Show that $\phi \circ f$ belongs to $L_p$ for each $f \in L_p$

Question

Let $(X, \mathbb{X}, \mu)$ be a finite measure space and let $1 \leq p< \infty$. Let $\phi$ be continuous on $\mathbb{R}$ to $\mathbb{R}$ and satisfy the condition: $(*)$ there esists $K>0$ such that $|\phi (t)|<K|t|$ for $t \geq K$. Show that $\phi \circ f$ belongs to $L_p$ for each $f \in L_p$ . Conversely , if $\phi$ does not satisfy $(*)$, then there is a function $f$ in $L_p$ on a finite measure space such that $\phi \circ f$ does not belong to $L_p$.

Any suggestion? I tried to use $|\phi (t)|<K|t|$ exchanging $t$ by $f(t)$ and I calculated $||\phi \circ f||_p$, but without success.

Answer 1

One has $\int_X \vert \phi \circ f \vert^p d\mu=\int_{f^{-1}([-K,K])} \vert \phi \circ f \vert^p d\mu + \int_{f^{-1}(\mathbb{R}\setminus [-K,K])}\vert \phi \circ f \vert^p d\mu$.

But $$\int_{f^{-1}([-K,K])} \vert \phi \circ f \vert^p d\mu \leqslant \mu(X) \max_{[-K,K]} (\vert \phi \vert ^p) < \infty.$$

And $$\int_{f^{-1}(\mathbb{R}\setminus [-K,K])}\vert \phi \circ f \vert^p d\mu \leqslant \int_{f^{-1}(\mathbb{R}\setminus [-K,K])} K^p \vert f \vert ^p d\mu \leqslant K^p \int_X \vert f \vert ^p d\mu < \infty .$$

Answer 2

As shown by AlexL the proof of necessity, i.e.

if $\phi\in\mathcal{C}(\mathbb{R})$ and there is $k$ such that $$\begin{align} |\phi(t)|\leq k|t| \qquad\text{for all}\quad |t|>k\tag{1} \label{one} \end{align} $$ then $\phi\circ f\in L_p$ for all $f\in L_p$

is straight forward since $\phi$ is bounded in $[-k,k]$, and $\mu(X)<\infty$.

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Here is a proof of sufficiency under the assumption that $(X,\mathscr{X},\mu)$ is non-atomic. Without loss of generality, we will assume that $\mu(X)=1$. Then, for any sequence $\{a_n:n\in\mathbb{N}\}\subset(0,\infty)$ such that $\sum_na_n\leq1$, there exists a countable collection $\{A_n:n\in\mathbb{N}\}$ of pairwise disjoint measurable sets such that $\mu(A_n)=a_n$ for each $n$.

Assume that $\phi$ is continuous, $\phi\circ f\in L_p(\mu)$ whenever $f\in L_p(\mu)$, but that the condition $\eqref{one}$ does to hold for any $k>0$. Then, for each $n\in\mathbb{N}$, there is $t_n$ with $|t_n|>n$, and such that $|\phi(t_n)|>n|t_n|$. Let $a_n=\frac{1}{Z_p|t_n|^pn^p}$, where $Z_p=\sum_n\frac{1}{n^p}$. As $\sum_na_n<1$, there exists a sequence of pairwise disjoint measurable sets $A_n$ such that $\mu(A_n)=a_n$.

The function $f(x)=\sum_n t_n\mathbb{1}_{A_n}$ satisfies $\|f\|^p_p=1$ and so $f\in L_p$. Since $$ |\phi(f)|\geq \sum_n nt_n\mathbb{1}_{A_n}, $$ $$ \|\phi(f)\|^p_p \geq \frac{1}{Z_p}\sum_n n^p|t_n|^p\mu(A_n)=\infty $$ in contradiction to the assumption that $\phi\circ f\in L_p$.

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Some remarks:

While continuity of $\phi$ is used in the proof of necessity, that condition is not used in the proof of sufficiency under the non-atomic condition for $\mu$.

I am not sure whether sufficiency holds for atomic measures. If so, maybe continuity plays a role there.

There is a related problem in Bartle's elements of integration. Here is a link

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