Integral inequality with partial derivative
$begingroup$
Fix $eta in Bbb R^n$, then for any $phi in C_c^infty(Bbb R^n)$, i.e. compactly supported smooth function,
$$ int_{Bbb R^n} frac{partial}{partial x_j} (phi(x) (x eta)) , dx le int_{Bbb R^n} |eta| Big| frac{partial}{partial x_j} (phi(x)x) Big| , dx $$
Here $x eta $ denotes dot product.
Any hints/ advice?
EDIT: I think this is correct.
Consider the bounded operator, $L_eta:Bbb R^n rightarrow Bbb R^n$, $x mapsto langle x, eta rangle $. Its norm is given by $||eta||_2$ by Hilbert space duality.By continuity, we have
$$| frac{partial}{partial x_j} L_eta (phi(x)x) | = | L_eta(frac{partial}{partial x_j} phi(x)x ) | le ||L_eta|| , || frac{partial}{partial x_j} phi(x)x ||_2 = ||eta||_2 || frac{partial}{partial x_j} phi(x)x ||_2 $$
real-analysis inequality integral-inequality
$endgroup$
add a comment |
$begingroup$
Fix $eta in Bbb R^n$, then for any $phi in C_c^infty(Bbb R^n)$, i.e. compactly supported smooth function,
$$ int_{Bbb R^n} frac{partial}{partial x_j} (phi(x) (x eta)) , dx le int_{Bbb R^n} |eta| Big| frac{partial}{partial x_j} (phi(x)x) Big| , dx $$
Here $x eta $ denotes dot product.
Any hints/ advice?
EDIT: I think this is correct.
Consider the bounded operator, $L_eta:Bbb R^n rightarrow Bbb R^n$, $x mapsto langle x, eta rangle $. Its norm is given by $||eta||_2$ by Hilbert space duality.By continuity, we have
$$| frac{partial}{partial x_j} L_eta (phi(x)x) | = | L_eta(frac{partial}{partial x_j} phi(x)x ) | le ||L_eta|| , || frac{partial}{partial x_j} phi(x)x ||_2 = ||eta||_2 || frac{partial}{partial x_j} phi(x)x ||_2 $$
real-analysis inequality integral-inequality
$endgroup$
add a comment |
$begingroup$
Fix $eta in Bbb R^n$, then for any $phi in C_c^infty(Bbb R^n)$, i.e. compactly supported smooth function,
$$ int_{Bbb R^n} frac{partial}{partial x_j} (phi(x) (x eta)) , dx le int_{Bbb R^n} |eta| Big| frac{partial}{partial x_j} (phi(x)x) Big| , dx $$
Here $x eta $ denotes dot product.
Any hints/ advice?
EDIT: I think this is correct.
Consider the bounded operator, $L_eta:Bbb R^n rightarrow Bbb R^n$, $x mapsto langle x, eta rangle $. Its norm is given by $||eta||_2$ by Hilbert space duality.By continuity, we have
$$| frac{partial}{partial x_j} L_eta (phi(x)x) | = | L_eta(frac{partial}{partial x_j} phi(x)x ) | le ||L_eta|| , || frac{partial}{partial x_j} phi(x)x ||_2 = ||eta||_2 || frac{partial}{partial x_j} phi(x)x ||_2 $$
real-analysis inequality integral-inequality
$endgroup$
Fix $eta in Bbb R^n$, then for any $phi in C_c^infty(Bbb R^n)$, i.e. compactly supported smooth function,
$$ int_{Bbb R^n} frac{partial}{partial x_j} (phi(x) (x eta)) , dx le int_{Bbb R^n} |eta| Big| frac{partial}{partial x_j} (phi(x)x) Big| , dx $$
Here $x eta $ denotes dot product.
Any hints/ advice?
EDIT: I think this is correct.
Consider the bounded operator, $L_eta:Bbb R^n rightarrow Bbb R^n$, $x mapsto langle x, eta rangle $. Its norm is given by $||eta||_2$ by Hilbert space duality.By continuity, we have
$$| frac{partial}{partial x_j} L_eta (phi(x)x) | = | L_eta(frac{partial}{partial x_j} phi(x)x ) | le ||L_eta|| , || frac{partial}{partial x_j} phi(x)x ||_2 = ||eta||_2 || frac{partial}{partial x_j} phi(x)x ||_2 $$
real-analysis inequality integral-inequality
real-analysis inequality integral-inequality
edited Dec 23 '18 at 8:34
CL.
asked Dec 23 '18 at 8:24
CL.CL.
2,2882925
2,2882925
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