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

for any smooth function with compact support in .

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 has compact support, there exists some sufficiently large such that vanishes outside of . Then we can write

and

.

Adding both inequalities we obtain

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

and

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

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Hi Anh!

I have a question about Ladyzhenskaya.

Suppose that and be bounded in one direction. So is Ladyzhenskaya’s inequality true in case the function .

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 in . Then you may replace by . If I am not wrong, then

and

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

But so 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

for unbounded domain and .

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

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

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

Why implies that has no compact support?

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

I guess if we have no info about the compact support of . If satisfies Neumann condition, maybe does not vanish on boundary.

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