How to construct such a function satisfying the following condition?

Question

I came across such a problem. It asked me to construct a function $f$ on $\mathbb{R}$ with the property that, for all $p \in (1, \infty)$,

$$\int^{\infty}_{-\infty}|f(x)|^pdx<\infty \iff p=4.$$

I have no clue where to start. I've tried something like $\frac{1}{x^{1/4} lnx^{1/4}}$, but it doesn't seem right. Does this problem involving functional analysis? Can anyone enlighten me? Thanks in advance.

It's not possible, cause if the integral converges for some $p$, then also converges for bigger $p$'s — k1.M, Aug 31 '17 at 17:47
What does the two-way implication mean? How can something that holds for all $p$ imply $p=4$? — Alex R., Aug 31 '17 at 17:48
I think that with your example you are almost there. Hints: modify the exponent of $ln x$ and modify the function near $x=1$. (Also remember that $lnx$ is not defined for $x\le 0$). — PozzPlot, Aug 31 '17 at 17:54
@k1.M That would be the case for series, but for an integral with respect to the Lebesgue measure, singularities at points of $\mathbb{R}$ can impose an upper bound on the exponent to remain integrable. — Daniel Fischer, Aug 31 '17 at 18:00
Related: Is it possible for a function to be in $L^p$ for only one $p$? — , Aug 31 '17 at 18:14

score 0 · Answer 1 · answered Aug 31 '17 at 18:12

To get a feeling about this, first consider sum of sequences: $$S_p = \sum_{n=1}^{\infty} |a_n|^p . $$ If this sum converges for any $p$ then elements $a_n$ need to go to zero as $n \to \infty$. Therefore the sequence is bounded, $|a_n| \leq S_p^{\frac{1}{p}}$ Therefore we have for $q>p$: $$ S_p \leq \sum_{n=1}^{\infty} S_p^{\frac{q-p}{p}} |a_n|^p = S_p^{\frac{q}{p}}. $$ The moral of this example is that for sequences the only reason for lack of convergence of a sum is low decay at infinity. Moreover, smaller exponents $p$ give better control over the decay of sequence at infinity (if q>p then statement $S_p < \infty$ is strictly stronger information than $S_q < \infty$, moreover knowing the value of the first gives you some estimate on the second). For functions we have another reason for lack of convergence: values of function going to infinity at some finite point. As an example take a function which vanishes outside of some finite interval, but blows up at some points. Moreover suppose that $$ I_p = \int |f(x)|^p $$ is finite for some $1 < p < \infty$. Take $1 \leq q <p$. We divide integral defining $I_q$ into subinterval where $|f|<1$ and that where $|f|>1$. In the first interval the integral is finite (we have a bounded function on finite interval). On the second interval we have $|f|^q < |f|^p$. Hence $I_q$ is finite. Therefore we see that in this case its larger exponents $p$ which give you better control over magnitude of $f$. Integrals of general functions may diverge for both reasons: lack of decay at infinity and singularities. Larger exponents give you better control over singularities, smaller exponents give you control over decay at infinity. To solve your problem you need to consider a function with both "problems". You may try playing with combinations of functions such as $x^{- \alpha}$, $(1+x^2)^{- \beta}$ for some exponents $\alpha, \beta$.

How to construct such a function satisfying the following condition?

1 Answers1