# Truncation in Lorentz spaces

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?

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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

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 $f_\delta\in L^p(X,M,\mu)$.

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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