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I'm trying to understand if it's true that: " if $f\in L^p\quad \forall p\in N\implies f\in L^\infty$"? My thoughts: Since $\int_R |f(x)|^p dx<\infty\quad\forall p\implies |f(x)|\to 0$? Can anyone help me please? Thank you.

Davide Giraudo
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user62138
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  • The implication "$f\in L^p\quad \forall p\in N\implies f\in L^\infty$" does not hold hence you might want to find a counterexample (by the way, "$\int_R |f(x)|^p dx<\infty\quad\forall p\implies |f(x)|\to 0$" does not hold either). – Did Aug 09 '13 at 08:21
  • what holds is the following: For ever $1\leq q\leq p$ let $||f||_p \leq C$ for $C$ independent of $p$. Then $f\in L^\infty$. The $q$ comes into play, since it suffices to start for large values of $p$. But at the moment I'm not sure if this also holds for unbounded domains :) – Quickbeam2k1 Aug 09 '13 at 08:30
  • What I wanted to prove is that if $\phi$ is a misurabile function in[0,1] and the linear trasformation $A\colon f\to \phi f$ maps $L^2[0,1]$ in $L^2[0,1]$ then $\phi\in L^\infty[0,1]$ – user62138 Aug 09 '13 at 08:35
  • I've managed to prove only that $\phi\in L^p\quad \forall p$ – user62138 Aug 09 '13 at 08:37
  • Do you habe a uniform $L^p$ bound on $\phi$ ? – Quickbeam2k1 Aug 09 '13 at 09:31
  • Re your "What I wanted to prove" comment: this is another question, neither implied by, nor implying the one you asked (and more interesting, if you ask me). – Did Aug 13 '13 at 08:43
  • And can you give me a hint of how to prove it – user62138 Aug 13 '13 at 15:06

3 Answers3

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Hint: Consider $$ f(x)=\left\{\begin{array}{} -\log(x)&\text{if }x\in(0,1)\\ 0&\text{otherwise} \end{array}\right. $$

robjohn
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Consider the function f in $R$ constructed as follows: Let M is a big natural number. Then let $m${x $\in R$:f(x)=M}= $\frac{1}{M}$. (where m is the lebesgue measure in $R$). Then $m${x $\in R$:f(x)=M+1} =$\frac{1}{(M+1)^2}$. Carry on this type of construction i.e in general $m${x $\in R$:f(x)=M+k} =$\frac{1}{(M+k)^{k+1}}$. Let {x $\in R$:f(x)=M+k}=$E_k$ $\forall k \in N$. Let $E_n$'s be such that each $E_n$ is an interval and they are mutually disjoint.(This type of construction is possible since $\sum (\frac{1}{M}+ \frac {1}{M+1^2} + ...)$ $< \infty$). And also f(x)=0 elsewhere. Now it is clear that f $\notin L^\infty$ and $|f(x)| \longrightarrow 0$ is not true. It is clear that f is a measurable function.We show that $f \in L^p \forall p \in N$. For some $k \in N$, $\int_{R}f^k dm$= $M^{k-1}+(M+1)^{k-2}+...+1+ \sum (\frac{1}{M+k}+\frac{1}{(M+k+1)^2}+...+\frac{1}{(M+k+n)^{n+1}}+...) < \infty$. Hence $f \in L^k \forall k \in N$.

Subhankar D.
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A Gaussian random variable has all finite moments, but is not bounded. You can use this to construct a countexample on $[0,1]$.

Neuromath
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