Image of $L^2M$ inside $L^1M$, for $M$ a von Neumann algebra Let $M$ be a factor (von Neumann algebra with trivial center), and let $L^1M:=M_*$ be its predual.
Let $\omega:M\to\mathbb C$ be a faithful normal state.
The Hilbert space $L^2M:=L^2(M,\omega)$ admits a distinguished vector $\omega^{1/2}$, defined as the image of $1\in M$ under the inclusion $M\hookrightarrow L^2M$.
There is also a ``multiplication'' map $L^2M \times L^2M \to L^1M$ given by
$\xi\cdot \eta := (x\mapsto \langle x \xi,J\eta\rangle)$, where $J$ denotes the modular conjugation. For $\xi\in L^2M$, we write $\lambda_\xi:L^2M \to L^1M$ for left multiplication by $\xi$.
For every $t\in\mathbb R$, consider the composite $F_t:L^2M \to L^1M$ of the modular flow $\Delta^{it}:L^2M\to L^2M$ with the map $\lambda_{\omega^{1/2}}$. Equivalently, $F_t$ is the composite of $\lambda_{\omega^{1/2}}$ with the the modular flow $\delta^{it}:M_*\to M_*$ on the predual.


Fact: the $1$-parameter family of maps $t\mapsto F_t$ admits an analytic continuation to the complex strip $0 \le Im(t) \le 1/2$, given by the formula
$$F_t:L^2M \to L^1M : \xi \mapsto \omega^{it+1/2} \xi \omega^{-it}.$$


Let me now assume that the state $\omega$ has trivial centralizer.


Let $\Omega:=\{t\in\mathbb C\mid0 \le Im(t) \le 1/2\}$, and let $$f:\Omega\to L^1M$$ be any holomorphic map (continuous on the closed strip and holomorphic in the interior) such that $f(s+t)=\delta^{it}(f(s))$. Does there exist $\xi\in L^2M$ such that $f(t)=F_t(\xi)$?


 A: No, such a $\xi \in L^2(M)$ need not exist.
Denote by $L$ the space of maps from $\Omega$ to $L^1(M)$ that are continuous on the closed strip $\Omega$, holomorphic on the interior of $\Omega$ and satisfy $f(s+t) = \delta^{it}(f(s))$ for all $s \in \Omega$ and $t \in \mathbb{R}$. One can identify $L$ with the subspace of $L^1(M)$ consisting of all $\mu \in L^1(M)$ that belong to the domain of the analytic continuation $\delta^{-1/2}$. One can then turn $L$ into a Banach space with norm
$$\|\mu\|_L = \|\mu\|_1 + \|\delta^{-1/2}(\mu)\|_1 \; .$$
As explained in the question, the map sending $\eta \in L^2(M)$ to $\theta_\eta \in L^1(M)$ given by $\theta_\eta(x) = \langle x \omega^{1/2},J \eta \rangle$ is a bounded, injective linear map from $L^2(M)$ to $L$. Assume that this map is surjective. By the open mapping theorem, we find $\kappa > 0$ such that $\|\eta\|_2 \leq \kappa \, \|\theta_\eta\|_L$ for all $\eta \in L^2(M)$.
Fix $a,b \in M$ and put $\eta = a J b^* \omega^{1/2}$. Then, $\theta_\eta(x) = \langle x J a^* \omega^{1/2},b^* \omega^{1/2} \rangle$ and
$$(\delta^{-1/2}\theta_\eta)(x) = \langle x a \omega^{1/2} , J b \omega^{1/2} \rangle$$
for all $x \in M$. Therefore,
$$\|\theta_\eta\|_L \leq \|a\|_2 \, \|b\|_2 + \|a^*\|_2 \, \|b^*\|_2 \; .$$
So if we assume moreover that $b$ belongs to domain $D(\sigma_{-i/2})$ of the analytic continuation $\sigma_{-i/2}$ of the modular automorphism group on $M$, we conclude that
$$
\|a \sigma_{-i/2}(b)\|_2 \leq \kappa \, \bigl( \|a\|_2 \, \|b\|_2 + \|a^*\|_2 \, \|b^*\|_2 \bigr) \; . \hspace{2cm} (\ast)
$$
Already at this stage, we should not expect that such an estimate can hold for all $a \in M$ and $b \in D(\sigma_{-i/2})$. Let's produce an explicit counterexample. So we need a factor $M$ with a normal faithful state $\omega$ with trivial centralizer such that we can easily make computations inside $(M,\omega)$. We use the free Araki-Woods factor associated with the regular representation of $\mathbb{R}$.
So, denote $H = L^2(\mathbb{R})$ and define the anti-unitary operator $J : H \to H$ and unitary representation $U_t$ on $H$ by
$$(J \xi)(x) = \overline{\xi(-x)} \quad\text{and}\quad (U_t \xi)(x) = \exp(itx) \xi(x) \; .$$
Denote by $S$ the closed densely defined antilinear involution on $H$ given by $S = J U_{-i/2}$, so that $(S \xi)(x) = \exp(-x/2) \overline{\xi(-x)}$ and with the domain of $S$ consisting of those $\xi \in L^2(\mathbb{R})$ with $S(\xi) \in L^2(\mathbb{R})$. Denote by $K \subset H$ the closed real subspace of all $\xi \in D(S)$ with $S(\xi) = \xi$.
Consider the full Fock space $\mathcal{F}(H)$ and denote for every $\xi \in H$ by $\ell(\xi)$ the left creation operator. The free Araki-Woods factor $M$ is the von Neumann algebra acting on the full Fock space $\mathcal{F}(H)$ generated by the operators $\ell(\xi) + \ell(\xi)^*$ with $\xi \in K$. This representation of $M$ on $\mathcal{F}(H)$ is standard, with the vacuum vector serving as the canonical implementation of the free quasi-free state $\omega$.
Choose a sequence of unit vectors $\xi_k \in K$ with support contained in $[-1/k,1/k]$. Define
$$a_k = (\ell(\xi_k) + \ell(\xi_k)^*)/2 \; .$$
Then, $a_k$ is a sequence of self-adjoint elements of $M$. We have $a_k \in D(\sigma_{-i/2})$ and, by our choice of support, $\|a_k - \sigma_{-i/2}(a_k)\|_2 \to 0$. Also, each $a_k$ is distributed, w.r.t.\ the vacuum vector, as a semicircular random variable $S_1$ with radius $1$. Applying $(\ast)$ to $a = a_k^n$ and $b = a_k^n$ and taking $k \to +\infty$, it follows that
$$\|S_1^{2n}\|_2 \leq 2 \kappa \, \|S_1^n\|_2^2 \; .$$
This means that
$$
E(S_1^{4n}) \leq (2\kappa)^2 \, E(S_1^{2n})^2 \quad\text{for all $n \in \mathbb{N}$.}\hspace{2cm} (\ast\ast)
$$
One knows that $E(S_1^{2n}) = 2^{-2n} C_n$, where $C_n$ is the $n$'th Catalan number. It follows that
$$E(S_1^{2n}) \sim \pi^{-1/2} n^{-3/2} \; .$$
This is incompatible with $(\ast\ast)$.
Concluding remark. The inequality $(\ast)$ shouldn't hold in any infinite-dimensional factor. The above concrete counterexample relies on sequences $b_k$ that are almost invariant under $\sigma_{-i/2}$. Such sequences always exist and it should then be possible to deduce from $(\ast)$ a contradiction in general.
