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Is there a constant $c>0$ for a sufficiently good function $f : \mathbb R^n \to \mathbb R$, $\| f^2 \|_2 \leqslant c\| f \|_2 ? $ If so, I'm wondering the sufficient condition of $f$. What about $\| f^p \|_2 \leqslant c \| f \|_2$ cases($p>1)$?

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    Bamily: Why on earth did you accept this answer?2012-07-26

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No; but of course, this depends on what exactly you mean by "sufficiently good." Consider the functions

$f_{n}(x) = \frac{\chi_{[1/n,1]}(x)}{\sqrt[4]x}$

That is, $f_n$ is one over the fourth root function on $[1/n,1]$, and zero otherwise. Then it's easy to check that $\|f_n\|_2$ is uniformly bounded by $2$ (since $2 = \int_0^1 dx / \sqrt x$), while

$\|f_n^2\|_2^2 = \int_{1/n}^1 \frac 1 {x} dx \to \infty$

as $n \to \infty$. This family can be easily modified to make each $f_n$ continuous (imagine a line segment connecting the peak of $f_n$ with the origin, perhaps), or even smooth. So even the restriction that each $f_n$ is smooth and compactly supported (hence in every $L^p$, $1 \le p \le \infty$) is insufficient to have such a universal bound.


In general, the problem is that raising $f$ to the power $p$ for any $p > 1$ enhances the "spikiness" of the function, leading it to have much worse integrability properties. So unless we impose extremely strong conditions (e.g. uniform $L^{\infty}$ boundedness), it's going to be very difficult to give a useful universal bound in this sense. In fact, this is similar to what Davide Giraudo pointed out in a comment: If we are allowed to rescale our functions by a number $a$, letting $a$ be large would break the inequality.