The Definition — Question 10

Question 10 – Solution

Strategy. Positivity of the Laplace weight gives bounds; continuity at the last observed point decides whether an extremum is actually attained.

Step 1: Bound the unknown tail. The observed contribution is $$L(s)=\int_0^1te^{-st}\,dt=\frac{1-(1+s)e^{-s}}{s^2}.$$ The remaining integral lies between zero and $\int_1^\infty e^{-st}\,dt=e^{-s}/s$. Therefore $$L(s)\le F(s)\le U(s),\qquad \boxed{U(s)=\frac{1-e^{-s}}{s^2}.}$$ All integrals converge absolutely because $0\le f\le 1$ and $s>0$.

Step 2: Resolve attainment. The upper bound is attained by $f(t)=\min\{t,1\}$. The lower bound is not attained: continuity and $f(1)=1$ force a strictly positive contribution on some interval immediately after one. For $\varepsilon>0$, instead continue linearly from $(1,1)$ to $(1+\varepsilon,0)$ and then stay zero. This continuous $f_\varepsilon$ satisfies $$0<F_\varepsilon(s)-L(s)\le\varepsilon e^{-s}\longrightarrow 0.$$ Thus $L(s)$ is the infimum, while $U(s)$ is a maximum.

Step 3: Give the full range at one. Substitution gives the interval $$\boxed{1-2/e<F(1)\le 1-1/e.}$$ For any target strictly above the lower endpoint, choose $\varepsilon$ small enough that $F_\varepsilon(1)$ is below it. A convex combination of $f_\varepsilon$ and the upper-bound signal remains continuous, in $[0,1]$, and agrees with the observed ramp. Splitting the defining integral shows its transform is the same convex combination. Choosing the weight appropriately reaches the target, proving no intermediate values are missing.

Step 4: Measure the uncertainty. The interval width is $W(s)=U(s)-L(s)=e^{-s}/s$, with $$W'(s)=-e^{-s}(s+1)/s^2<0.$$ Since $W(3)=e^{-3}/3\approx 0.01660$ and $W(4)=e^{-4}/4\approx 0.00458$, the least positive integer meeting the tolerance is $\boxed{s=4}$. Increasing $s$ reduces sensitivity to the unknown late-time portion; it does not supply missing pointwise information about that future.