_j=- infty^infty Psi(fracx2^j) =1 ,forall x neq 0$. This can be choosen radial and non-negative.
Then defining $Psi_j(x):= Psi(fracx2^j)$ for $Psi$ as above and observing that each $Psi_j$ is supported in a dyadic annulus of size $2^j$ and given $j in mathbbZ, m_j(xi):=Psi_j(xi)m(xi)$
How the The Fourier multiplier of the above $m_j(xi)$ is $m_j(D)$
How do we obtain $m(D)f = sum_jm_j(D)f$ with convergence in $L^2$ (I know we have to use Plancherel and the Dominated Convergence Theorem somehow)
fourier-analysis proof-explanation lp-spaces harmonic-analysis
edited Jul 16 at 20:03
Hugocito
1,6451019
asked Jul 16 at 19:49
VBACODER
748
The Mikhlin multiplier states the following:
Let $m : mathbbR^n backslash 0 rightarrow mathbbC$ satisfy the following:
$$|partial^alpham(xi)|leq C_0|xi|^, forall alpha in mathbbN_0^n text i.e. alpha is a multi-index with |alpha| leq n+2.$$ $$text Then, for all 1 < p < infty, exists B=B(m,n,p) > 0 text such that ||T_mf||_L^p leq B||f||_L^p, forall f in mathcalS(mathbbR^n).$$
In proving this one considers a partition of unity as follows $Psi in C_c^infty(mathbbR^n),$ with support of $Psi subseteq mathbbR^n backslash 0$ and $ sum_j=- infty^infty Psi(fracx2^j) =1 ,forall x neq 0$. This can be choosen radial and non-negative.
Then defining $Psi_j(x):= Psi(fracx2^j)$ for $Psi$ as above and observing that each $Psi_j$ is supported in a dyadic annulus of size $2^j$ and given $j in mathbbZ, m_j(xi):=Psi_j(xi)m(xi)$
How the The Fourier multiplier of the above $m_j(xi)$ is $m_j(D)$
How do we obtain $m(D)f = sum_jm_j(D)f$ with convergence in $L^2$ (I know we have to use Plancherel and the Dominated Convergence Theorem somehow)
fourier-analysis proof-explanation lp-spaces harmonic-analysis
edited Jul 16 at 20:03
Hugocito
1,6451019
asked Jul 16 at 19:49
VBACODER
748
edited Jul 16 at 20:03
Hugocito
1,6451019
edited Jul 16 at 20:03
Hugocito
1,6451019
edited Jul 16 at 20:03
Hugocito
1,6451019
1,6451019
asked Jul 16 at 19:49
VBACODER
748
asked Jul 16 at 19:49
VBACODER
748
asked Jul 16 at 19:49
VBACODER
748
748
Who is the operator $T_m$?
– Hugocito
Jul 16 at 19:55
1
@Hugocito Given a function $m : mathbbR^n rightarrow mathbbC$ the operator $(T_mf)(x):= int_mathbbR^nm(xi)f^hat(xi)e^2 pi i x xid xi$ where $f in mathcalS(mathbbR^n)$ For ease of notation we denote $T_m = m(D) $
– VBACODER
Jul 16 at 20:02
As your hints say: Use the Plancherel theorem to express both $m(D)f$ and the sum of the $m_i(D) f$ as functions in the frequency domain. By the identity $sum_j Psi(x 2^j) = 1$ you have pointwise convergence, Use the dominated convergence theorem to pass from a pointwise limit to a integral one.
– Adrián González-Pérez
Jul 17 at 11:02
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet.
– VBACODER
Jul 17 at 18:56
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet. So $m(D)f = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ $ m_j(D)f = int_mathbbR^nm_j(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n Psi_j(xi)m(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ Therefore $m(D)f = sum_jm_j(D)f$ For the convergence I need to use the $L^p$ version of the dominated convergence theorem with $p=2.$
– VBACODER
Jul 17 at 19:02
 |Â
show 3 more comments
Who is the operator $T_m$?
– Hugocito
Jul 16 at 19:55
1
@Hugocito Given a function $m : mathbbR^n rightarrow mathbbC$ the operator $(T_mf)(x):= int_mathbbR^nm(xi)f^hat(xi)e^2 pi i x xid xi$ where $f in mathcalS(mathbbR^n)$ For ease of notation we denote $T_m = m(D) $
– VBACODER
Jul 16 at 20:02
As your hints say: Use the Plancherel theorem to express both $m(D)f$ and the sum of the $m_i(D) f$ as functions in the frequency domain. By the identity $sum_j Psi(x 2^j) = 1$ you have pointwise convergence, Use the dominated convergence theorem to pass from a pointwise limit to a integral one.
– Adrián González-Pérez
Jul 17 at 11:02
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet.
– VBACODER
Jul 17 at 18:56
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet. So $m(D)f = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ $ m_j(D)f = int_mathbbR^nm_j(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n Psi_j(xi)m(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ Therefore $m(D)f = sum_jm_j(D)f$ For the convergence I need to use the $L^p$ version of the dominated convergence theorem with $p=2.$
– VBACODER
Jul 17 at 19:02
Who is the operator $T_m$?
– Hugocito
Jul 16 at 19:55
Who is the operator $T_m$?
– Hugocito
Jul 16 at 19:55
1
1
@Hugocito Given a function $m : mathbbR^n rightarrow mathbbC$ the operator $(T_mf)(x):= int_mathbbR^nm(xi)f^hat(xi)e^2 pi i x xid xi$ where $f in mathcalS(mathbbR^n)$ For ease of notation we denote $T_m = m(D) $
– VBACODER
Jul 16 at 20:02
@Hugocito Given a function $m : mathbbR^n rightarrow mathbbC$ the operator $(T_mf)(x):= int_mathbbR^nm(xi)f^hat(xi)e^2 pi i x xid xi$ where $f in mathcalS(mathbbR^n)$ For ease of notation we denote $T_m = m(D) $
– VBACODER
Jul 16 at 20:02
As your hints say: Use the Plancherel theorem to express both $m(D)f$ and the sum of the $m_i(D) f$ as functions in the frequency domain. By the identity $sum_j Psi(x 2^j) = 1$ you have pointwise convergence, Use the dominated convergence theorem to pass from a pointwise limit to a integral one.
– Adrián González-Pérez
Jul 17 at 11:02
As your hints say: Use the Plancherel theorem to express both $m(D)f$ and the sum of the $m_i(D) f$ as functions in the frequency domain. By the identity $sum_j Psi(x 2^j) = 1$ you have pointwise convergence, Use the dominated convergence theorem to pass from a pointwise limit to a integral one.
– Adrián González-Pérez
Jul 17 at 11:02
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet.
– VBACODER
Jul 17 at 18:56
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet.
– VBACODER
Jul 17 at 18:56
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet. So $m(D)f = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ $ m_j(D)f = int_mathbbR^nm_j(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n Psi_j(xi)m(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ Therefore $m(D)f = sum_jm_j(D)f$ For the convergence I need to use the $L^p$ version of the dominated convergence theorem with $p=2.$
– VBACODER
Jul 17 at 19:02
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet. So $m(D)f = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ $ m_j(D)f = int_mathbbR^nm_j(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n Psi_j(xi)m(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ Therefore $m(D)f = sum_jm_j(D)f$ For the convergence I need to use the $L^p$ version of the dominated convergence theorem with $p=2.$
– VBACODER
Jul 17 at 19:02
 |Â
show 3 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853792%2fmikhlin-multiplier-theorem%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Who is the operator $T_m$?
– Hugocito
Jul 16 at 19:55
1
@Hugocito Given a function $m : mathbbR^n rightarrow mathbbC$ the operator $(T_mf)(x):= int_mathbbR^nm(xi)f^hat(xi)e^2 pi i x xid xi$ where $f in mathcalS(mathbbR^n)$ For ease of notation we denote $T_m = m(D) $
– VBACODER
Jul 16 at 20:02
As your hints say: Use the Plancherel theorem to express both $m(D)f$ and the sum of the $m_i(D) f$ as functions in the frequency domain. By the identity $sum_j Psi(x 2^j) = 1$ you have pointwise convergence, Use the dominated convergence theorem to pass from a pointwise limit to a integral one.
– Adrián González-Pérez
Jul 17 at 11:02
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet.
– VBACODER
Jul 17 at 18:56
@AdriánGonzález-Pérez I've had an attempt although I'm not very confident with these calculations yet. So $m(D)f = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ $ m_j(D)f = int_mathbbR^nm_j(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n Psi_j(xi)m(xi)f^hat(xi)e^2 pi i x xid xi = int_mathbbR^n m(xi)f^hat(xi)e^2 pi i x xid xi$ Therefore $m(D)f = sum_jm_j(D)f$ For the convergence I need to use the $L^p$ version of the dominated convergence theorem with $p=2.$
– VBACODER
Jul 17 at 19:02