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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 $\mathcal{I}$.

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  • $\begingroup$ Dear @user64076, please use markdown instead of mathjax to format text. For example, instead of $mathbf{some text}$, you can use **some text** to produce bold text. Mathjax is meant for mathematical displays alone. Thank you. $\endgroup$ – Ricardo Andrade Dec 17 '14 at 19:14
  • $\begingroup$ Do you conjecture that $f_1$ and $f_2$ are solutions to the functional equation? Or do you conjecture that $f_1$ and $f_2$ are the only solutions to the functional equation? In any event you need to assume more properties of $f$ than that it is real-valued and even: if it takes negative values, is not measurable, or is measurable but very far from being integrable, the convolution written above does not make sense. $\endgroup$ – Ian Morris Dec 18 '14 at 10:45
  • $\begingroup$ Ok, fair enough.. The functions belong to the probability simplex defined over the interval $\mathcal{I}$, i.e., they are non-negative Lebesgue absolute-integrable functions. And by the way, yes, I am conjecturing that $f_1$ and $f_2$ are the only solutions.. I hope this clarifies the question a bit.. @IanMorris $\endgroup$ – user64076 Dec 18 '14 at 16:58
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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.

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