Convolution Integrals — Question 4

Question 4 – Solution

Strategy. A sign-changing kernel can reinforce a suitably chosen sign-changing input. Its signed area loses that information.

Step 1: Recover the kernel. The zero-data transform gives $Y=F/[(s+1)^2+1]$, so $$h(t)=e^{-t}\sin t,\qquad y=h*f,\qquad \int_0^\infty h(t)\,dt=\tfrac 12.$$ For example, the last integral follows from the antiderivative $-e^{-t}(\sin t+\cos t)/2$.

Step 2: Sum the absolute areas. Put $q=e^{-\pi}$. The first positive lobe has area $A_0=(1+q)/2$. Each later absolute lobe is $q$ times its predecessor, so $$\boxed{M=\int_0^\infty|h|=\frac{1+q}{2(1-q)}.}$$ For every admissible input, $|y(t)|\le\int_0^t|h(t-u)|\,du\le M$.

Step 3: Prove sharpness and distinguish attainment. For any $T>0$, set $$f_T(u)=\operatorname{sgn}(\sin(T-u))\quad(0\le u\le T), \qquad f_T(u)=0\quad(u>T),$$ with arbitrary values in $[-1,1]$ at zeros. This input is admissible and $y(T)=\int_0^T|h(T-u)|\,du=\int_0^T|h(v)|\,dv$. These values approach $M$ as $T\to\infty$, so any smaller proposed constant fails for some finite $T$. However, the absolute kernel has a strictly positive tail after every finite $T$, so $|y(T)|<M$ for every input at every finite time. The universal supremum is sharp but is not attained.

Step 4: Refute the signed-area bound explicitly. Take $f=-1$ on $(0,\pi)$, $f=1$ on $(\pi,2\pi)$ and zero afterward. This is the preceding construction for $T=2\pi$. Therefore $$y(2\pi)=A_0(1+q)=\boxed{\frac{(1+q)^2}{2}>\frac 12}.$$ The signed integral describes cancellation within the kernel, as for a constant input’s limiting response. An arbitrary bounded input can reverse its own sign so that both lobes contribute positively; the absolute area is the appropriate worst-case quantity.