# 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

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

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

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.

1. Hi Anh!
Suppose that $\Omega \subset \mathbb R^2$ and be bounded in one direction. So is Ladyzhenskaya’s inequality true in case the function $u\in H^1(\Omega)$.

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