An inequality for $C^2$ functions
One more problem... It's simpler than the previous I posted in this section (at least, you had not to write down any boring formula 
).
Problem:
Let $Omega \in RR^n$ be open and $phi \in C^2(Omega)$.
Since $phi$ is smooth enough in $Omega$, it makes sense to consider the Hessian matrix and the laplacian of $phi$, i.e $D^2phi(x):=((\partial^2 phi)/(\partial x_i\partial x_j)(x))$ and $Delta phi(x):=\sum_(i=1)^n (\partial^2 phi)/(\partial x_i^2)(x)$.
Suppose that $D^2phi(x)$ is positive definite for all $x in Omega$ and show that the following inequality:
(I) $\quad \{ det D^2 phi(x)\}^(1/n)<= 1/n*Delta phi(x)$
holds again for all $x \in Omega$.
).Problem:
Let $Omega \in RR^n$ be open and $phi \in C^2(Omega)$.
Since $phi$ is smooth enough in $Omega$, it makes sense to consider the Hessian matrix and the laplacian of $phi$, i.e $D^2phi(x):=((\partial^2 phi)/(\partial x_i\partial x_j)(x))$ and $Delta phi(x):=\sum_(i=1)^n (\partial^2 phi)/(\partial x_i^2)(x)$.
Suppose that $D^2phi(x)$ is positive definite for all $x in Omega$ and show that the following inequality:
(I) $\quad \{ det D^2 phi(x)\}^(1/n)<= 1/n*Delta phi(x)$
holds again for all $x \in Omega$.
Risposte
Correct!
Easy but nice.
Easy but nice.
Please consider what I write cum grano salis...
If $phi in C^2(Omega)$ then Schwarz's theorem holds:
$(del)/(del x_j) ((del phi (x))/(del x_i))= (del^2 phi(x))/(del x_j x_i) = (del^2 phi(x))/(del x_i x_j) =(del)/(del x_i) ((del phi (x))/(del x_j))$,
so that $D^2 phi(x)$ is symmetric. The trace of $D^2 phi(x)$ is the laplacian of $phi(x)$: $"tr" D^2 phi(x) = sum_(i=1)^n (del^2 phi(x))/(del x_i^2)=Delta phi(x)$.
As the hint says, we can reduce the problem to a simpler one by diagonalization of $D^2 phi(x)$: there exists an invertible matrix (representing a coordinate change in $RR^n$) $T$ s.t. $E=T^(-1) * D^2phi(x) * T$ is diagonal.
The elements on the main diagonal of $E$ will be its eigenvalues:
$E=((lambda_1(x),0,ldots,0),(0,lambda_2(x),ldots,0),(vdots,vdots,ddots,vdots),(0,0,ldots,lambda_n(x)))$
By algebraic similarity, $det D^2 phi(x)=det E$ and $"tr"D^2 phi(x)="tr"E$. Furthermore, being $D^2 phi(x)$ positive definite,$lambda_i(x)>0$ $forall i$, so AM-GM inequality holds:
$ root(n) (prod_(i=1)^n lambda_i(x) ) <= 1/n sum_(i=1)^n lambda_i (x) quad quad hArr quad quad root(n) (det E ) <= 1/n * "tr"E quad quad hArr quad quad root(n) (det D^2 phi(x) ) <= 1/n * "tr" D^2 phi(x) = 1/n * Delta phi(x)$.
If $phi in C^2(Omega)$ then Schwarz's theorem holds:
$(del)/(del x_j) ((del phi (x))/(del x_i))= (del^2 phi(x))/(del x_j x_i) = (del^2 phi(x))/(del x_i x_j) =(del)/(del x_i) ((del phi (x))/(del x_j))$,
so that $D^2 phi(x)$ is symmetric. The trace of $D^2 phi(x)$ is the laplacian of $phi(x)$: $"tr" D^2 phi(x) = sum_(i=1)^n (del^2 phi(x))/(del x_i^2)=Delta phi(x)$.
As the hint says, we can reduce the problem to a simpler one by diagonalization of $D^2 phi(x)$: there exists an invertible matrix (representing a coordinate change in $RR^n$) $T$ s.t. $E=T^(-1) * D^2phi(x) * T$ is diagonal.
The elements on the main diagonal of $E$ will be its eigenvalues:
$E=((lambda_1(x),0,ldots,0),(0,lambda_2(x),ldots,0),(vdots,vdots,ddots,vdots),(0,0,ldots,lambda_n(x)))$
By algebraic similarity, $det D^2 phi(x)=det E$ and $"tr"D^2 phi(x)="tr"E$. Furthermore, being $D^2 phi(x)$ positive definite,$lambda_i(x)>0$ $forall i$, so AM-GM inequality holds:
$ root(n) (prod_(i=1)^n lambda_i(x) ) <= 1/n sum_(i=1)^n lambda_i (x) quad quad hArr quad quad root(n) (det E ) <= 1/n * "tr"E quad quad hArr quad quad root(n) (det D^2 phi(x) ) <= 1/n * "tr" D^2 phi(x) = 1/n * Delta phi(x)$.
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