Convolution of a function and a density distribution

Question

Let $f$ be a probability density function on $\mathbb R$. How can we find $g(x)\ge 0$ such that $$\forall x<1:\ g(x)=\int_{-\infty}^1g(y)f(y-x)dy+\int_1^\infty f(y-x)dy?$$ I thought about putting $$h(x)=\begin{cases} g(x)\ \ x<1 \\ 1\ \ x\ge 1\end{cases}$$ in order to find $$h(x)=\int_{-\infty}^\infty h(y)f(y-x)dy$$ but the last equation has no solution, in general