Ngô Quốc Anh

October 11, 2010

The Agmon type inequality in 1D

Filed under: Uncategorized — Tags: , , — Ngô Quốc Anh @ 18:14

In this note, we prove a very interesting inequality known as the Agmon type inequality in space dimension 1.

\displaystyle |g(x){|^2} \leqslant {\left( {\int_\mathbb{R} {|g(\xi ){|^2}d\xi } } \right)^{\frac{1}{2}}}{\left( {\int_\mathbb{R} {|g'(\xi ){|^2}d\xi } } \right)^{\frac{1}{2}}},\quad \forall x \in \mathbb{R}

for any smooth function g with compact support in \mathbb R.

The proof is standard and classical. The trick is to use the integral representation for functions that we have already discussed when we talk about the Poincare inequality.

Proof. Since g has compact support, there exists some L>0 sufficiently large such that g vanishes outside of (-L,L). Then we can write

\displaystyle g{(x)^2} = 2\int_{ - L}^x {g(\xi )g'(\xi )d\xi } \leqslant 2\int_{ - L}^x {|g(\xi )||g'(\xi )|d\xi }

and

\displaystyle g{(x)^2} = -2\int_x^{ L}{g(\xi )g'(\xi )d\xi } \leqslant 2\int_x^{L}{|g(\xi )||g'(\xi )|d\xi }.

Adding both inequalities we obtain

\displaystyle g{(x)^2} \leqslant \int_{ - L}^L {|g(\xi )||g'(\xi )|d\xi } = \int_\mathbb{R} {|g(\xi )||g'(\xi )|d\xi }

Then using the Cauchy-Schwarz inequality we find the desired inequality.

I will show some application of this equality, precisely, I will derive a proof of the Ladyzhenskaya inequalities

\displaystyle \int_{{\mathbb{R}^2}} {|u(x){|^4}dx} \leqslant \left( {\int_{{\mathbb{R}^2}} {|u(x){|^2}dx} } \right)\left( {\int_{{\mathbb{R}^2}} {|\nabla u(x){|^2}dx} } \right)

and

\displaystyle \int_{{\mathbb{R}^3}} {|u(x){|^4}dx} \leqslant {\left( {\int_{{\mathbb{R}^3}} {|u(x){|^2}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{{\mathbb{R}^3}} {|\nabla u(x){|^2}dx} } \right)^{\frac{3}{2}}}

which plays an important role in the theory of Navier-Stokes equations.

11 Comments »

  1. Hi Anh!
    I have a question about Ladyzhenskaya.
    Suppose that \Omega \subset \mathbb R^2 and be bounded in one direction. So is Ladyzhenskaya’s inequality true in case the function u\in H^1(\Omega).
    Thank for your answer!

    Comment by Dang Son — April 24, 2014 @ 15:16

    • My guess is that it still holds. For example, consider the trip \Omega = [0,1] \times \mathbb R in \mathbb R^2. Then you may replace u=u(x,y) by v(x,y)=u(x,y)\chi_{[0,1]}(x). If I am not wrong, then

      \displaystyle \int_{\mathbb R^2} |v|^p dxdy = \int_\Omega |u|^p dxdy

      and

      \displaystyle \int_{\mathbb R^2} |\nabla v|^2 dxdy = \int_\Omega |\nabla u|^2 dxdy.

      Comment by Ngô Quốc Anh — April 24, 2014 @ 15:29

      • But u\in H^1(\Omega) so u has no compact support and we can not apply this inequality.

        Comment by Dang Son — April 24, 2014 @ 16:30

      • So what happens with non-zero constant functions?

        Comment by Ngô Quốc Anh — April 24, 2014 @ 16:34

      • In the proof, we need the function has compact support. When we integral by part, the term on boundary vanish.

        Comment by Dang Son — April 24, 2014 @ 17:01

      • I guess for non-zero constant functions, the LHR is positive while the RHS vanishes.

        Comment by Ngô Quốc Anh — April 24, 2014 @ 17:02

      • What are LHR and RHS?

        Comment by Dang Son — April 24, 2014 @ 17:08

      • I guess we are talking about the following inequality

        \displaystyle \int_\Omega {|u(x){|^4}dx} \leqslant \left( {\int_\Omega {|u(x){|^2}dx} } \right)\left( {\int_\Omega {|\nabla u(x){|^2}dx} } \right)

        for unbounded domain \Omega \subset \mathbb R^2 and u\in H^1(\Omega).

        Comment by Ngô Quốc Anh — April 24, 2014 @ 17:10

      • Yes, and \Omega is bounded in one direction. I don’t know whether it still holds.

        Comment by Dang Son — April 24, 2014 @ 17:15

      • Why u\in H^1(\Omega) implies that u has no compact support?

        Comment by Ngô Quốc Anh — April 24, 2014 @ 17:21

      • I guess if u\in H^1(\Omega) we have no info about the compact support of u. If u satisfies Neumann condition, maybe u does not vanish on boundary.

        Comment by Dang Son — April 24, 2014 @ 17:30


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