Complete solution set of a Convolutional Equation? Here is a problem that am I stuck and I appreciate any help. In essence, I am trying to show that the only solutions for the described problem are the ones provided below. Best.. 
Setup: In what follows, we consider the functions defined over the half-closed symmetric interval $\mathcal{I}\triangleq [-\frac{1}{2},\frac{1}{2}]$  with the usual inner product defined as
\begin{align}
\langle f \mid g \rangle = \int_{x\in\mathcal{I}} f(x)\, g(x) \, \mathrm{d}x
\end{align}
Consider orthonormal basis for the the even-symmetric functions $\{\varphi_k\}_{k=0}^\infty$ that are defined for all $x\in\mathcal{I}$ as
\begin{align}
\varphi_0(x)&=1\\
\varphi_k(x)&=\sqrt{2}\cos(2\pi x),\quad k=1,2,\ldots.
\end{align}
Conjecture: Among all real valued even-symmetric functions defined over the interval $\mathcal{I}$, 
the following functional equation (for the given fixed real pair ($\alpha_0\neq 0, \alpha_1$),
\begin{align}
f(x) \circledast \ln(f(x)) = \alpha_0+\alpha_1\varphi_1(x)
\end{align}
is satisfied  either by
\begin{align}
f_1 (x) = \beta_0+\beta_1\varphi_1(x),
\end{align}
or
\begin{align}
f_2 (x) = \gamma_0 \exp\left(\gamma_1 \varphi_1(x)\right),
\end{align}
for some appropriately selected constants $(\beta_n,\gamma_n)$, for $n=0,1$. Here, the convolutions are in the circular sense over the interval $\mathcal{I}$.
 A: Well, I would say that at least for some (degenerate) $\alpha_0, \alpha_1$ there will be more solutions. Consider $f_2$ of the form $f_2(x)=\gamma_0+\gamma_1 \cos x + \gamma_2 \cos 2x$. If your equation to never has a solution of a form other than the two above, then the convolution $f_2* \log f_2$ would never be a degree one trigonometric polynomial. 
Though, this convolution is surely a degree at most two trigonometric polynomial, as $f_2$ is. Hence, to construct an example that becomes of degree one, you need only to eliminate the second degree term. To do so, you need $\log f_2$ to have zero $\cos 2x$-Fourrier coefficient. And it is just one relation on $\gamma_0$, $\gamma_1$ and $\gamma_2$. For instance, once you check that this coefficient can be (for different values of $\gamma_i$'s) both positive and negative, you are done.
It seems to me that this is easy to be checked: for instance, taking $\gamma_0=1$, $\gamma_1$ to be small positive and $\gamma_2$ to be small negative, one gets $$
\log(1+\gamma_1\cos x +\gamma_2 \cos 2x)= \gamma_1 \cos x - \frac{1}{4} \gamma_1^2 (1+\cos 2x) + \gamma_2 \cos 2x  + o(\gamma_1^2+|\gamma_2|).$$
So one can both ensure positivity and negativity of the $\cos 2x$-coefficient: the former by taking $\gamma_1=0$, $\gamma_2$ small, the latter by taking $\gamma_2=0, \gamma_1$ small.
This leads, surely, to a degenerate example. But it shows that even if for generic $\alpha$'s your conjecture holds, it should be handled by more elaborate arguments.
