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Brezis–Gallouët inequality

${\displaystyle \displaystyle \|u\|_{L^{\infty }(\Omega )}\leq C\|u\|_{H^{1}(\Omega )}\left(1+{\Bigl (}\log {\bigl (}1+{\frac {\|u\|_{H^{2}(\Omega )}}{\|u\|_{H^{1}(\Omega )}}}{\bigr )}{\Bigr )}^{1/2}\right).}$

• ${\displaystyle P}$ is a bounded operator from ${\displaystyle H^{1}(\Omega )}$to${\displaystyle H^{1}(\mathbb {R} ^{2})}$;
• ${\displaystyle P}$ is a bounded operator from ${\displaystyle H^{2}(\Omega )}$to${\displaystyle H^{2}(\mathbb {R} ^{2})}$;
• the restriction to ${\displaystyle \Omega }$of${\displaystyle Pu}$ is equal to ${\displaystyle u}$ for all ${\displaystyle u\in H^{2}(\Omega )}$.

Let ${\displaystyle u\in H^{2}(\Omega )}$ be such that ${\displaystyle \|u\|_{H^{1}(\Omega )}=1}$. Then, denoting by ${\displaystyle {\widehat {v}}}$ the function obtained from ${\displaystyle v=Pu}$ by Fourier transform, one gets the existence of ${\displaystyle C>0}$ only depending on ${\displaystyle \Omega }$ such that:

• ${\displaystyle \|(1+|\xi |){\widehat {v}}\|_{L^{2}(\mathbb {R} ^{2})}\leq C}$,
• ${\displaystyle \|(1+|\xi |^{2}){\widehat {v}}\|_{L^{2}(\mathbb {R} ^{2})}\leq C\|u\|_{H^{2}(\Omega )}}$,
• ${\displaystyle \|u\|_{L^{\infty }(\Omega )}\leq \|v\|_{L^{\infty }(\mathbb {R} ^{2})}\leq C\|{\widehat {v}}\|_{L^{1}(\mathbb {R} ^{2})}}$.

For any ${\displaystyle R>0}$, one writes:

${\displaystyle {\begin{aligned}\displaystyle \|{\widehat {v}}\|_{L^{1}(\mathbb {R} ^{2})}&=\int _{|\xi |<R}|{\widehat {v}}(\xi )|{\rm {d}}\xi +\int _{|\xi |>R}|{\widehat {v}}(\xi )|{\rm {d}}\xi \\&=\int _{|\xi |<R}(1+|\xi |)|{\widehat {v}}(\xi )|{\frac {1}{1+|\xi |}}{\rm {d}}\xi +\int _{|\xi |>R}(1+|\xi |^{2})|{\widehat {v}}(\xi )|{\frac {1}{1+|\xi |^{2}}}{\rm {d}}\xi \\&\leq C\left(\int _{|\xi |<R}{\frac {1}{(1+|\xi |)^{2}}}{\rm {d}}\xi \right)^{\frac {1}{2}}+C\|u\|_{H^{2}(\Omega )}\left(\int _{|\xi |>R}{\frac {1}{(1+|\xi |^{2})^{2}}}{\rm {d}}\xi \right)^{\frac {1}{2}},\end{aligned}}}$

owing to the preceding inequalities and to the Cauchy-Schwarz inequality. This yields

${\displaystyle \|{\widehat {v}}\|_{L^{1}(\mathbb {R} ^{2})}\leq C(\log(1+R))^{\frac {1}{2}}+C{\frac {\|u\|_{H^{2}(\Omega )}}{1+R}}.}$

The inequality is then proven, in the case ${\displaystyle \|u\|_{H^{1}(\Omega )}=1}$, by letting ${\displaystyle R=\|u\|_{H^{2}(\Omega )}}$. For the general case of ${\displaystyle u\in H^{2}(\Omega )}$ non identically null, it suffices to apply this inequality to the function ${\displaystyle u/\|u\|_{H^{1}(\Omega )}}$.

Noticing that, for any ${\displaystyle v\in H^{2}(\mathbb {R} ^{2})}$, there holds

${\displaystyle \int _{\mathbb {R} ^{2}}{\bigl (}(\partial _{11}^{2}v)^{2}+2(\partial _{12}^{2}v)^{2}+(\partial _{22}^{2}v)^{2}{\bigr )}=\int _{\mathbb {R} ^{2}}{\bigl (}\partial _{11}^{2}v+\partial _{22}^{2}v{\bigr )}^{2},}$

one deduces from the Brezis-Gallouet inequality that there exists ${\displaystyle C>0}$ only depending on ${\displaystyle \Omega }$ such that, for all ${\displaystyle u\in H^{2}(\Omega )}$ which is not a.e. equal to 0,

${\displaystyle \displaystyle \|u\|_{L^{\infty }(\Omega )}\leq C\|u\|_{H^{1}(\Omega )}\left(1+{\Bigl (}\log {\bigl (}1+{\frac {\|\Delta u\|_{L^{2}(\Omega )}}{\|u\|_{H^{1}(\Omega )}}}{\bigr )}{\Bigr )}^{1/2}\right).}$

The previous inequality is close to the way that the Brezis-Gallouet inequality is cited in.^[2]

See also[edit]

• Ladyzhenskaya inequality
• Agmon's inequality

References[edit]