Unbounded operator with no bounded extension. 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 $f \mapsto fg$. Suppose that $T_g$ is injective but not surjective and show that $T_g^{-1}: \text{range}(T_g) \rightarrow L^p(X)$ is not bounded and that there is no bounded extension of $T_g^{-1}$ to $L^p(X)$.
I know that we can prove the first bit by showing that the $\text{range}(T_g)$ is a dense and proper subset of $L^p(X)$ and this will show that $T_g^{-1}$ is not bounded. However, I am not sure how to show that it is dense. I have looked at questions similar to this part on here but none seem to use the argument I am going for here. The proper subset is given to us by assumption. As for the existence of a bounded extension, how does one usually go about these arguments? Thanks in advance!
 A: *

*Let $X,Y$ be Banach spaces and $T\in \mathcal{B}(X,Y)$. Let $T^\star$ be the adjoint of $T$.
Then

(1) $T$ is an isomorphism into $Y$ if and only if $T^\star$ is onto.
(2) $T^\star$ is an isomorphism into $X^\star$ if and only if $T$ is onto.
(3) $T$ is an isomorphism onto if and only if $T^\star$ is an isomorphism onto.
For a proof, see, for example, Banach Space Theory Exercise 2.49 on p. 78 (note that (1)+(2) imply (3)).


*If $1\leq p<+\infty$. Suppose that $T_g^{-1}:gf\in T_g(L^p(X))  \mapsto f\in L^p(X)$ is bounded. Then
$T_g$ is an isomorphism from $L^p(X)$ into $T_g(L^p(X))$. According to step 1-(1),
$T_g^\star$ is onto, but
$$T_g^\star : h\in L^q(X)\mapsto gh\in L^q(X),$$
where $q$ is the conjugate
exponent of $p$.
It follows that $\forall f\in L^q(X)$, $\exists h\in L^q(X)$ such that $f=gh$. Note that
$g\neq 0$ a.e. (Indeed, if $\mu(\{g=0\})>0$, since $\mathbb{R}^d$ is $\sigma$-finite,
we can find a set $A\subset \{g=0\}$ with $0<\mu(A)<+\infty$, let $f= 1_A\in L^q(X)$, then
there doesn't exist an $h\in L^q$ such that $f=gh$, a contradiction.) Hence,
For any $f\in L^q(X)$, $h= \frac{f}{g}$ is unique. Hence, $T_g^\star$ is bijective.
According to step 1-(3), $T_g$ is surjective, a contradiction.


*If $p=+\infty$.
Define $$\widetilde{T}_g: f\in L^1(X)\mapsto gf\in L^1(X).$$
Then $\widetilde{T}_g^\star =T_g$. The proof is entirely analogous.

Another solution:
$T_g^{-1}$ is continuous iff there is a constant $c>0$ such that $\|T_gf\|_{L^p(X)} \geq c \|f\|_{L^p(X)}$ for all $f\in L^p(X)$,
to put it another way,
$T_g^{-1}$ is not continuous iff there is a sequence $\{f_n\}_{n\geq 1}\subset L^p(X)$ with $\|f_n\|_{L^p(X)}=1$ such that
$T_gf_n = gf_n \to 0$ in $L^p(X)$. So, our purpose is to find such a sequence $\{f_n\}_{n\geq 1}$.
Since $T_g$ is injective, we obtain that $g\neq 0$ a.e. on $X$.
Note that $\frac{1}{g}\notin L^\infty(X)$, otherwise $T_g$ is bijective. That means, there doesn't exist a constant $c>0$ such that
$\left|\frac{1}{g}\right|\leq c$ (or $|g|\geq \frac{1}{c}$) a.e. on $X$. Hence, for each $n$, we can find
a measurable subset $S_n \subset X$ with
$\mu(S_n)>0$ such that $|g|\leq \frac{1}{n}$ a.e. on $S_n$. Since $\mathbb{R}^d$ is $\sigma$-finite,
we can assume that $\mu(S_n)<+\infty$.
For each $n$, we can find a measurable function $f_n \in L^p(S_n)$ such that $\|f_n\|_{L^p(S_n)}=1$ (this is possible since we can use those simple functions).
Extend $f_n$  by $0$ outside $S_n$. Also,
$$\|T_gf_n\|_{L^p(X)}= \|gf_n\|_{L^p(X)}= \|gf_n\|_{L^p(S_n)}\leq \frac{1}{n} \cdot \|f_n\|_{L^p(S_n)}=\frac1n\to 0 \quad \text{as $n\to\infty$}.$$
