Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading a paper, whose author state the following: if $f \in L^{(q,\infty)}(\mathbb{R}^N)$, then $f_\delta \in L^p(\mathbb{R}^N)$ for every $p \in [1,q)$, where $\delta > 0$ and $$ f_\delta = f\; \boldsymbol{1}_{\{x\in X: f(x) \geq \delta\}}. $$ But is this really true?

share|cite|improve this question
The author states that there exist $\delta>0$ such that... ? – Norbert Jun 17 '12 at 16:10
Where in what paper? – Jonas Meyer Jun 17 '12 at 17:11
@Norbert: I'd say "for every $\delta >0$ sufficiently small." – Siminore Jun 17 '12 at 17:26
@Siminore So here is an answer. – Norbert Jun 17 '12 at 17:26
@JonasMeyer It is a preprint. Actually my guess is that the stated property is true for the particular case analyzed in the paper. Anyway, the authors seem to justify it as a universal property of Lorentz spaces. – Siminore Jun 17 '12 at 17:27
up vote 2 down vote accepted

I will use the following fact

Lemma. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space and $p\in[1,+\infty)$. Then for measurable non-negative function $f$ and measurable set $A\in M$ we have $$ \int\limits_A f(x)^p d\mu(x)=p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt $$ where $F_{f,A}(t)=\mu(\{x\in A:f(x)>t\})$

Proof. Using Fubini theorem for positive functions we can say that $$ \int\limits_A f(x)^p d\mu(x)= \int\limits_A \int\limits_{(0,f(x))}pt^{p-1}dt d\mu(x)= \int\limits_A \int\limits_{(0,+\infty)}pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)dt d\mu(x)= $$ $$ \int\limits_{(0,+\infty)}\int\limits_A pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt = \int\limits_{(0,+\infty)}pt^{p-1}\int\limits_A \boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt = $$ $$ \int\limits_{(0,+\infty)}pt^{p-1}\mu(\{x\in A:f(x)>t\}) dt = \int\limits_{(0,+\infty)}pt^{p-1}F_{f,A}(t) dt $$

Theorem. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space, and $f\in L^{(q,\infty)}(X,M,\mu)$. Then for all $\delta>0$ the function $f_\delta = f\; \boldsymbol{1}_{\{x\in X:f(x)>\delta\}}$ is in $L^p(X,M,\mu)$.

Proof. From definition of quasi-norm in Lorentz space we see that there exist $C>0$ such that $$ \mu(\{x\in X:|f(x)|>t\})\leq \frac{C}{t^q} $$ Denote $A=\{x\in X: f(x)>\delta\}$. Since $\delta>0$, then the function $f_\delta$ is non-negative and measurable. Then from previous lemma for $p\in[1,q)$ we have $$ \Vert f_\delta\Vert_p^p= \int\limits_{X} |f_\delta(x)|^pd\mu(x)= \int\limits_A f(x)^pd\mu(x)= p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt $$ Note that for $t\in(0,\delta)$ we have $\mu(\{x\in A:f(t)>t\})=0$ so $$ \Vert f_\delta\Vert_p^p= p\int\limits_{[\delta,+\infty)}t^{p-1}F_{f,A}(t)dt\leq p\int\limits_{[\delta,+\infty)}t^{p-1}F_{f,X}(f)dt\leq p\int\limits_{[\delta,+\infty)}t^{p-1}\frac{C}{t^q}dt= Cp\int\limits_{[\delta,+\infty)}\frac{1}{t^{q+1-p}}dt $$ Since $p<q$ the last integral is convergent, so $\Vert f_\delta\Vert_p <+\infty$ and $f_\delta\in L^p(X,M,\mu)$.

share|cite|improve this answer
It seems convincing. At a first reading, I thought this were true under the additional assumption that $f$ vanishes at infinity (as this is the particular case of the paper). Indeed, I thought that $\{x \mid f(x) \geq \delta\}$ should have been of finite measure (or even a bounded set), in order to apply the alternative definition of the quasi-norm as a supremum over subsets of finite measure. – Siminore Jun 17 '12 at 17:33
It seems that all the post pre-statement of the theorem can be replaced by a one-line proof based on Fubini theorem for nonnegative functions. – Did Aug 4 '12 at 21:34
@did, thanks I'll edit my answer – Norbert Aug 4 '12 at 21:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.