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$begingroup$

Let $M$ be a smooth compact riemannian manifold with Levi Civita connection and consider a smooth function $f: M to mathbbR$. Then the Laplacian of $f$
$$ Delta f = textdiv ( textgrad f) $$
end the Hessian of $f$
$$ nabla^2 f (X,Y) = langle nabla_X textgrad f , Y rangle $$
are well defined. I want to prove the inequality
$$ | Delta f|_infty le C| nabla^2 f |$$
for some positive constant $C$. I'm using
$$ | Delta f|_infty = sup_x in M |Delta f| quad quad |nabla^2 f|_infty = sup_x in M Biggl ( sup_u,v in T_xM setminus 0 frac Biggr ) $$

How can I prove it?

Looking at the problem in an abstract way, maybe I can consider a general $tau in T_2^0(M)$ and since the laplacian is the trace of the hessian I have

$$ |texttr(tau)|_infty = sup_x in M |texttr(tau(x))|= sup_x in M tau_ij(x)g^ij(x) le sup_x in M biggl ( sum_i,jtau_ij(x) max_ij g^ij(x) biggr ) =$$
$$ sup_x in M biggl ( max_ijg^ij(x) sum_i,j tau_ij(x) biggr ) = biggl ( sup_x in M max_i,j g^ij(x) biggr ) sup_x in M sum_i,jtau_ij(x) lesssim biggl ( sup_x in M max_i,j g^ij(x) biggr ) |tau|_infty$$
where the last inequality is due to the fact that all norms are equivalent in finite dimensional spaces. Moreover the quantity
$$ biggl ( sup_x in M max_i,j g^ij(x) biggr )$$
is finite being it the supremum of a continuous function on a compact set.

differential-geometry laplacian

share|cite|improve this question

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

asked 15 hours ago

Bremen000Bremen000

448210

$endgroup$

add a comment | 

0

$begingroup$

Let $M$ be a smooth compact riemannian manifold with Levi Civita connection and consider a smooth function $f: M to mathbbR$. Then the Laplacian of $f$
$$ Delta f = textdiv ( textgrad f) $$
end the Hessian of $f$
$$ nabla^2 f (X,Y) = langle nabla_X textgrad f , Y rangle $$
are well defined. I want to prove the inequality
$$ | Delta f|_infty le C| nabla^2 f |$$
for some positive constant $C$. I'm using
$$ | Delta f|_infty = sup_x in M |Delta f| quad quad |nabla^2 f|_infty = sup_x in M Biggl ( sup_u,v in T_xM setminus 0 frac Biggr ) $$

How can I prove it?

Looking at the problem in an abstract way, maybe I can consider a general $tau in T_2^0(M)$ and since the laplacian is the trace of the hessian I have

$$ |texttr(tau)|_infty = sup_x in M |texttr(tau(x))|= sup_x in M tau_ij(x)g^ij(x) le sup_x in M biggl ( sum_i,jtau_ij(x) max_ij g^ij(x) biggr ) =$$
$$ sup_x in M biggl ( max_ijg^ij(x) sum_i,j tau_ij(x) biggr ) = biggl ( sup_x in M max_i,j g^ij(x) biggr ) sup_x in M sum_i,jtau_ij(x) lesssim biggl ( sup_x in M max_i,j g^ij(x) biggr ) |tau|_infty$$
where the last inequality is due to the fact that all norms are equivalent in finite dimensional spaces. Moreover the quantity
$$ biggl ( sup_x in M max_i,j g^ij(x) biggr )$$
is finite being it the supremum of a continuous function on a compact set.

differential-geometry laplacian

share|cite|improve this question

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

asked 15 hours ago

Bremen000Bremen000

448210

$endgroup$

add a comment | 

$begingroup$

Let $M$ be a smooth compact riemannian manifold with Levi Civita connection and consider a smooth function $f: M to mathbbR$. Then the Laplacian of $f$
$$ Delta f = textdiv ( textgrad f) $$
end the Hessian of $f$
$$ nabla^2 f (X,Y) = langle nabla_X textgrad f , Y rangle $$
are well defined. I want to prove the inequality
$$ | Delta f|_infty le C| nabla^2 f |$$
for some positive constant $C$. I'm using
$$ | Delta f|_infty = sup_x in M |Delta f| quad quad |nabla^2 f|_infty = sup_x in M Biggl ( sup_u,v in T_xM setminus 0 frac Biggr ) $$

How can I prove it?

Looking at the problem in an abstract way, maybe I can consider a general $tau in T_2^0(M)$ and since the laplacian is the trace of the hessian I have

$$ |texttr(tau)|_infty = sup_x in M |texttr(tau(x))|= sup_x in M tau_ij(x)g^ij(x) le sup_x in M biggl ( sum_i,jtau_ij(x) max_ij g^ij(x) biggr ) =$$
$$ sup_x in M biggl ( max_ijg^ij(x) sum_i,j tau_ij(x) biggr ) = biggl ( sup_x in M max_i,j g^ij(x) biggr ) sup_x in M sum_i,jtau_ij(x) lesssim biggl ( sup_x in M max_i,j g^ij(x) biggr ) |tau|_infty$$
where the last inequality is due to the fact that all norms are equivalent in finite dimensional spaces. Moreover the quantity
$$ biggl ( sup_x in M max_i,j g^ij(x) biggr )$$
is finite being it the supremum of a continuous function on a compact set.

differential-geometry laplacian

share|cite|improve this question

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

asked 15 hours ago

Bremen000Bremen000

448210

$endgroup$

share|cite|improve this question

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

asked 15 hours ago

Bremen000Bremen000

448210

share|cite|improve this question

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

asked 15 hours ago

Bremen000Bremen000

448210

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

edited 12 hours ago

Rodrigo de Azevedo

13.1k41960

asked 15 hours ago

Bremen000Bremen000

448210

asked 15 hours ago

Bremen000Bremen000

448210