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($p>1)$?
When does $\| f^2 \|_2 \leqslant c \| f \|_2 $ hold?
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4If we replace $f$ by $af$, $a>0$, we see that if $f$ "is good", $af$ cannot "be good" for all $a$. – 2012-07-22
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0@DavideGiraudo : I am not getting what u mean by "good"? do u mean to say that the norm may not be finite or something ? – 2012-07-22
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0@theorem In fact I don't understand what is meant by good in the OP. The class of functions $C(c)$ for which such a property can be true has to be contained in $L^4$ and for a $f$, the set $\{a,af\in C(c)\}$ is bounded. – 2012-07-22
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7The point is that the inequality does not scale properly. For any $f \in L^4$, $\|f^2\|_2$ is finite and $c = \|f^2\|_2/\|f\|_2$ works, but there is no $c$ that works for all multiples of any given nonzero $f$. – 2012-07-22
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12Bamily: Why on earth did you accept this answer? – 2012-07-26
1 Answers
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.