Let \begin{equation*} U = \big\{f\in C(\mathbb{R}) \mid f=f\mathbb{I}_{[0,1]}\big\}, \end{equation*} where $\mathbb{I}_{[0,1]}$ is the indicator function of the interval $[0,1]$. Let \begin{equation*} V = \bigg\{f\in U \mid \big(f\ge 0\big)\land\bigg(\int_0^1 f(t)\operatorname{d}t = 1\bigg)\bigg\}. \end{equation*} Let \begin{equation*} W = \bigg\{f\in U \mid \int_0^1 f(t)\operatorname{d}t = 0\bigg\}. \end{equation*} Let \begin{equation*} I:\mathbb{R} \to \mathbb{R}, x\mapsto x. \end{equation*} Define \begin{equation*} \Gamma\,\colon V\to W, f\mapsto (f*I)f, \end{equation*} where $*$ is the convolution operator, i.e. $g*h(x)=\int_{\mathbb{R}}g(t)h(x-t)\operatorname{d}t$.
I'm trying to solve the equation $\Gamma(f) = \varphi$, i.e. I'm looking for a characterization of the image of the operator $\Gamma$.
Taking Fourier transform the equation becomes \begin{equation*} \hat{f}'=\hat{f}'(0)\hat{f}-2\pi i\hat{\varphi} \end{equation*} so a viable strategy seems to solve the following linear ODE with a constraint on the derivative in $0$ \begin{equation*} \begin{cases} u'&=au-2\pi i\hat{\varphi} \\ u'(0) &= a \end{cases} \end{equation*} then, anti-trasforming the equation $\hat{f} = u$, and finally see for which $\varphi \in W$ this procedure ends with an element in $V$.
However - especially the last step - this seems a little bit messy... is it clear when this is so? Or maybe has anyone a simpler/more elementary idea about how to tackle this problem?
As mentioned in the title, $V$ is the collection of continuous pdfs of random variables (rv) supported on $[0,1]$. Let $f \in V$ be the pdf of the rv $X$. Then,
$$f\ast I(x) = \int_0^1 f(t)(x-t)\,dt = x\int_0^1 f(t)\,dt - \int_0^1 tf(t)\,dt = x - \mathbb{E}[X].$$
Let $\phi_f = \Gamma f = (f\ast I)f$. Also for any $\phi$ in the image of $\Gamma$, let $f_{\phi} = \Gamma^{-1}\phi$ whenever the inverse is unique and associate it with the rv $X_{\phi}$.
First, suppose that $f > 0$ on $(0,1)$. Then $\phi_f(t) = 0$ precisely when $t \in \{0,\mathbb{E}[X],1\}$. So if the unique zero of $\phi$ in $(0,1)$ is some $y \in (0,1)$, then we know that $\mathbb{E}[X_{\phi}] = y$. Then since $\phi(t) = (t-y)f_{\phi}(t)$, it follows that,
$$f_{\phi}(t) = \frac{\phi(t)}{t - y}.$$
So in the class of functions $\phi \in W$ such that $\phi$ has a unique zero, $y\in (0,1)$, $\phi$ is in the image of $\Gamma$ if and only if $\frac{\phi(t)}{t-y} \in V$ and this candidate solution has the right expectation. That is,
In particular, this shows that although $W$ is closed under scalar multiplication, $\Gamma(V)$ is not.
Now we consider the general case. For each $y \in (0,1)$, define
$$f_{\phi,y}(t) = \frac{\phi(t)}{t-y}.$$
Suppose $\phi$ has multiple zeros in $(0,1)$. Then it should be easy to see that for each such zero $y$, if $f_{\phi,y}$ satisfies the four points above, then $f_{\phi,y} \in V$ and $\Gamma f_{\phi,y} = \phi$. It should also be clear by now that if there does not exist such a $y$, then there is no $f \in V$ such that $\Gamma f = \phi$. Also, if there exist $y_1 \neq y_2$ satisfying the 4 conditions, then $\Gamma f_{\phi,y_1} = \Gamma f_{\phi,y_2} = \phi$. However, I'm not sure if this ever happens or not.
Edit: I thought of one other thing that might help in practice. Suppose $\phi$ has multiple zeros. If $y$ and $y'$ are both zeros satisfying condition 1, then it must follow that $\phi(t) = 0$ for all $t \in [y,y']$. Thus, the candidate zeros will always fall in some interval $[a,b]\subset (0,1)$ such that $\phi|_{[a,b]} \equiv 0$. Given some $\phi$ which may or may not be in the image of $\Gamma$, there exists an interval $[a,b]$ on which condition 1 easily holds. Condition 2 trivially holds on the interval $(a,b)$ and should be simple to verify or reject on the boundaries. The integrals in conditions 3 and 4 may be viewed as equations in terms of $y$ which can be solved for on the interval $[a,b]$.
Edit 2: Looking back at this today I realize I made a mistake with condition 2. We need to verify that $f_{\phi}$ is continuous at $y$, not $0$. Condition 2 should actually be $\lim_{t \to y} \frac{\phi(t)}{t-y}$ exists which is equivalent to stating that $\phi$ is differentiable at $y$.