Problem 13 Milestone — Exponential boundary continuous abel solution

For every $b₀,c\in\mathbb R$ with $b₀>0$ and $c>0$, there exists at least one total function $f:\mathbb R\to\mathbb R$, possibly depending on both parameters, that is continuous on the open positive half-line and satisfies $f(t)\ge 0$ for every $t>0$. Writing $B(u)=b₀\exp(-cu)$ and $\varphi(x)=(2\pi)^{-1/2}\exp(-x^2/2)$, this same $f$ must satisfy, for every real $t>0$, $\displaystyle \int_0^t \frac{1}{\sqrt{t-s}}\,\varphi\!\left(\frac{B(t)-B(s)}{\sqrt{t-s}}\right)f(s)\,ds=\frac{1}{\sqrt t}\,\varphi\!\left(\frac{B(t)}{\sqrt t}\right)$, where the integral is the Lebesgue interval integral in the variable $s$. For positive $t$, its effective integration domain is $(0,t]$, equivalently $[0,t]$ up to a null endpoint. At $s=t$, the apparently singular expression is interpreted using total real division: $\sqrt{t-s}=0$, both $1/0$ and $0/0$ are defined as $0$, and hence the integrand’s endpoint value is $0$; this singleton value does not affect the integral. No separate integrability hypothesis is stated: under the totalized integral convention a nonintegrable integrand has integral $0$, while the displayed right-hand side is strictly positive under the stated assumptions, so the required equality can hold only when the integrand is integrable for each positive $t$. The declaration imposes no condition at $t=0$ or at negative arguments: $f(0)$ and all values $f(t)$ for $t<0$ are unrestricted, continuity need not extend to $0$, and no boundedness near $0$ is asserted. It also asserts neither uniqueness nor strict positivity of $f$, and requires no normalization, elementary-expression or closed-form representation, or additional regularity.