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Question above proof of Marcinkiewicz Interpolation Theorem
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I am reading through the proof in Tao's lecture notes and I have a question about justifying a "without loss of generality" statement.
Statement: Let $T$ be a sublinear operator taking functions on $(X,mathcalB_X,mu)$ to functions on $(Y,mathcalB_Y,nu)$. Let $0 < p_o, p_1, q_0, q_1 leq infty$ and $A_0, A_1 > 0$ be such that $|Tf|_q_i,infty lesssim_p_i,q_i A_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i = 1,2$. Then for any $theta in (0,1)$ and $r in [1,infty]$ with $q_theta > 1$, we have
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta A_theta |f|_p_theta,r$$
for all simple function $f$ with finite measure support where $frac1p_theta := frac1-thetap_0 + fracthetap_1, frac1q_theta := frac1-thetaq_0 + fracthetaq_1,$ and $A_theta := A_0^1-thetaA_1^theta$.
My question is about the following reduction. We observe the following symmetries. First, we can multiply $T$ (and the $A_theta$) by an
arbitrary constant. Moreover, we may multiply the underlying measure $mu$ by a constant $C$ and $A_theta$ by $C^frac-1p_theta$. Similarly, we may multiply the underlying measure $nu$ by a constant $C$ and $A_theta$ by $C^frac1q_theta$. Therefore, we may assume $A_0 = A_1 = 1$ (and therefore $A_theta = 1$).
Usually, when I am trying to justify a "without loss of generality" claim, I assume the more specific claim is true, i.e., the claim is true for $A_0 = A_1 = 1$, and try to generalize. So assume it's true in the special case. I can apply the special case of the theorem to the operator $aT$ using the measures $bmu$ and $cnu$ for some parameters $a,b,c$. Then I have $|aTf|_q_i,infty lesssim_p_i,q_i ab^frac1p_ic^frac-1q_iA_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i=0,1$. Now in order to apply the special case of theorem, I need to select $a,b,c$ such that $ab^frac1p_ic^frac-1q_iA_i = 1$. It seems to me that scaling $T$ by $a$ is unnecessary. If we just select $a=1$ and $b,c$ such that $b^frac1p_ic^frac-1q_iA_i = 1 ,,, (i=1,2)$, then we can use the special case of the theorem to conclude
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta |f|_p_theta,r$$
Now if we rescale to the original measures, $bmu mapsto mu$ and $cnu mapsto nu$ we get
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta b^frac-1p_thetac^frac1q_theta|f|_p_theta,r = A_theta|f|_p_theta,r$$
Is this reasoning correct? Is the homogeneity with repsect to scaling $T$ unnecessary?
real-analysis lp-spaces harmonic-analysis interpolation-theory
asked Jul 14 at 21:14
H_R
11010
add a comment |Â
up vote
0
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I am reading through the proof in Tao's lecture notes and I have a question about justifying a "without loss of generality" statement.
Statement: Let $T$ be a sublinear operator taking functions on $(X,mathcalB_X,mu)$ to functions on $(Y,mathcalB_Y,nu)$. Let $0 < p_o, p_1, q_0, q_1 leq infty$ and $A_0, A_1 > 0$ be such that $|Tf|_q_i,infty lesssim_p_i,q_i A_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i = 1,2$. Then for any $theta in (0,1)$ and $r in [1,infty]$ with $q_theta > 1$, we have
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta A_theta |f|_p_theta,r$$
for all simple function $f$ with finite measure support where $frac1p_theta := frac1-thetap_0 + fracthetap_1, frac1q_theta := frac1-thetaq_0 + fracthetaq_1,$ and $A_theta := A_0^1-thetaA_1^theta$.
My question is about the following reduction. We observe the following symmetries. First, we can multiply $T$ (and the $A_theta$) by an
arbitrary constant. Moreover, we may multiply the underlying measure $mu$ by a constant $C$ and $A_theta$ by $C^frac-1p_theta$. Similarly, we may multiply the underlying measure $nu$ by a constant $C$ and $A_theta$ by $C^frac1q_theta$. Therefore, we may assume $A_0 = A_1 = 1$ (and therefore $A_theta = 1$).
Usually, when I am trying to justify a "without loss of generality" claim, I assume the more specific claim is true, i.e., the claim is true for $A_0 = A_1 = 1$, and try to generalize. So assume it's true in the special case. I can apply the special case of the theorem to the operator $aT$ using the measures $bmu$ and $cnu$ for some parameters $a,b,c$. Then I have $|aTf|_q_i,infty lesssim_p_i,q_i ab^frac1p_ic^frac-1q_iA_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i=0,1$. Now in order to apply the special case of theorem, I need to select $a,b,c$ such that $ab^frac1p_ic^frac-1q_iA_i = 1$. It seems to me that scaling $T$ by $a$ is unnecessary. If we just select $a=1$ and $b,c$ such that $b^frac1p_ic^frac-1q_iA_i = 1 ,,, (i=1,2)$, then we can use the special case of the theorem to conclude
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta |f|_p_theta,r$$
Now if we rescale to the original measures, $bmu mapsto mu$ and $cnu mapsto nu$ we get
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta b^frac-1p_thetac^frac1q_theta|f|_p_theta,r = A_theta|f|_p_theta,r$$
Is this reasoning correct? Is the homogeneity with repsect to scaling $T$ unnecessary?
real-analysis lp-spaces harmonic-analysis interpolation-theory
asked Jul 14 at 21:14
H_R
11010
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
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I am reading through the proof in Tao's lecture notes and I have a question about justifying a "without loss of generality" statement.
Statement: Let $T$ be a sublinear operator taking functions on $(X,mathcalB_X,mu)$ to functions on $(Y,mathcalB_Y,nu)$. Let $0 < p_o, p_1, q_0, q_1 leq infty$ and $A_0, A_1 > 0$ be such that $|Tf|_q_i,infty lesssim_p_i,q_i A_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i = 1,2$. Then for any $theta in (0,1)$ and $r in [1,infty]$ with $q_theta > 1$, we have
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta A_theta |f|_p_theta,r$$
for all simple function $f$ with finite measure support where $frac1p_theta := frac1-thetap_0 + fracthetap_1, frac1q_theta := frac1-thetaq_0 + fracthetaq_1,$ and $A_theta := A_0^1-thetaA_1^theta$.
My question is about the following reduction. We observe the following symmetries. First, we can multiply $T$ (and the $A_theta$) by an
arbitrary constant. Moreover, we may multiply the underlying measure $mu$ by a constant $C$ and $A_theta$ by $C^frac-1p_theta$. Similarly, we may multiply the underlying measure $nu$ by a constant $C$ and $A_theta$ by $C^frac1q_theta$. Therefore, we may assume $A_0 = A_1 = 1$ (and therefore $A_theta = 1$).
Usually, when I am trying to justify a "without loss of generality" claim, I assume the more specific claim is true, i.e., the claim is true for $A_0 = A_1 = 1$, and try to generalize. So assume it's true in the special case. I can apply the special case of the theorem to the operator $aT$ using the measures $bmu$ and $cnu$ for some parameters $a,b,c$. Then I have $|aTf|_q_i,infty lesssim_p_i,q_i ab^frac1p_ic^frac-1q_iA_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i=0,1$. Now in order to apply the special case of theorem, I need to select $a,b,c$ such that $ab^frac1p_ic^frac-1q_iA_i = 1$. It seems to me that scaling $T$ by $a$ is unnecessary. If we just select $a=1$ and $b,c$ such that $b^frac1p_ic^frac-1q_iA_i = 1 ,,, (i=1,2)$, then we can use the special case of the theorem to conclude
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta |f|_p_theta,r$$
Now if we rescale to the original measures, $bmu mapsto mu$ and $cnu mapsto nu$ we get
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta b^frac-1p_thetac^frac1q_theta|f|_p_theta,r = A_theta|f|_p_theta,r$$
Is this reasoning correct? Is the homogeneity with repsect to scaling $T$ unnecessary?
real-analysis lp-spaces harmonic-analysis interpolation-theory
asked Jul 14 at 21:14
H_R
11010
I am reading through the proof in Tao's lecture notes and I have a question about justifying a "without loss of generality" statement.
Statement: Let $T$ be a sublinear operator taking functions on $(X,mathcalB_X,mu)$ to functions on $(Y,mathcalB_Y,nu)$. Let $0 < p_o, p_1, q_0, q_1 leq infty$ and $A_0, A_1 > 0$ be such that $|Tf|_q_i,infty lesssim_p_i,q_i A_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i = 1,2$. Then for any $theta in (0,1)$ and $r in [1,infty]$ with $q_theta > 1$, we have
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta A_theta |f|_p_theta,r$$
for all simple function $f$ with finite measure support where $frac1p_theta := frac1-thetap_0 + fracthetap_1, frac1q_theta := frac1-thetaq_0 + fracthetaq_1,$ and $A_theta := A_0^1-thetaA_1^theta$.
My question is about the following reduction. We observe the following symmetries. First, we can multiply $T$ (and the $A_theta$) by an
arbitrary constant. Moreover, we may multiply the underlying measure $mu$ by a constant $C$ and $A_theta$ by $C^frac-1p_theta$. Similarly, we may multiply the underlying measure $nu$ by a constant $C$ and $A_theta$ by $C^frac1q_theta$. Therefore, we may assume $A_0 = A_1 = 1$ (and therefore $A_theta = 1$).
Usually, when I am trying to justify a "without loss of generality" claim, I assume the more specific claim is true, i.e., the claim is true for $A_0 = A_1 = 1$, and try to generalize. So assume it's true in the special case. I can apply the special case of the theorem to the operator $aT$ using the measures $bmu$ and $cnu$ for some parameters $a,b,c$. Then I have $|aTf|_q_i,infty lesssim_p_i,q_i ab^frac1p_ic^frac-1q_iA_iHW^frac1p_i$ for all sub-step functions of height $H$ and width $W$ and $i=0,1$. Now in order to apply the special case of theorem, I need to select $a,b,c$ such that $ab^frac1p_ic^frac-1q_iA_i = 1$. It seems to me that scaling $T$ by $a$ is unnecessary. If we just select $a=1$ and $b,c$ such that $b^frac1p_ic^frac-1q_iA_i = 1 ,,, (i=1,2)$, then we can use the special case of the theorem to conclude
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta |f|_p_theta,r$$
Now if we rescale to the original measures, $bmu mapsto mu$ and $cnu mapsto nu$ we get
$$|Tf|_q_theta,r lesssim_p_0,p_1,q_0,q_1,r,theta b^frac-1p_thetac^frac1q_theta|f|_p_theta,r = A_theta|f|_p_theta,r$$
Is this reasoning correct? Is the homogeneity with repsect to scaling $T$ unnecessary?
real-analysis lp-spaces harmonic-analysis interpolation-theory
asked Jul 14 at 21:14
H_R
11010
asked Jul 14 at 21:14
H_R
11010
asked Jul 14 at 21:14
H_R
11010
asked Jul 14 at 21:14
H_R
11010
11010
add a comment |Â
add a comment |Â
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