Harnack inequality under scaling

Thm: Suppose {\Omega} is domain in {\mathbb{R}^n}. {u} is a harmonic function in {\Omega}, {u\geq 0}. For any subdomain {\Omega'\subset\subset \Omega}, there exists a constant {C} such that

\displaystyle \sup_{\Omega'}u\leq C(n,\Omega,\Omega')\inf_{\Omega'}u

To prove this Harnack inequality, there is an intermediate step

\displaystyle \sup_{B_R}u\leq 3^n\inf_{B_R}u

whenever {B_{4R}\subset\Omega}.

Here the constant is independent of {R}, actually the constant {3^n} is not so important. We can use the above Thm to give another proof.

Suppose {v(x)=u(Rx)} for {x\in B_1}. Let {\Omega=B_2} and {\Omega'=B_1}, applying the thm

\displaystyle \sup_{B_1}v\leq C(n)\inf_{B_1}v

convert back to {u}

\displaystyle \sup_{B_R}u\leq C(n)\inf_{B_R}u

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: