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Let $X$ be a measurable subset of $\mathbb{R^d}$, $g \in L^\infty(X)$, and $p \in [1,\infty]$. Define the operator $T_g:L^p(X) \rightarrow L^p(X)$ by $f \mapsto fg$. I want to find necessary and sufficient conditions to determine when $T$ is a bijection.

  1. Injection: I have read some posts on here and can see that if $g \neq 0$ almost everywhere, then $T_g$ is an injection. But how do you show the other direction? If $T_g$ is an injection, then $T_g(f)=T_g(h)$ would mean that $f=h$. I cannot seem to get a condition here for $g$.

  2. Surjection: Suppose that for every $\hat{f} \in L^p(X)$ we can find $f \in L^p(X)$ such that $T_g(f)=\hat{f}$. So we have $fg=\hat{f}$ which tells us that $f = \hat{f}/g$ provided that $g \neq 0$ almost everywhere. So here I think it is enough for $1/g \in L^\infty(X)$. But I am still unsure how to prove this without handwaving my argument.

The surjection implies that $g \neq 0$ almost everywhere, so is it enough for $M_\phi$ to be surjective to conclude that it is a bijection?

BingerBiter
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    Are you clear why $g\neq 0$ and $1/g\in L^{\infty}$ is sufficent and are wondering if it is nessciary? For this point do you know that a linear map given by function multiplication from $L^p\to L^p$ must be in $L^{\infty}$ (https://math.stackexchange.com/questions/2675165/linear-map-between-lp-and-lq) – Pax Jul 28 '21 at 07:26
  • @pax: I still do not see why this shows why being surjective implies that $1/g \in L^\infty$. – BingerBiter Jul 28 '21 at 23:04

1 Answers1

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Regarding injectivity.

Let $m(S)$ denote the measure of a measurable set $S.$

Let $E=g^{-1}\{0\}.$

(1). If $m(E)\ne 0$ then take a measurable $E'\subset E$ with $\infty >m(E')>0.$ Then $T_g$ maps both the constant function $0$ and the characteristic (indicator) function $\chi_{E'}$ to $0,$ so $T_g$ is not injective.

(2). The members of $L^p$ are not actually functions but equivalence classes of functions. So $T_g$ maps the equivalence class $[f]$ of the function $f$ to the equivalence class $[fg]$ of the function $fg.$ And for $[f_1],[f_2]\in L^p$ we have $[f_1]=[f_2]\iff (f_1= f_2\;$ a.e.)

Now if $m(E)=0$ then $$T_g([f_1])= T_g([f_1])\iff [gf_1]=[gf_2]\iff$$ $$\iff (gf_1=gf_2 \;a.e.)\iff$$ $$\iff 0=m(\{x: f_1(x)=f_2(x)\lor x\in E\})= m(\{x:f_1(x)=f_2(x)\})\iff$$ $$\iff f_1=f_2 \; a.e. \iff [f_1]=[f_2].$$

DanielWainfleet
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  • Thank you for your answer. – BingerBiter Jul 28 '21 at 22:38
  • I think that the comment by @pax that $1/g\in L^{\infty}$ iff $T_g$ is surjective is correct. If $g\not\in L^{\infty}$ I think you can find some $f\in L^p$ such that $T_g(h)\ne f$ for any $h\in L^p$ by using the infinite set ${n\in \Bbb N: m({x: 1/|g(x)|\in [n,n+1),})>0,}.$ – DanielWainfleet Jul 29 '21 at 05:18