Convolution Integrals — Question 8

Question 8 – Solution

Strategy. Inversion differentiates the output. Check the initial-value restriction before using that derivative formula.

Step 1: Derive and prove the inverse rule. Differentiation under the integral gives $z'=f-z$, $z(0)=0$. Therefore a $C^1$ target is obtainable exactly when it starts at zero, and its only possible continuous input is $\boxed{f=z'+z}$. Conversely, for this input, the target and the convolution solve the same IVP $w'+w=f$, $w(0)=0$. Uniqueness proves they agree.

Step 2: Check two targets. For $z_0=te^{-t}$, $z_0'+z_0=e^{-t}$, so the input is $f_0=e^{-t}$. The target $e^{-t}$ is impossible because its value at zero is $1$. Although its derivative plus itself is zero, a zero input with zero initial state produces the zero output. Differentiation discards the missing initial condition.

Step 3: Differentiate the perturbation. The two exponential-derivative terms cancel, leaving $$\boxed{f_{\varepsilon,N}=e^{-t}+\varepsilon N e^{-t}\cos(Nt).}$$ All these inputs are continuous, and the target outputs vanish at zero. The uniform output error is at most $\varepsilon$ because $e^{-t}|\sin Nt|\le 1$. The input error at zero is $\varepsilon N$; in fact this is its uniform norm since $e^{-t}|\cos Nt|\le 1$.

Step 4: Exhibit the instability. Take $\varepsilon_N=N^{-1/2}$. Then $$\|z_{\varepsilon_N,N}-z_0\|_\infty\le N^{-1/2}\to 0, \qquad\|f_{\varepsilon_N,N}-f_0\|_\infty=\sqrt N\to\infty.$$ Thus the inverse is not continuous in the uniform output norm, even on smooth attainable outputs. Small high-frequency measurement errors can produce large reconstruction errors. The figure uses $\varepsilon=0.05$, $N=12$; each panel compares its own exact and perturbed functions.