Solution verification : Truncation of $L^p$ functions

Question

Suppose $f$ is an unsigned measurable function in $L^p \, ,$ $1<p< \infty$. For $t>0,$ let $$E_t = \{x : |f(x)| > t\}$$.

Prove that for each $t>0 \ ,$ the horizontal truncation $1_{E_t}f \in L^q$ for all $1<q \leq p.$

I considered $$\int_X (1_{E_t}f)^q \,d\mu \leq \int_{E_t} \, f^q \,d\mu = \int_{E_t} \, (f^p)^{q/p} \,d\mu $$ $$ \leq \int_{E_t} \, (f^p) \,d\mu < \int_X \, (f^p) \, d \mu < \infty$$

Hence, $1_{E_t}f \in L^q$ for all $1 < q \leq p$. Is my apporach correct?.I didn't use the finiteness of the set $E_t$ here.

Answer 1

Provided you have shown that $\mu(E_t)<\infty$ (follows from Chebyshev's inequaility), you can use Hölder's inequality to obtain: \begin{align*} \int_{X} (1_{E_t} |f|)^q d\mu &\leq \Big(\int_{X} 1_{E_t}^{\frac{p}{p-q}}d\mu\Big)^{\frac{p-q}{p}} \big(\int_{X} |f|^p d\mu\Big)^{\frac{q}{p}}\\ & = \mu(E_t)^{\frac{p-q}{p}} ||f||_p^{q} < \infty \end{align*}

Answer 2

The key step is $\mu(E_t)=\mu(|f|>t)\leq t^{-p}\int_{E_t}|f|\,d\mu\leq t^{-p}\|f\|^p_p$, which means that $\mathbb{1}_{E_t}\in L_s(\mu)$ for all $s>0$.

The rest, as has been point out, is to apply Hölder's inequality. Thus, if $0<q\leq p$, $r:=\tfrac{p}{q}>1$ and so, $|f|^p\in L_r(\mu)$. Then with $r'=\frac{r}{r-1}$ (the convex conjugate of $r$)

$$ \|\mathbb{1}_{E_t}f\|_q=\int \mathbb{1}_{E_t}|f|^q\leq \|m(E_t)\|^{\tfrac{r-1}{r}}\|f\|^{p/r}_p\leq t^{-p\tfrac{r-1}{r}}\|f\|^{p}=t^{-(p-q)}\|f\|^p_p$$

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A much simpler solution, along the lines of the reasoning in the OP is as follows:

$$\int \mathbb{1}_{\{|f|>t\}}|f|^q=t^{-(p-q)}\int t^{p-q}\mathbb{1}_{\{|f|>t\}}|f|^p\leq t^{-(p-q)}\int \mathbb{1}_{\{|f|>t\}}|f|^p\leq t^{-(p-q)}\|f\|^p_p$$