Banguizb

Answered

2022-07-07

Is there some inequality which relates

for some $f\in {L}^{2}\left(\mathrm{\Omega }\right)\cap {L}^{2}\left(\mathrm{\partial }\mathrm{\Omega }\right)$?

Answer & Explanation

lywiau63

Expert

2022-07-08Added 13 answers

Let $\mathrm{\Omega }$ be the unit disk in ${\mathbf{R}}^{2}$. The function $f\left(x,y\right)=\left(1-{x}^{2}-{y}^{2}{\right)}^{1/2}$ satisfies
${\int }_{\mathrm{\Omega }}|f{|}^{2}=\frac{\pi }{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\int }_{\mathrm{\partial }\mathrm{\Omega }}|f{|}^{2}=0.$
On the other hand the functions ${f}_{n}\left(x,y\right)=\left({x}^{2}+{y}^{2}{\right)}^{n/2}$ satisfy
${\int }_{\mathrm{\Omega }}|{f}_{n}{|}^{2}=\frac{\pi }{n+1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\int }_{\mathrm{\partial }\mathrm{\Omega }}|{f}_{n}{|}^{2}=2\pi$
so that in general the ratio
$\frac{{\int }_{\mathrm{\Omega }}|f{|}^{2}}{{\int }_{\mathrm{\partial }\mathrm{\Omega }}|f{|}^{2}}$
is bounded neither above nor below.

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