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If $f(k)\leq K$ a.e in $\Omega$.

$|\Omega|= +\infty$ ,$K$ is constant,and $f \in L^2$.

Why we can say $K\geq 0$? (from Haim Brezis functional analysis sobolev space and partial differential equations, P308, chapter 9.)

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  • $\begingroup$ It is not true for example f=-1 on (0,1) is such that $f\leq -1$ but $k=-1\leq 0$ $\endgroup$ – Federico Fallucca Mar 31 at 15:11
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$$ f(x)\le K < 0, \mu(\Omega) = \infty\implies \infty = \int_\Omega K^2\,d\mu\le\int_\Omega f^2\,d\mu. $$

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  • $\begingroup$ oh, i see, but if $|\Omega|= +\infty$ $\endgroup$ – Ben Mar 31 at 15:14
  • $\begingroup$ thank you, but i think the proof is not rigorous.if suppose $f\leq 0 \leq K,$ also can get the right result.? $\endgroup$ – Ben Mar 31 at 15:24
  • $\begingroup$ @Ben, with $f\le 0\le K$ we can't deduce $f^2\le K^2$ (counterexample: $-2\le0 \le 1$ but $(-2)^2\not\le 1^2)$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 31 at 15:26
  • $\begingroup$ get it, thanks a lot! $\endgroup$ – Ben Mar 31 at 15:31
  • $\begingroup$ @Ben, then you can accept the answer. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 31 at 15:32
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$L^2(\mathbb{R})$ contains unbounded functions.

Consider the piecewise constant function $f$ defined as follows. For any positive integer $n$ put $f(x) = n$ for $x\in(n, n+1/n^4)$, elsewhere we set $f(x)=0$.

Now $$\int|f(x)|^2\,dx =\sum_{n=1}^\infty \frac{n^2}{n^4}=..$$

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