Unitary Operators on

Question

Let $L^2_{\mathbb{P}}(\Omega)$ be a separable Hilbert space and $(\Omega,\Sigma,\mathbb{P})$ be a probability space. Given two $f,g \in L^2_{\mathbb{P}}(\Omega)$ is there a unitary (or self-adjoint) operator $U$ on $L^2_{\mathbb{P}}(\Omega)$ satisfying $$ U(f)=g\qquad \mathbb{P}-a.e? $$

Answer 1

Let $\phi:\Omega \rightarrow \Omega $ be a bijective measurable function preserving the $\mathbb{P}$-size of each set; ie: $$ \mathbb{P}\left( \phi^{-1}(B) \right)=\mathbb{P}(B)\qquad (\forall B \in \Sigma). $$ Then the composition operator $C_{\phi}:f\mapsto f\circ \phi \in L\left(L^2_{\mathbb{P}}(\Omega)\right)$ (where $L(L^2_{\mathbb{P}}(\Omega))$ is the set of continuous linear maps from $L^2_{\mathbb{P}}(\Omega)$ to itself) is unitary.

Well-defined

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Using the push-forward property of measure $$ \int_{\Omega}|C_{\phi}f|^2d\mathbb{P} = \int_{\Omega}|f\circ \phi|^2d\mathbb{P}= \int_{\phi(\Omega)}|f\circ \phi|^2d\phi_{\#}\mathbb{P}= \int_{\Omega}|f\circ \phi|^2d\mathbb{P}<\infty; $$ where $\phi_{\#}\mathbb{P}$ is the $\phi$-push-forward of $\mathbb{P}$.

Unitary: Let $f,g \in L^2_{\mathbb{P}}(\Omega)$ and note that: $$<C_{\phi}f,g> = \int_{\Omega}f\circ \phi g d\mathbb{P}= \int_{\Omega}f g\circ \phi^{-1} d\mathbb{P}= <f,C_{\phi^{-1}}g>; $$ whence $C_{\phi}$ is unitary with adjoint $C_{\phi^{-1}}$!

Similarly, we compute that: $C_{\phi}\circ C_{\phi^{-1}}=C_{\phi^{-1}}\circ C_{\phi}=1_{L^2_{\mathbb{P}}(\Omega)}$.

A Fun Little Example...But and important One: The Fourier Transform

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The Fourier Transform $f\mapsto \int_{\omega \in\Omega} e^{-2 \pi \lambda \omega} f(\omega) d\mathbb{P}(\omega)$ is a particularly important case. I leave this exercise to you though

Then as Kavi stated, you just need to choose $\phi$ that maps $f$ into $g$.

Answer 2

For a unitary operator you will need the condition $\|f\|=\|g\|$. W.l.o.g. assume $\|f\|=\|g\|=1$. Extend $f$ to an orthonormal basis $\{f_1,f_2,...\}$ with $f_1=f$ and extend $g$ to an orthonormal basis $\{g_1,g_2,...\}$ with $g_1=g$ and define $U(\sum a_n f_n)=\sum a_n g_n$. This gives a unitary map with $Uf=g$.