# $f$ in $L^2$ space imply $\sup{f} > 0$?

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.)

• It is not true for example f=-1 on (0,1) is such that $f\leq -1$ but $k=-1\leq 0$ – Federico Fallucca Mar 31 at 15:11

$$f(x)\le K < 0, \mu(\Omega) = \infty\implies \infty = \int_\Omega K^2\,d\mu\le\int_\Omega f^2\,d\mu.$$

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

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