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Suppose $f, g\in L^{p}(\mathbb R), (1\leq p < \infty).$

For simplicity, let us assume that, $g,f:\mathbb R\to \mathbb R$ so that composition of $f$ and $g$, namely, $f\circ g(x)= f(g(x)); (x\in \mathbb R)$ is well-defined.

My Question is: Given real-valued functions $f,g \in L^{p}(\mathbb R).$ Can we expect its composition $f\circ g$ is again in $L^{p}(\mathbb R)$ ? If not, under what condition on $f$ one can expect that, $f\circ g\in L^{p}(\mathbb R)$ for all $g\in L^{p}(\mathbb R).$ ?

Thanks,

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    $\begingroup$ Basically only if $f$ is Borel measurable with $|f(x)| \leq C \cdot |x|$ for all $x$ with some constant $C \geq 0$. I will post a proof later if noone else does. $\endgroup$ – PhoemueX Aug 30 '14 at 6:30
  • $\begingroup$ Can we find c and d such that |f(x)|≤c + d⋅|x|. If f∈Lp(R) and g∈Lp(E) and f∘g∈Lp(E) where E is set of non zero finite measure? @PhoemueX $\endgroup$ – Sushil Nov 24 '15 at 21:38
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It is very easy to see that $f\circ g\in L^{p}\left(\mathbb{R}\right)$ holds for all $g\in L^{p}\left(\mathbb{R}\right)$ if $f$ is Borel measurable with $\left|f\left(x\right)\right|\leq C\cdot\left|x\right|$ for all $x\in\mathbb{R}$ and some (fixed) $C\geq0$, so I will only show the converse (and I will not go into Borel-measurability, because that is a rather natural condition).

To this end, first observe that $f\left(0\right)=0$, because otherwise $f\circ0\notin L^{p}\left(\mathbb{R}\right)$, where $0$ is the constant function $x\mapsto0$. Now assume (for a contradiction) that there is no such $C\geq0$. We will construct pairwise distint numbers $x_{1},\dots,\in\mathbb{R}$ with $\left|f\left(x_{i}\right)\right|>i\cdot\left|x_{i}\right|$ for all $i\in\mathbb{N}$. This in particular implies $x_{i}\neq0$ for all $i$, because of $f\left(0\right)=0$.

For $i=1$, observe that by assumption $C=1$ is no valid choice, so that there is some $x_{1}\in\mathbb{R}$ with $\left|f\left(x_{1}\right)\right|>1\cdot\left|x_{1}\right|$.

Now, assume that $x_{1},\dots,x_{n}$ have already been constructed. By assumption, there is some $x_{n+1}\in\mathbb{R}$ with $$ \left|f\left(x_{n+1}\right)\right|>\max\left\{ n+1,\frac{\left|f\left(x_{1}\right)\right|}{\left|x_{1}\right|},\dots,\frac{\left|f\left(x_{n}\right)\right|}{\left|x_{n}\right|}\right\} \cdot\left|x_{n+1}\right|\geq\left(n+1\right)\cdot\left|x_{n+1}\right|, $$ because otherwise, $C=\max\left\{ n+1,\frac{\left|f\left(x_{1}\right)\right|}{\left|x_{1}\right|},\dots,\frac{\left|f\left(x_{n}\right)\right|}{\left|x_{n}\right|}\right\} $ would be a valid choice. Note that the above inequality shows that $x_{n+1}\notin\left\{ x_{1},\dots,x_{n}\right\} $, so that the $\left(x_{n}\right)_{n\in\mathbb{N}}$ are indeed pairwise distinct.

Now define $y_{n}:=\sum_{i=1}^{n}\frac{1}{i^{2}\left|x_{i}\right|^{p}}$ for $n\in\mathbb{N}_{0}$, so that $\left(y_{n}\right)_{n\in\mathbb{N}_{0}}$ is strictly increasing, which entails that the intervals $\left(\left(y_{n-1},y_{n}\right)\right)_{n\in\mathbb{N}}$ are pairwise disjoint, and set $$ g:=\sum_{n=1}^{\infty}\chi_{\left(y_{n-1},y_{n}\right)}\cdot x_{n}. $$ This implies $$ \left\Vert g\right\Vert _{p}^{p}=\sum_{n=1}^{\infty}\int_{y_{n-1}}^{y_{n}}\left|x_{n}\right|^{p}\, dx=\sum_{n=1}^{\infty}\left(y_{n}-y_{n-1}\right)\cdot\left|x_{n}\right|^{p}=\sum_{n=1}^{\infty}\frac{\left|x_{n}\right|^{p}}{n^{2}\cdot\left|x_{n}\right|^{p}}<\infty. $$ But we also have (because the intervals are pairwise disjoint) $$ \left(f\circ g\right)\left(x\right)=\sum_{n=1}^{\infty}\chi_{\left(y_{n-1},y_{n}\right)}\cdot f\left(x_{n}\right) $$ and hence \begin{eqnarray*} \left\Vert f\circ g\right\Vert _{p}^{p} & = & \sum_{n=1}^{\infty}\int_{y_{n-1}}^{y_{n}}\left|f\left(x_{n}\right)\right|^{p}\, dx\geq\sum_{n=1}^{\infty}\int_{y_{n-1}}^{y_{n}}n^{p}\cdot\left|x_{n}\right|^{p}\, dx\\ & = & \sum_{n=1}^{\infty}\frac{n^{p}\cdot\left|x_{n}\right|^{p}}{n^{2}\cdot\left|x_{n}\right|^{p}}=\sum_{n=1}^{\infty}n^{p-2}=\infty, \end{eqnarray*} because of $p\geq1$, which entails $p-2\geq-1$.

But by assumption, $f\circ g\in L^{p}\left(\mathbb{R}\right)$. This contradiction shows that $\left|f\left(x\right)\right|\leq C\cdot\left|x\right|$ must indeed be true for all $x\in\mathbb{R}$ and some $C\geq0$.

Note that basically the same proof works for $p > 0$, because one can take any power $\alpha \in (1,2)$ in the definition of $y_n$, i.e. one can take $y_{n}:=\sum_{i=1}^{n}\frac{1}{i^{\alpha}\left|x_{i}\right|^{p}}$ and will still have $g \in L^p$.

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  • $\begingroup$ Wow, thanks lot; for such a wonderful answer; $\endgroup$ – Inquisitive Aug 30 '14 at 7:28
  • $\begingroup$ Just for curiosity; can you just indicate where do we need Borel measurable ? (I guess, as composition of two measurable function need not be measurable; but here to make sense $f\circ g$ as an element of $L^{p};$ we need first it is measurable, so some where here we need Borel measurable, or am I missing something, please correct me, If I am wrong some where ); thanks, $\endgroup$ – Inquisitive Aug 30 '14 at 7:38
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    $\begingroup$ A composition of two Borel measurable functions is always Borel measurable. The same is in general not true of the composition of two Lebesgue measurable functions and also not of $f\circ g$ for $f$ Lebesgue measurable and $g$ Borel measurable. Finally note that if $g=h$ almost everywhere, this easily implies $f\circ g=f\circ h$ almost everywhere, so that everything is well-defined on the level of elements of $L^p$. Now, every $g \in L^p$ has a Borel-measurable representant $h$, so that $f\circ h$ is Borel (and hence Lebesgue) measurable. This shows that $f\circ g$ is Lebesgue measurable. $\endgroup$ – PhoemueX Aug 30 '14 at 7:46
  • $\begingroup$ sorry but why argument given in comment how it prove f will be Borel measurable. Can you please explain $\endgroup$ – Sushil Nov 24 '15 at 21:57

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