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Consider the unitary operator $U$ on the Hilbert space $H:=L^2([0,\frac\pi2])$, that takes $\cos((2k+1)x)$ to $\sin((2k+1)x)$, for $k\in\mathbb N$ (both are orthogonal basis). How can we explicitly write it in exponential form, $U=e^{iA}$ for a bounded self-adjoint operator $A$?

By the spectral theorem for normal operators, every unitary operator on a complex Hilbert space $H$ can be written this way. For instance, since $\text{ spec } U\subset \partial B_\mathbb{C}(0,1)$, such $A$ can be given by the Borel functional calculus. Nevertheless, in most concrete cases like the above, I have no clue about how to compute explicitly the exponential form.
I would be also glad to see similar examples, so feel free to change the data, if it helps you (but I'd prefer to accept an answer to the original problem).

note. In the above case, one can split $H$ as $H_+\oplus H_-$, where $H_+$ and $H_-$ are the closed subspaces spanned resp. by $\{\cos((4j+1)x)\}_{j\in\mathbb N}$, $\{\cos((4j+3)x)\}_{i\in\mathbb N}$.

Then $ Uf(x) = f(\frac\pi2-x)$ for all $f\in H_+ $ and $ Uf(x) = -f(\frac\pi2-x) $ for all $ f\in H_-. $ However, this doesn’t seem to help, since the decomposition is not $U$-invariant.

(edit Jan 11, 2025) After suggestion of Christian Remling, here is the (infinite) matrix of $U$ w.r.to the orthonormal basis $ \phi_k:=\frac2{\sqrt \pi}\cos((2k+1)x)$. It turns out to be a curious superposition of a $\color{blue}{\text {Hankel}}$ matrix and a $\color{red}{\text {Toeplitz}}$ matrix. Note that each column is a permutation of $\big(\frac1{2m+1}\big)_{m\in\mathbb Z}$, and each column is obtained from the preceding by switching pairs of adjacent coefficients, so that negative coefficients slowly migrate to infinity.

$$U=\frac2\pi\,\begin{bmatrix} \color{blue}1 & \color{red}1 & \color{blue}{\frac13} & \color{red}{\frac13}& \color{blue}{\frac15} & \color{red}{\frac15}&\color{blue}{\frac17} &\dots \\ \color{red}{-1} & \color{blue}{\frac13} & \color{red} 1 & \color{blue}{\frac15} & \color{red}{\frac13} & \color{blue}{\frac17}& \color{red}{\frac15}&\dots \\ \color{blue}{\frac13} & \color{red}{-1} & \color{blue}{\frac15} & \color{red}{1}& \color{blue}{\frac17} & \color{red}{\frac13}&\color{blue}{\frac19} &\dots \\ \color{red}{-\frac13} & \color{blue}{\frac15} & \color{red} {-1} & \color{blue}{\frac17} & \color{red}1 & \color{blue}{\frac19}&\color{red}{\frac13}&\dots \\ \color{blue}{\frac15} & \color{red}{-\frac13} & \color{blue}{\frac17} & \color{red}{-1}& \color{blue}{\frac19} & \color{red}{1}&\color{blue}{\frac1{11}}&\dots \\ \color{red}{-\frac15} & \color{blue}{\frac17} & \color{red} {-\frac13} & \color{blue}{\frac19} & \color{red}{-1} & \color{blue}{\frac1{11}}&\color{red}{1} &\dots \\ \color{blue}{\frac17} & \color{red}{-\frac15} & \color{blue}{\frac19} & \color{red}{-\frac13}& \color{blue}{\frac1{11}} & \color{red}{-1}& \color{blue}{\frac1{13}}&\dots \\ \color{red}{-\frac17} & \color{blue}{\frac19} & \color{red} {-\frac15} & \color{blue}{\frac1{11}} & \color{red}{-\frac13} &\color{blue}{\frac1{13}}&\color{red}{-1} &\dots \\ \vdots & \vdots & \vdots & \vdots& \vdots& \vdots & \vdots & \ddots \end{bmatrix}$$

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    $\begingroup$ just to make sure I have understood the setup correctly: is your convention that $0\in {\mathbb N}$? $\endgroup$ Commented Jan 10 at 12:45
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    $\begingroup$ Yes, so the normalized basis is $\frac2{\sqrt\pi}\cos(nx)$, for all odd positive integers $n$ $\endgroup$ Commented Jan 10 at 13:34
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    $\begingroup$ Since on $[0,\pi/2)$ $1=\frac4\pi\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\cos((2n+1)x)$ I find $U1=\frac4\pi\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\sin((2n+1)x)=\frac1\pi\log\big(\frac{1+\sin(x)}{1-\sin(x)}\big)$ $\endgroup$ Commented Jan 11 at 1:01
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    $\begingroup$ Along similar lines, I was actually also wondering if we can even say what type of spectrum $U$ has (pure point, purely absolutely continuous, something more complicated?). $\endgroup$ Commented Jan 11 at 21:39
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    $\begingroup$ $A=(\pi/2) (1-V)$ works for any such $V$ (unitary, self-adjoint, idempotent), because then $A=\pi P$, the projection onto the eigenspace of the eigenvalue $-1$. $\endgroup$ Commented Jan 12 at 18:16

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The unitary displacement operator $U=e^{iA}$ with Hermitian $A=(i\pi/2)d/dx$ transforms $\cos[(2k+1)x]$ into $\cos[(2k+1)(x-\pi/2)]=(-1)^k\sin[(2k+1)x]$.

Alternatively, decompose a function into components ${f\choose g}$, with $f$ in the subspace spanned by $\cos[(2k+1)x]$ and $g$ in the subspace spanned by $\sin[(2k+1)x]$, $k\in\mathbb{N}$. These two subspaces span the whole Hilbert space. Choose as basis states ${{\cos[(2k+1)x]}\choose 0}$ and ${0\choose{\sin[(2k+1)x]}}$ The unitary $U=e^{iB}$ with $B={{\;0\;\;i\pi/2}\choose{-i\pi/2\;0}}$ transforms $\cos[(2k+1)x]\mapsto\sin[(2k+1)x]$.

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  • $\begingroup$ So in this case $U$ is the operator $f\mapsto f(\frac\pi2-x)$... Do you see how to treat the original one (without $(-1)^k$ in front of $\sin(2k+1)x) $) ? $\endgroup$ Commented Jan 8 at 15:21
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    $\begingroup$ This A is not bounded. $\endgroup$ Commented Jan 8 at 16:33
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    $\begingroup$ While it's true that $e^{i d/dx}$ is a shift on $L^2(\mathbb R)$, I don't think this is clear on $L^2(0,\pi/2)$. (You also need to specify boundary conditions, $i d/dx$ on $C_0^{\infty}$, say, has many self-adjoint extensions.) $\endgroup$ Commented Jan 8 at 21:47
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    $\begingroup$ In fact, it's straightforward to check that this isn't working. For example, if we try the boundary conditions $f(0)=f(\pi/2)$, then the eigenfunctions of $A$ are $e^{-4inx}$, with eigenvalues $2\pi n$, so $e^{iA}=1$. $\endgroup$ Commented Jan 8 at 22:00
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    $\begingroup$ I think in L²([-π/2,π/2]) it is true that the union of the two sequences is an orthogonal basis and the two subspaces are an orthogonal decomposition (odd and even functions). But note that here H=L²([0,π/2]), both {cos((2k+1)x): k ≥0} and {sin((2k+1)x), k≥0} are basis and span the whole H $\endgroup$ Commented Jan 9 at 0:05

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