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Let $y_n=⟨e_n,y⟩$. Using the phase of $u_n$ in the singular value system, the Moore–Penrose inverse of an operator becomesThis gives the admissible data for a periodic convolution inverse. Because the range is dense and the kernel is zero here, the domain of the inverse is exactly the range, not all of $L^2$.

For $y(x)=e^{\alpha x}$ and $\alpha \notin i\mathbb{Z}$,The formal expression would consequently beHowever, the high-frequency obstruction for nonperiodic exponential data prevents this from being a Hilbert-space solution. The numerator is nonzero, $|y_n|$ is asymptotic to a nonzero constant divided by $|n|$, and part (c) proved $|c_n|=o(1/|n|)$. Hence $|y_n/c_n|\to \infty$, so the Picard criterion fails. For $\alpha \notin i\mathbb{Z}$, $A^{†}y$ is not defined in the specified Hilbert space. The formal series is not a convergent generalized solution.

If $\alpha =im$ with $m\in \mathbb{Z}$, the data are a single periodic Fourier mode: $y_n=\sqrt{2\pi }{\delta }_{nm}$. Then there is an exact unique solution,In particular, if the intended parameter is real, only $\alpha =0$ is admissible, giving $f^{†}=1/c_0$.

For the inadmissible cases the inverse problem still has approximate solutions $f_N={\sum }_{|n|\le N}(y_n/c_n)e_n$: their images are Fourier projections converging to $y$ in $L^2$, while their norms diverge. Thus the least-squares residual has infimum zero but no minimizer. This distinguishes an undefined exact inverse from a regularized truncated reconstruction; it is the necessary qualification to the question's unrestricted constant $\alpha$.