Ngô Quốc Anh

November 20, 2007

1 BĐT khó nhưng có nhiều áp dụng trong bài toán p-Laplacian

Filed under: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 4:48

Let p \geq 2, then for all x_1, x_2 \in \mathbb R^N,

\displaystyle\left| {x_2 } \right|^p \geqslant \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right) + \frac{{\left| {x_2 - x_1 } \right|^p }}{{2^{p - 1} - 1}}.

If 1<p<2, then

\displaystyle\left| {x_2 } \right|^p \geqslant \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right) + c\left( p \right)\frac{{\left| {x_2 - x_1 } \right|^p }}{{\left( {\left| {x_1 } \right| + \left| {x_2 } \right|} \right)^{2 - p} }}.

Proof. We consider two cases separately.

Case 2 \leq p. The strict convexity of x \mapsto \left| x \right|^p implies that for any x_1, x_2 \in \mathbb R^N

\displaystyle\left| {x_2 } \right|^p \geqslant \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right).

Writing \frac{{x_2 + x_1 }}{2} instead of x_2 we obtain

\displaystyle\left| {\frac{{x_2 + x_1 }}{2}} \right|^p \geqslant \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {\frac{{x_2 + x_1 }} {2} - x_1 } \right)

or

\displaystyle\left| {\frac{{x_2 + x_1 }}{2}} \right|^p \geqslant \left| {x_1 } \right|^p + \frac{1}{2}p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right).

Using Clarkson inequality

\displaystyle\left| {x_1 } \right|^p + \left| {x_2 } \right|^p \geqslant 2\left| {\frac{{x_2 + x_1 }} {2}} \right|^p + 2\left| {\frac{{x_1 - x_2 }} {2}} \right|^p

we arrive at

\displaystyle\left| {x_2 } \right|^p \geqslant \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right) + 2\left| {\frac{{x_2 - x_1 }}{2}} \right|^p

or

\displaystyle\left| {x_2 } \right|^p \geqslant \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right) + \frac{1}{{2^{p-1} }}\left| {x_2 - x_1 } \right|^p.

Repeating this procedure we ca replace the constant

\displaystyle\frac{1} {{2^{ p - 1} }}

by

\displaystyle\frac{1}{{2^{p - 1} }} + \frac{1} {{4^{p - 1} }}.

By iteration one obtains the constant

\displaystyle\frac{1}{{2^{p - 1} }} + \frac{1}{{4^{p - 1} }} + \frac{1}{{8^{p - 1} }} + ... = \frac{1}{{2^{p - 1} - 1}}.

Case 1<p<2. Fix x_1, x_2 and expand the real function

\displaystyle f\left( t \right) = \left| {x_1 + t\left( {x_2 - x_1 } \right)} \right|^p .

Using Taylor formula

\displaystyle f\left( t \right) = \left| {x_1 + t\left( {x_2 - x_1 } \right)} \right|^p .

Then, provided f(t) \ne 0 for all 0 \leq t \leq 1,

\displaystyle \left| {x_2 } \right|^p = \left| {x_1 } \right|^p + p\left| {x_1 } \right|^{p - 2} x_1 \left( {x_2 - x_1 } \right) + \int_0^1 {\left( {1 - t} \right)f''\left( t \right)dt} .

At the same time

\displaystyle f''\left( t \right) = p\left( {p - 2} \right)\left| {x_1 + t\left( {x_2 - x_1 } \right)} \right|^{p - 4} \left( {\left( {x_1 + t\left( {x_2 - x_1 } \right)} \right)\left( {x_2 - x_1 } \right)} \right)^2 + p\left| {x_1 + t\left( {x_2 - x_1 } \right)} \right|^p \left| {x_2 - x_1 } \right|^2.

Schwartz inequality yields

\displaystyle f''\left( t \right) \geqslant p\left( {p - 1} \right)\left| {x_1 + t\left( {x_2 - x_1 } \right)} \right|^{p - 2} \left| {x_2 - x_1 } \right|^2 .

Now,

\displaystyle\int\limits_0^1 {\left( {1 - t} \right)f''\left( t \right)dt} \geqslant \frac{3}{4}\int\limits_0^{\frac{1}{4}} {f''\left( t \right)dt}.

Lastly

\displaystyle c\left( p \right) = \frac{3}{{16}}p\left( {p - 1} \right).

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